Mathematicianadda

Inspired by Does this sequence reach infinity?, I propose the following generalization:

$F_b(n)$ replaces consecutive repeating digits of $n$ in base $b$, with just one digit.

Find smallest $m\in\mathbb N$ such that sequence $a(0)=1, a(k)=F_b(m\cdot a(k-1))\to\infty$.

For example, $F_2(17)=F_2(2^4+1)=F_2(10001_2)=101_2=2^2+1=5$.

E.g. $a(k)$ for $b=10,m=2$ does not diverge to $\infty$, because it loops $a(68)=a(4)=16$.

(b=10, m=2) => a(k) = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 6536, 13072, 2614, 528, 1056, 212, 424, 848, 1696, 392, 784, 1568, 3136, 6272, 1254, 2508, 5016, 1032, 2064, 4128, 8256, 16512, 3024, 6048, 12096, 24192, 48384, 96768, 193536, 387072, 7414, 14828, 29656, 59312, 18624, 37248, 7496, 1492, 2984, 5968, 1936, 3872, 74, 148, 296, 592, 184, 368, 736, 1472, 294, 58, 16, ...

I couldn't find such $m$ for base $b=10$, so I've tried looking at smaller number bases $b$.

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Base $2$

The $m=17=10001_2$ diverges in base $2$, because it follows the following increasing pattern:

$$ a(k)=F_2(10\color{red}{11}\color{blue}{\overline{01}}\dots\color{blue}{\overline{01}}\color{red}{00}101_2) $$

after $k\ge 2$. By applying $F_2$ and evaluating $a(k)$, we see $a(k)=4 a(k-1) + 1$ diverges to $\infty$.

That is, in this case, the "overlined blue digit substrings $\color{blue}{\overline{01}}$" repeats $k-2$ times.

The proof is inductive. Notice $m=17=10001_2$ and assume $a(k)=1010(10)_{k-2}101_2$. Then

$$\begin{align} a(k+1) &= F_2(m\cdot a(k)) \\ &= F_2(10001_2\cdot 1010(10)_{k-2}101_2) \\ &= F_2(1010(10)_{k-2}1010000_2 + 1010(10)_{k-4}1010101_2) \\ &= F_2(1011(01)_{k-2}0100101_2) \\ &= 101(01)_{k-2}010101_2 \\ &= 1010(10)_{k-2}10101_2 \\ &= 1010(10)_{k-1}101_2 \end{align}$$

as expected, and the base case is $$a(0)=1, a(1)=5\implies a(2)=85=1010101_2=1010(10)_{0}101_2.$$

Base $3$

The $m=14=112_3$ diverges in base $3$, because it cycles over $12$ increasing patterns:

$$\begin{align} a(12k+0) &= F_3(10\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}102\color{red}{00}20101_3)\\ a(12k+1) &= F_3(120\color{red}{22}\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}\color{red}{22}0\color{red}{22}012_3) \\ a(12k+2) &= F_3(\color{red}{22}01202\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}10121_3)\\ a(12k+3) &= F_3(10\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}1020120120\color{red}{22}_3)\\ a(12k+4) &= F_3(120\color{red}{22}\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}102\color{red}{00}20101_3)\\ a(12k+5) &= F_3(\color{red}{22}01202\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}\color{red}{11}0\color{red}{22}012_3)\\ a(12k+6) &= F_3(10\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}10121_3)\\ a(12k+7) &= F_3(120\color{red}{22}\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}1020120120\color{red}{22}_3)\\ a(12k+8) &= F_3(\color{red}{22}01202\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}102\color{red}{00}20101_3)\\ a(12k+9) &= F_3(10\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}\color{red}{11}0\color{red}{22}012_3)\\ a(12k+10) &= F_3(120\color{red}{22}\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}10121_3)\\ a(12k+11) &= F_3(\color{red}{22}01202\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}1020120120\color{red}{22}_3)\\ \end{align}$$

after $(12k+0)\ge 24$, where all patterns diverge to $\infty$. This can be proven by induction.

Base $4$

The $m=2046=133332_4$ diverges in base $4$, because it cycles over $2$ increasing patterns:

$$\begin{align} a(2k+0)&=F_3(323\color{red}{00}\color{blue}{\overline{103230}}\dots\color{blue}{\overline{103230}}10\color{red}{33}2301\color{red}{00}_4)\\ a(2k+1)&=F_3(1312\color{red}{00}\color{blue}{\overline{213120}}\dots\color{blue}{\overline{213120}}21312\color{red}{00}21320_4) \end{align}$$

after $(2k+0)\ge 4$, where both patterns diverge to $\infty$. This can be proven by induction.

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I haven't found such $m$'s for $b\ge 5$, yet.

Since finding the smallest $m$ for a given base $b$ looks hard, I am asking just about the existence:

Does for every number base $b$, there exists at least one $m$ such that $a(k)\to\infty$?

That also might be a hard question. Can we at least prove or disprove this for small bases $b\le 10$ ?

E.g. for $b=5$, all $m\le 4000$ get stuck in a loop (do not diverge to $\infty$), so far. Does such $m$ exist?

from Hot Weekly Questions - Mathematics Stack Exchange
Vepir

Let $p$ and $q$ be distinct odd primes and $a$ be a positive integer with $a<p<q$. I need to prove that if $pq \ | \ \phi(n) $ then $n$ is prime. The proof for the trivial case when $a=2$ is given below.

Proof. Let $n=2pq+1$. Assume $pq \ | \ \phi(n)$. We may write $\phi(n) = cpq$ for some positive integer $c \le 2$. We know that $\phi(n)$ is even for all $n>2$ therefore we must have $c=2$ otherwise $\phi(n) $ is odd. $\phi(n) =2pq=n-1$ which shows that $ n $ is prime. This completes the proof for the trivial case $a=2$.

How do I prove that the proposition holds for an arbitrary positive integer $a$?

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David Jones

Algebra Word Problems Worksheet

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I think the answer is that there are only 2 such functions: the zero function and the identity function, but I'm not able to prove it.

A few findings:

1. $f(0)=0$ and thus $f(a^2)=f(a)^2$.

2. If $f(1)=0$, then $f(2^n)=0$ for all $n\in \mathbb N_0$; if $f(1)=1$, then $f(2^n)=2^n$ for all $n\in \mathcal N_0$ (can be proven by PMI).

3. For any functions satisfying the condition, say $f,g$, $f\circ g$ also satisfies the condition.

Source of this problem: https://www2.math.binghamton.edu/p/pow/problem2f20

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Yuqiao Huang

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Two well-known results in number theory are:

Fermat's $4n+1$ theorem: Every prime of the form $4n+1$ can be represented as $a^2+b^2 (a,b \in\mathbb{N})$.

Euler's $6n+1$ theorem: Every prime of the form $6n+1$ can be represented as $a^2+3b^2 (a,b \in\mathbb{N})$.

Looking at the Mathworld entries on these theorems here and here, I notice that representation of primes of the form $4n+1$ is stated to be unique (up to order), but that there is no mention of uniqueness in respect of representation of primes of the form $6n+1$. Uniqueness does however seem to hold at least for small primes of this form.

Question: Is the representation of any prime of the form $6n+1$ as $a^2+3b^2$ essentially unique?

If this is the case then a reference to a proof would be appreciated.

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Adam Bailey

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     Recently I found and enjoyed the article "Scientists Take On Poetry," an article by Katherine Wright in Physics  --  a free, online magazine from the American Physical Society.  After the following lead-in:

Stuck with how to present your latest scientific project? Try a poem.

Wright's article tells of numerous scientists who have been poets and offers visual poetry by Stephany Mazon and Manjula Silva.  The article quotes Sam Illingworth, a poet and geoscientist at the University of Australia, "Poetry is a great tool for interrogating and questioning the world."  Illingworth heads the Editorial Team of an online journal, Consilience -- a newish journal that describes itself as "the online poetry journal exploring the spaces where the sciences and the arts meet."  The current issue has the theme "uncertainty" and offers 19 poems; one of these is "Heisenberg's uncertainty principle" by Alicia Sometimes -- and it begins with these words:

       The reality we can put into words is never reality itself

       we cannot measure
       the position (x) and the momentum (p)
       of a particle with absolute precision
         . . .

This link leads to the rest of Sometimes' poem and to others offered in Consilience.

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noreply@blogger.com (JoAnne Growney)

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Here’s a round-up of some mathematical news stories from this month.

European Women in Mathematics shares COVID-19 Open Letter

European Women in Mathematics, an international association of women working in mathematics in Europe, has written an open letter encouraging universities and institutions to take action to lessen the disproportionate impact of the COVID pandemic on women in mathematics. Advocating a flexible approach in these uncertain times, the letter seeks signatures from anyone who supports their aim – to ‘shape smart policy to recruit and retain a diverse group of talented young scientists.’

EWM Open Letter

Breakthrough Prizes in Mathematics

This year’s Breakthrough Prize in Mathematics is awarded to Martin Hairer @imperialcollege, who has tamed a large class of equations that can describe processes from the wind on the waves to the jitters of the stock market. https://t.co/u9JIYwmejL pic.twitter.com/8sRBC52sVb

— Breakthrough (@brkthroughprize) September 10, 2020

The Breakthrough Prizes (if you’d forgotten, those are the ones founded and awarded by a collection of incredibly rich people to recognise achievements in maths and science by giving people ludicrous piles of money) for 2021 have been announced. Mathematician and Fields medalist Martin Hairer has been awarded the Breakthrough Prize in Mathematics, which comes with a cash gift of $3m and a nice trophy, for his “transformative contributions to the theory of stochastic analysis, particularly the theory of regularity structures in stochastic partial differential equations”.

The board also awards six New Horizons Prizes of $100,000 to early career researchers in maths and physics. The maths winners this year were Bhargav Bhatt of the University of Michigan, Aleksandr Logunov from Princeton University and Song Sun of UC Berkeley, who are working in areas including algebra and geometry.

From this year Breakthrough are also awarding three Maryam Mirzakhani New Frontiers prizes of $50,000 to women mathematicians for early-career achievements, which went to Nina Holden of ETH Zurich, Urmila Mahadev at Caltech and Lisa Piccirillo from MIT, for work in geometry, quantum computing and knot theory.

Winners of the 2021 Breakthrough Prizes in Life Sciences, Fundamental Physics and Mathematics Announced, on the Breakthrough website

Four from MIT awarded 2021 New Horizons in Physics and New Frontiers in Mathematics prizes, on the MIT News website

John Barrow has died

We are so deeply sad that John Barrow, MMP Director since 1999, has died. He was an inspirational, imaginative and generous colleague, and his energy, positivity and humour will be hugely missed. Our thoughts are with his wife, Elizabeth, and family at this terribly sad time.

— MMP – maths.org (@MMPmaths) September 28, 2020

Mathematician, cosmologist and theoretical physicist John Barrow passed away this month from colon cancer. As well as a distinguished research career, publishing more than 500 journal articles, he worked extensively in communicating maths, through lectureships, public talks – including at 10 Downing Street, Windsor Castle and the Vatican – a long list of popular maths and astrophysics books, and his work with the MMP. He was recognised for this work with, among many other honours, the IMA’s Zeeman Medal and the Royal Society’s Faraday Prize.

Seeking your favourite Ron Graham quotes

Ron Graham had many great quotes which are now scattered across the internet. His wife, Fan Chung, would like to collect them.

I am swamped with teaching, but if you want to help Fan out, reply to this tweet with quotes you know or find.

(Or RT to spread the word!)

Thank you!

— Jay Cummings (@LongFormMath) September 23, 2020

The late Ron Graham, who passed away in July, was a fountain of mathematical wisdom. His widow, Fan Chung, is looking to collect some of these – if you know of any that have inspired you, add them to this thread.

Vicky Neale has a book out soon

I'm excited to say that my new book will be out soon! Why Study Mathematics? is designed for students considering a maths degree, their families and their teachers. It's available to preorder now (with free P&P). Please tell your friends! https://t.co/2OnYJumDcw

— Vicky Neale (@VickyMaths1729) September 23, 2020

Mathematician and author Vicky Neale is following up her brilliant book Closing the Gap: the quest to understand prime numbers with a second, Why Study Mathematics?. Aimed at students considering a maths degree, it gives an insight into what’s involved in a maths degree and why it’s a useful thing to have. It’s available to preorder now on the LPP website.

Mathemalchemy Exhibition to proceed in 2021

Mathemalchemy is a mathematical art collaboration between a group of 23 mathematicians/artists, as outlined in this teaser video:

The brainchild of mathematician Ingrid Daubechies and fibre artist Dominique Ehrmann, the project was planned to take place this year, but due to COVID, workshops planned in March and August were cancelled. The project will still go ahead, with plans for a physical installation at the end of the summer of 2021.

Mathemalchemy website

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Katie Steckles

How much food do you need to carry to hike the Appalacian Trail?

The AT is a hiking trail that begins in Georgia and ends on Mount Katadin, Maine. Fall is a beautiful time to be outdoors or on the Trail.  This activity is about calculating the needs and rates involved in enduring this 2,168 mile trek.

Hopefully you and your students are still enjoying the outdoors during this cooler, more colorful time of year.  Use this enthusiasm to look at the rate calculations necessary for an Appalachian Trail thru-hiker.

Appalachian-trail.pdf     Updated!

For members we have an editable Word doc and solutions.

Appalachian-trail.docx      Appalachian-trail-solutions.pdf

CCSS: 6.RP1, 6.RP.2, 6.RP.3, 7.RP.1

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Is a 6% road grade too steep?    

Steepness and Fall Hiking -  Let them combine what they are doing with their families and friends with a little mathematics about percent grade change.  Why are there warnings about a 6% road grade change ahead?  Why does everyone talk about the vertical gain on a trail that is only 3 miles long?  Enjoy using slope and the Pythagorean Theorem in a new way.

CCSS: 7.EE.3, 8.G.7, 8.G.8, 8.EE.5, 8.EE.6, HSF.IF.B.6, HSG.MG.A.1, MP1, MP2, MP6

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Leslie

Let $n=am+1$ where $a $ and $m>1$ are positive integers and let $p$ be the least prime divisor of $m$. Prove that if $a<p$ and $ m \ | \ \phi(n)$ then $n$ is prime.

This question is a generalisation of the question at Let $n=apq+1$. Prove that if $pq \ | \ \phi(n)$ then $n$ is prime.. Here the special case when $m$ is a product of two distinct odd primes has been proven. The case when $m$ is a prime power has also been proven here https://arxiv.org/abs/2005.02327

How do we prove that the proposition holds for an arbitrary positive integer integer $m>1 $? ( I have not found any counter - examples).

Note that if $n=am+1$ is prime, we have $\phi(n)= n-1=am$. We see that $m \ | \ \phi(n) $. Its the converse of this statement that we want to prove i.e. If $m \ | \ \phi(n) $ then $n$ is prime.

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David Jones

I was watching this video by Flammable Maths about why $$ \begin{pmatrix} 3 &4\\ 6&8 \end{pmatrix}^2 = \begin{pmatrix} 33 &44\\ 66&88 \end{pmatrix} $$ In the video, it is left as a challenge for the viewer to see if you can generalize the result as follows:

Given some $k \in \mathbb{N}\cap[2,\infty), $ can you find a matrix $A\in \mathcal{M}_{n \times n} (\mathbb{N})$ such that $A^k =\left( \sum_{i=1}^{k}10^{i-1} \right)A$?

I attempted a solution to this problem and did the following. I supposed (with the intention of hopefully simplifying calculations) that $A$ is diagonalizable. This means that I can write the equation we want as $$ PD^{k} P^{-1}= \left( \sum_{i=1}^{k}10^{i-1} \right)PD P^{-1} $$ Now, taking the determinant on both sides I get that \begin{align*} &|P||D|^k|P^{-1}| = \left( \sum_{i=1}^{k}10^{i-1} \right)^n |P||D| |P^{-1}|\\ \implies & \left(\prod_{j=1}^n \lambda_j\right)^k = \left( \sum_{i=1}^{k}10^{i-1} \right)^n\left(\prod_{j=1}^n \lambda_j\right)\\ \implies & \prod_{j=1}^n \lambda_j^{k-1} = \left( \sum_{i=1}^{k}10^{i-1} \right)^n \end{align*} where the $\lambda_j$'s are the eigenvalues of $A$. From here I think that if I find a set of eigenvalues that satisfy the above equation I can reconstruct a matrix which satisfies our original intended equation, however, I'm not sure if this is a good way to approach this problem.

Does anyone know a better way to solve this? Or does anyone have any other ideas on how to tackle it? Ideally, I would like to find some patter or family of matrices which satisfy the desired property, buy any and all suggestions would be greatly appreciated. Thank you very much!

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Robert Lee

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Practice Test in Algebra

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The source for this problem is this 3b1b video. I understand this:

$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{\pi^2}{6}$$

Now he alters the series to include only primes and powers of primes (eg. 4 and 8 are included because they are powers of 2, which is prime) while scaling down the powers of primes by a factor of the exponent, as in:

$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2(2)}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{8^2(3)}+\frac{1}{9^2(2)}+...$$

This happens to equate to: $$ \ln\left(\frac{\pi^2}{6}\right)$$

I tried to search the proof of this for a while, but could not find anything. I apologize if this happens to be a very trivial question for this site, but would be delighted to see an elementary explanation because I am just an interested layman in this matter.

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Bipasha

Substitution Method Solving Systems of Equations - Solved Problems

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Could anybody give a hint me how to construct a measurable function $f:[0,1]\rightarrow [0,1]$ such that:

$$\forall_{0\leq a < b \leq 1}: f(a,b)=[0,1]$$

I tried to define a sequence of linear functions $g_n$ that would "vertically cover" more and more of $[0,1]$ like so:

$g_0(x)=x$

$g_1(x) = \begin{cases}2x \text{, for } x\in[0,\frac{1}{2}]\\ 2x-1 \text{, for } x\in[\frac{1}{2},1]\end{cases}$

$...$

$g_n(x) = \begin{cases}2^nx \text{, for } x\in[0,\frac{1}{2^n}]\\ 2^nx-1 \text{, for } x\in[\frac{1}{2^n},\frac{2}{2^n}]\\ ...\\ 2^nx - (2^n-1)\text{, for } x\in[\frac{2^n-1}{2^n},1]\end{cases}$

Then I hoped for such a sequence, of clearly measurable functions, to converge pointwise, so that I could define $f(x)=\lim_{n\rightarrow\infty}g_n(x)$. It seems however, that $g_n$ will not converge and I am stuck looking for a different sequence. Or perhaps my approach is flawed from the very beginning? I will be gratefull for any inights or suggestions.

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Sioux

Angle Pairs and Two Step Equations - Concept - Solved Examples

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Angle Pairs and Two Step Equations Worksheet

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If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$

I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result: Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.

This means that we are saying that $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$ Consider $\tan (A+B+C)$: $$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$ So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.

Now consider $\tan (2A+2B+2C)$: $$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$ $$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$ $$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$ $$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$ as required.

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My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)

Thank you for your help.

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A-level Student

Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of $$P = (a - b)(b - c)(c - a)$$

My solution:

• We have:

$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$ $$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$ $$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$

• Using AM-GM Inequality, we have:

$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$ $$\implies 3 \sqrt[3]{P^2} \leq 12$$ $$\implies -8 \leq P \leq 8$$

• Therefore, $\min P = -8$

Is this solution correct? If not, then why?

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windows prime

If I want to distribute 50 identical candies to 100 children, what is the expected number of candies a child with at least one candy has?

For example, if I give 24 candies to child A and 26 candies to child B, and don't give any candy to other 98 children, since only two children has nonzero candies, if this way the only way how I can distribute 50 identical candies to 100 children, the quantity I'm looking for would be 25.

I tried for an hour, and in the end came up with the following "solution":

Let $Q(k)$ be the average number of candies a child with at least one candy has provided that we distribute the candies only to $k$ children. Then I assumed that the average number of candies a child from this set has $50/k$, and there are $\binom{100}{k}$ different ways of selecting these set of children, so doing a weighted average, I got

$$ \frac{ \sum_{k=1}^{50} 50*(100!) / (k * (k!) * (100-k)!)} { \sum_{k=1}^{50} 100! / ((k!) * (100-k)!)} \approx 1.08481. $$

Is my solution correct? If not, could you provide me with an detailed answer about how you solved it?

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onurcanbkts

When trying to derive, from first principles, the fact that exponential functions $a^x$ (where $a>1$ is real) are differentiable, we easily see that $$ \lim_{h\to0} \frac{a^{x+h}-a^x}h = a^x \lim_{h\to0} \frac{a^h-1}h, $$ provided the latter limit exists. It's even pretty easy to see that $$ \lim_{h\to0} \frac{a^h-1}h = ( \log_b a ) \lim_{h\to0} \frac{b^h-1}h $$ for any other real $b>1$, provided the latter limit exists. (And then one can define $e$ to be the number such that $\lim_{h\to0} \frac{e^h-1}h = 1$ and continue.)

So my question, which doesn't seem to have an answer on this site (though I'd be happy to be proved wrong) nor in the textbooks I've consulted: how can one justify the existence of any limit of the form $\lim_{h\to0} \frac{b^h-1}h$ $(b>1)$, without using the as-yet-underived fact that $b^x$ is differentiable? (Edited to add: I also want to avoid infinite series.)

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Greg Martin

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Dividing 2 Digit Decimals by Whole Numbers Worksheet

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Let $n=d_m...d_2d_1$ where $d_i$ is digit of $n$ .

Define function $F(n)$, If there are any digits of $n$ being repeated consecutively, they convert those digits to that digit and generate new number.

Example: $F(10225)=1025$ because 2 repeat itself

$F(10000)=10$, $F(223335300)=23530$, $F(23)=23$.

Define recurrence relation $a_k= F(2\cdot a_{k-1})$ for $k\ge 1$ and $a_0=1$

$\{a_0,a_1,a_2,...\}=\{1 , 2 , 4 , 8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,6536,13072,2614,528,1056,212,424,848,1696,392,784,1568,3136,6272,1254,2508,5016,1032,2064,4128,8256,16512,3024,6048,12096,24192,48384,96768,19356,387072,...\}$

Problem: Does above sequence will ends with loop or is it approaches to infinity?

I haven't finished programming. It is very possible that you will get some loop. Thanks

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Pruthviraj

Is the proof given below acceptable?

Claim. In any triangle $ABC$ construct isosceles triangles $\Delta ACE$ and $\Delta BDC$ on sides $AC$ and $BC$ , with apices at points $A$ and $B$ , such that $\angle EAC+\angle CBD=180^{\circ}$ holds true. Let points $F$ and $G$ divide legs $AE$ and $BD$ respectively in the same arbitrary ratio . The midpoint $H$ of the line segment that connects points $F$ and $G$ is independent of the location of $C$ .

GeoGebra applet that demonstrates this claim can be found here.

The following proof is inspired by this answer to my previous question.

Proof. Consider $A$, $B$ , $C$ as complex numbers and choose a $\lambda \in \mathbb{R}$. Denote $\angle EAC=\alpha$ and $\angle CBD=\beta$ . Then, $$F=A+\lambda(E-A)=A+\lambda(\cos \alpha +i \sin \alpha)(C-A)$$ $$G=B+\lambda(D-B)=B+\lambda(\cos (-\beta) +i \sin (-\beta))(C-B)$$ $$H=\frac{1}{2}(F+G)=$$ $$\frac{1}{2}(A+\lambda(\cos \alpha +i \sin \alpha)(C-A)+B+\lambda(-\cos \alpha -i \sin \alpha)(C-B))=$$ $$\frac{1}{2}(A+\lambda(\cos \alpha + i\sin \alpha)C-\lambda(\cos \alpha + i\sin \alpha)A+ $$$$B-\lambda(\cos \alpha + i\sin \alpha)C+\lambda(\cos \alpha + i\sin \alpha)B)=$$ $$\frac{1}{2}(A(1-\lambda(\cos \alpha + i\sin \alpha))+B(1+\lambda(\cos \alpha + i\sin \alpha)))$$

This shows that $H$ is independent of the location of $C$.

Q.E.D.

EDIT

It is possible to generalize this claim even further.

Claim.In any triangle $ABC$ construct triangles $\Delta ACE$ and $\Delta BDC$ on sides $AC$ and $BC$ such that $\frac{AC}{AE}=\frac{BC}{BD}$ and $\angle EAC+\angle CBD=180^{\circ}$ hold true. Let points $F$ and $G$ divide sides $AE$ and $BD$ respectively in the same arbitrary ratio . The midpoint $H$ of the line segment that connects points $F$ and $G$ is independent of the location of $C$ .

The proof for this claim is essentially the same as the proof given above.

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Peđa Terzić

Dividing Decimals by Whole Numbers Worksheet

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We are almost midway through Hispanic Heritage Month (September 15 – October 15)!  This month marks a national holiday in the United States that began as a way to promote the history, contributions, and culture of Hispanic-Americans. The month wouldn’t … Continue reading →

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Vanessa Rivera-Quinones

Dividing Decimals by Whole Numbers

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Dividing Decimals Online Practice

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Find coefficient of $x^3y^4z^5$ in polynomial $(x + y + z)^8(x + y + z + 1)^8$

It is pretty easy to see that our goal is to choose from each multiplier in this polynomial $x,y,z,1$ in certain amounts. Since sum of powers equals $12$ and we have $16$ multipliers, we have to choose four $1$'s. There are $\displaystyle {8 \choose 4}$ ways to choose $1$ as part of our product (actually, four $1$'s). But how do we handle what has left? It seems like in such solution we have to consider many cases.

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math-traveler

Practice on Dividing Decimals

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A sheaf (French faisceau) is called 層 in Japanese. Literal meaning of 層 would be a layer, a stratum, or a story (as in a building,) so the terminology is a bit different from what sheaf or faisceau might suggest. Korean and Chinese translations are the same, but Japaneses are likely the first to use the translation, per the early contribution of Japanese mathematicians like Oka Kiyoshi.

Who was the first to introduce the translation 層, and why?

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puzzlet

Dividing of 4 Digit Numbers by 1 Digit Number Practice

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Dividing 4 Digit Numbers by 1 Digit

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Division Worksheet for Grade 3

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Dividing Two Numbers Worksheet

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Problem $1$: Let $M$ be the non-compact manifold obtained from $\Bbb R^2$ removing $n$-distinct points of $\Bbb R^2$. Suppose $f:M\to M$ is a homotopy-equivalence, i.e. there is a map $g:M\to M$ such that both $f\circ g$ and $g\circ f$ are homotopic to $\text{Id}_M$. Is it true that $f:M\to M$ is homotopic to a homeomorphism of $\psi:M\to M$?

Motivation: A closed topological manifold $X$ is called topological rigid if any homotopy equivalence $F : Y → X$ with some manifold $Y$ as source and $X$ as target is homotopic to a homeomorphism. It is well-known that, any homotopy equivalence of closed surfaces deforms to a homeomorphism. Also, there are rigidity theorems, like Mostow's rigidity theorem, Bieberbach's Theorem, etc, but these manily deal with closed-manifolds, and in some cases dimensions higher than $2$.

Thoughts: Here I am considering most elementary non-compact surface, namely punctured plane $\Bbb R^2-0$. Note that, any two self-maps of $\Bbb R^2$ are homotopic as $\Bbb R^2$ is convex, so $\Bbb R^2$ excluded. Now, any homeomorphism is a proper map, so I have to find an invariant of proper map that is fixed or fully stable under ordinary homotopy. The only fact I know is, the set of regular values of a proper map is open and dense. But, my guess is : it is not a fully stable property.

My second thought is to use compactly supported cohomology, we can also consider de-Rham type cohomology as we have enough smooth maps for approximation. Note that $H^2_{\text{c}}(\Bbb R^2-0)=\Bbb R$ and, we can consider the degree of a map between compactly supported chomology groups induced by a proper map, and by checking degrees of two proper maps we can say they are properly homotopic or not. But the homotopy equivalence may not necessarily homotopic to a proper homotopy equivalence. And, this thought gives me another question written below.

Problem $2$: Is every proper self-homotopy equivalence of punctured plane properly homotopic to a self-homeomorphism of punctured plane? What about if I replace the term "punctured plane" by $M$?

My third thought is to construct an explicit homotopy equivalence of punctured plane not homotopic to a homeomorphism. Here I am tring to construct a homotopy-equivalence $f:\Bbb R^2-0\longrightarrow \Bbb R^2-0$ with $f(z)=z$ for $1<|z|<2$ and $f$ is "bad-enough" near $0$ or $\infty$ so that it is far from being homotopic to a proper map. Maybe annuls fixing property is not necessary, I am considering just because to induce an self-isomorphsim of $\pi_1(\Bbb R^2-0)=\Bbb Z$.

As you can see, I could not complete my thoughts, so, any help, comment, reference will be highly appreciated. Thanks in advance.

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MathLover

The einstein problem asks about the existence of a single prototile that by itself forms an aperiodic set of prototiles, that is, a shape that can tessellate space, but only in a nonperiodic way. (This Numberphile video has more context).

Now there are pentagonal shapes that tile the plane in a non-periodic way.

Why are these shapes no solution to the einstein problem?

I suspect that I misunderstand some criteria of Aperiodic set of prototiles, but i cannot figure out what it is.

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Mario Krenn

Integers are perhaps one of the easiest things to work with in mathematics. We start working with integers from the time we start school, maybe even sooner. It begins with basic counting, then addition and subtraction, multiplication, division, and beyond. Without integers, we would not have math as we know it.

Multiplying and dividing integers is a stepping stone when working with these numbers. It opens the door to more advanced functions and shows us how complex an integer is! So let's dive into integers, what they are, and how to multiply and divide them.

Properties of Integers

An integer is a whole number. It can be either positive or negative. All integers less than 0 have a minus sign ( - ) before the numbers. Positive integers have a plus sign ( + ). However, for positive numbers, the plus sign is usually omitted.

An integer is not a fraction, and it is not a decimal. Integers are a subset of all rational numbers, Q, and rational numbers are a subset of all real numbers, R. When you want to represent a set of integers, we use the symbol, Z.

Using set notation, we write integers as {…-3, -2, -1, 0, 1, 2, 3, …}. The ellipsis means that the numbers continue infinitely in both directions.

Properties of integers are presented using algebraic symbols.

Closure

Closure states that the sum or product of two or more integers will always be an integer.

a + b is an integer

a * b is an integer.

Distributive Law

Multiplying a group of added numbers is like adding each number in the group with the multiplier.

a ( b + c ) = ( a * b ) + ( a * c )

( a + b ) * c = ( a * c ) + ( b * c )

Associative Law

When you multiply or add, the order doesn't affect the sum or result.

f + ( b + c ) = ( f + b ) + c

( f * b ) * c = f * ( b * c )

Commutative Law

Like associative law, when you multiply or add, the order doesn't matter.

g + b = b + g

g * b = b * g

Rules for Multiplying and Dividing Integers

Issues arise when multiplying or dividing integers when the numbers used are not natural numbers, i.e., not positive whole numbers. The following rules show you how to multiply and divide these integers.

Multiplying Two Integers with the Same Sign

The result is always a positive integer when you multiply two integers with the same sign.

a * b = ab

( -a ) *  ( -b ) = ab

OR

+ * + = +

( - ) * ( - ) = +

So when multiplying two negative integers, multiply the numbers as usual and remove the minus sign.

Multiplying a Positive and a Negative

When you multiply two integers, and the signs are not the same, the result is always a negative integer.

( -a ) * b = -ab

a * ( -b ) = -ab

OR

( - ) * + = ( - )

+ * ( - ) = ( - )

Dividing Two Integers with the Same Sign

You will always get a positive quotient when you divide two integers with the same sign.

a / b = ab

( -a ) /  ( -b ) = ab

OR

+ / + = +

( - ) / ( - ) = +

So when dividing two negative integers, the quotient will be positive.

Dividing a Positive and a Negative

When you divide two integers, and the signs are not the same, the result is always a negative integer.

( -a ) / b = -ab

a / ( -b ) = -ab

OR

( - ) / + = ( - )

+ / ( - ) = ( - )

Now that we know the rules of multiplying and dividing integers, let us learn how to use them in the following examples.

Example 1

Multiply the numbers below.

6 * 4

The signs for the two numbers are not there, which means that they are positive integers.

6 * 4 = 24.

The answer is positive because both numbers are positive.

Example 2

Multiply the numbers below.

4 * ( - 5 )

The first number is positive, while the second is negative.

4 * ( - 5 ) = -20

The result is negative because the signs of the two numbers are not the same.

Example 3

Multiply the numbers below.

( - 6 ) * ( - 4 )

Both numbers are negative. They have the same sign.

( - 6 ) * ( - 4 ) = 24

The result is positive because both numbers have the same sign.

Example 4

Divide the numbers below.

12 / 3

Both numbers have the same sign.

12 / 3 = 4

The result is positive, the same as it was for multiplication.

Example 5

Divide the numbers below.

( - 12 ) / 3

The signs are not the same.

( - 12 ) / 3 = - 4

The result is negative.

Example 6

Divide the numbers below.

( - 2 ) / ( - 2 )

Both numbers have the same sign.

( - 2 ) / ( - 2 ) = 1

The result is positive because both signs of the numbers are the same.

All the examples above involve only two numbers. Let's see what happens when there's more than two numbers.

Example 7

Multiply the numbers below.

5 * 2 * ( - 3 )

The first two numbers are positive, while the last one is negative. When there are more than two numbers, the best approach is to multiply two numbers at a time.

Let's start with the first two numbers.

5 * 2 = 10

Both numbers are positive.

The next step is to multiply the answer with the last number remaining ( - 3 ).

10 * ( - 3 ) = - 30

The answer is negative because the signs of 10 and 3 are not the same.

Example 8

Look at the expression below.

7 * ( - 2 ) * 3 * ( - 1 )

We have four numbers now so we can group them into two groups. We will handle the first two, and then the last two. We will then multiply the results.

7 * ( - 2 ) = - 14

3 * ( - 1 ) = - 3

Multiply the two answers.

( - 14 ) * ( - 3 ) = 42

When there are more than two numbers in a problem, group them and multiply. Let's see how to handle division when there are multiple numbers.

Example 9

Divide the numbers below.

12 / ( - 2 ) / 3

As with multiplication, let's first look at the first two numbers.

12 / ( - 2 ) = - 6

We can now divide the answer with the remaining number.

( - 6 ) / 3 = - 2

The final result is negative.

No matter where you are in your mathematics education, whether you're just starting with multiplication and division or need a refresher course, these rules stand the test of time. No educational standard changes can change these integer facts. Memorizing them takes practice, but they are hard to forget.

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Kathleen Knowles

A polygon is any closed plane figure. It comes from the Latin poly meaning "many" and gōnia, meaning "angle." "Closed," in this context, means that the sides form a complete circuit. This definition does not exclude shapes such as an hourglass or a star where sides cross each other. When sides do not cross each other, we call them "simple polygons." For this article, we will only be using simple polygons.

Polygons must have at least three sides as a two-sided shape cannot be closed. The point where two sides meet is called a "vertex" (plural: "vertices"). When you draw a line segment drawn from one vertex to another, it is called a "diagonal."

A polygon is "convex" if any diagonal is drawn in the exterior of the polygon. The Red Cross logo is convex because the diagonal from one corner to the next is on the exterior. The "angles" of the polygon are all interior angles. Thus, the four angles where the cross pieces meet are 270° rather than 90°.

Naming Polygons

Naming polygons are generally based on the number of sides or number of angles. For example, an "equilateral" triangle has three equal sides, and an "equiangular" triangle has three equal angles. (Of course, they are the same.) Usually, however, names refer to the number of sides a polygon has. A hexagon has six sides, an octagon has eight, etc. The general name "n-gon" is a polygon with n sides. So, a hexagon is a 6-gon.

When a polygon's sides are the same length and angles are the same degree, we call it a "regular" polygon. A square is regular. All sides and all angles are equal. The Pentagon in Washington D.C. is a regular 5-gon.  A cutaway of most pencils is a regular hexagon, and stop signs are usually regular octagons. A rectangle has four equal (90°) angles, and a rhombus has four equal sides, but they are not "regular."

Angle Measurements

Sum of All Angles

The sum of the angles of any triangle will always equal 180° no matter how big it is.

You can divide larger polygons into triangles by drawing diagonals between vertices, quadrilaterals (4-sided polygons) into two triangles, pentagons (5-sided polygons) into three triangles, etc. You can divide any n-gon into n-2 triangles. Each triangle has a sum of 180°.

Thus, the sum of the angles of any polygon is:

S = (n – 2) * 180

For example, the sum of all eight angles of an octagon is:  S = (8 – 2) * 180  =  1080°.

This formula works whether or not the polygon is regular and even works if the polygon is convex. The Red Cross symbol is a convex 12-gon. It has four 270° angles where the cross pieces meet and eight 90° angles on the outside corners.

4(270) + 8(90) = 1800°.

Using our polygon formula, (12 – 2) * 180 = 1800°. Isn't that amazing?

Individual Angles

This formula can be used to find individual angles if the polygon is regular. For a regular octagon, such as a stop sign, the sum of all eight angles is 1080°, so each angle must be 1080/8 = 135°. Each angle in a regular hexagon is (6 – 2) * 180 / 6 = 120°.

For irregular polygons, if you know all angles except one, you can find the missing angle.

A gardener has walkways that form an almost pentagon—almost because one of the corners is covered by a fishpond. The shape has two right angles, and he measures the other two at 65° and 58°. The sum of the known angles is 303°.  The sum of all angles of a pentagon is 540°, so the angle under the fishpond is 540 – 303 = 237°.  This interior angle is greater than 180°.  Most likely, the gardener wants the exterior angle, which is 237 – 180 = 57°.

Side Lengths

There aren't many rules for finding the lengths of sides of polygons, but this usually isn't a problem. Side lengths are much easier to measure than angles, especially if you're working with a regular polygon. All sides are equal on regular polygons. If you measure one side, you'll know the length of the rest. In rectangles, opposite sides are equal.

Pythagorean Theorem

One way to calculate the sides of a right triangle is with the Pythagorean Theorem. A right triangle is a triangle that has a right angle (90°) made from two legs. The "hypotenuse" is the side across from the right angle. If you square (multiply a number by itself) the length of the two legs and then add the sums together, you will get the result of squaring the hypotenuse. If the leg lengths are represented by a and b, and the length of the hypotenuse is c, then the equation is a2 + b2 = c2

Here is a simple example.  An ordinary sheet of plywood is 4 ft wide by 8 ft long.  How long is the diagonal from one corner to the opposite corner?

42 + 82 = c2

16 + 64 = c2

80 = c2

√(80) = c ≈ 8.94

You would express this in carpenter math as: 0.94 ft = 0.94 x 12 in/ft = 11.28 inches. The length of the diagonal is about 8'-11¼".

Trigonometric Relations

Trigonometric relations are useful for right triangle ratios. They are based on an observation of similar triangles (same shape but not the same size). If two triangles have the same three angles, then the ratio of two sides of the first triangle will equal the ratio of the corresponding sides of the second triangle.

Here's how trig functions work. Consider a triangle with a right angle on one corner and a 31° angle on another corner. The third angle must be 180 – 90 – 31 = 59°.  All 31-59-90 triangles are similar, and the ratio of two sides of one will equal the ratio of the corresponding sides of all others.

Remember, the sides that form the right angle are called "legs" (usually designated a and b), and the side opposite the right angle is the "hypotenuse" (usually designated c).  More specifically, the side opposite the designated angle (in this case, 31°) is a, and the side adjacent to it is b, and the angle measures are α and β, respectively.

The ratio a/b or opposite/adjacent is given the name "tangent." You can determine the value of the tangent by measuring a and b and then dividing. But since it will be the same for every 31-59-90 triangle, you can find the value using trig tables or on scientific or online calculators.

The Pythagorean Theorem and Trig Functions only apply to right triangles, but very often, you can break down more complex polygons into several right triangles. You can then use known values to calculate the unknown ones.

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Kathleen Knowles

In mathematics, an equation is a statement where two mathematical expressions are equal to each other. Because this is algebra, there must be a variable in the equation. A variable is an unknown number, and we end up mostly solving these variables to prove the equation true. In some cases, we'll have to solve an equation that uses more than one variable and one equation. There are plenty of established methods for solving these equations, but one of the more common ways is by using elimination.

Let's first review some key points about equations.

Equation Review

An equal sign separates the two mathematical expressions of an algebraic equation. One expression is on the right-hand side of the equal sign, and the other expression is on the left-hand side of the equal sign.

An example of an equation is:

2x + 3 = 12

The left-hand side, which is 2x + 3, is equal to the right-hand side, 12.

When dealing with equations, you'll often come across these other terms:

• Term: the part(s) of the equation usually separated by a plus or minus sign. In the example given, the terms in the equation are 2x, 3, and 12.
• Coefficient: a number associated with a variable by multiplication. In the example above, the coefficient of x is 2. Only variables have coefficients.
• Constants: terms of an equation that do not have an associated variable.
• Solving the Equation: finding the values of the variables that make the equation true.

Elimination Method

Some equations are very simple, and you can solve them without needing elaborate methods, like y = 3 or x + 1 = 3.

However, some equations are complex and require an established method for finding the solution. The elimination method is used for solving equations that have more than one variable and more than one equation. In the elimination method, you eliminate one of the variables to solve for the remaining one. Once you have solved for that variable's value, you can substitute the value into any of the equations to find the other variable.

Generally, if an equation contains two unknown variables, you need at least two equations to solve for the two unknown variables. This is called system equations.

2x + 3y = 7 …. eqn 1

x – y = 2 …. eqn 2

Eqn 1 and Eqn 2 form a system equation. The two unknown variables in the two equations are x and y.

Example 1: Two Equations

Solve the system equation below using the elimination method.

3x – 2y = 9 ….. (eqn 1)

6x – y = 27 ….. (eqn 2)

Let’s call the first equation Eqn 1 and the second equation Eqn 2.

Step 1

The first step is to choose which variable to eliminate. I am going to eliminate x. Before you can eliminate, the coefficients of the variable in the two equations must be the same. The coefficient of x in eqn 1 must be the same as the coefficient of x in eqn 2. But they are not the same, so we have to make them the same. You can change the coefficients of variables by multiplying the equation with constants. So let's multiply eqn 1 by 2.

2 * (3x – 2y = 9)

6x – 4y = 18 …. (eqn 3)

As you can see, we multiplied all the terms of the equation by 2. You can also choose to divide an equation by a constant if you prefer.

Our eqn 1 is now eqn 3.

Step 2

Since the coefficients of x are now the same, we can proceed with the elimination.

Let's subtract eqn 3 from eqn 2.

     6x – y = 27

  – 6x – 4y = 18     

    0  + 3y = 9  …… (eqn 4)

A few notes on this subtraction:

• The coefficient of x is now 0.
• -y – (- 4y) = -y + 4y = 3y

We now have eqn 4, which is 3y = 9.

Step 3

Solve the resulting equation to find the remaining variable.

3y = 9

y = 9 / 3

y = 3.

We now know the value of y is 3.

Step 4

Substitute the value of y = 3 into eqn 2 to find the value of x.

6x – y = 27

Replace y with 3.

6x – 3 = 27

6x = 30

x = 30 / 6

x = 5.

We have solved the system of equations to arrive at x = 5 and y = 3.

Let us look at another example.

Example 2: Three Equations

Solve the systems of equations below.

2x – y + 3z = 14 .... (eqn 1)

x + 5y – 2z = 20 …. (eqn 2)

x – y = 4 …. (eqn 3)

The above system equations contain three variables x, y, and z. The third equation does not have the z variable. This only means that the coefficient of z in eqn 3 is 0. So if you are to subtract, you will simply include 0z in eqn 3.

Step 1

By looking at the three equations, subtracting any two equations won't leave us with only one variable, because there are three variables. If we eliminate one, we still have two variables left. There is something else we can do, though.

Consider eqn 3. It has only two variables, but we can express y in terms of x.

x – y = 4

x = 4 + y …. (eqn 4)

By moving y to the right side of the equation, we have a new equation to help us solve the problem.

Step 2

Substitute eqn 4 into eqn 1. This means we will replace the x in eqn 1 with 4 + y

2x – y + 3z = 14

2 (4 + y) – y + 3z = 14

8 + 2y – y + 3z = 14

y + 3z = 14 – 8

y + 3z = 6 …. (eqn 5)

We also have a new equation.

Step 3

Substitute eqn 4 into eqn 2.

x + 5y – 2z = 20

4 + y + 5y – 2z = 20

6y – 2z = 20 – 4

6y – 2z = 16 ….. (eqn 6)

This makes eqn 6, where there are now two variables.

Step 4

The next step is to eliminate y. But we first need to make the coefficient of y in eqn 5 the same as in eqn 6. So we multiply eqn 5 by 6.

6 * (y + 3z = 6)

6y + 18z = 36 …. (eqn 7)

We can now subtract eqn 6 from eqn 7.

    6y + 18z = 36

– (6y – 2z = 16) 

0 + 20z = 20 …. (eqn 8)

Eqn 8 only contains one variable.

Step 5

Solve for z.

20z = 20

z = 1

Now substitute z = 1 into eqn 5.

y + 3z = 6

y + 3 * 1 = 6

y = 3.

Now substitute y = 3 into eqn 4.

x = 4 + y

x = 4 + 3

x = 7.

The solution to the system equations is x = 7, y = 3 and z = 1.

The elimination method is not difficult to learn, but you must stay organized. Variables and substitutions can get pretty messy and confusing if you don't lay them out on the paper correctly. And, as you can see, some equations take more than a few steps to complete. Just keep your pencil handy and have plenty of scrap paper to show your work.

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Kathleen Knowles

As is usually the case in mathematics and physics, formulas and experiments usually begin with curiosity --- even if it's something profoundly simple. What caused the apple to fall from the tree? Why did the ball move faster when it rolled downhill? These simple questions evolve into more complex questions. How do we measure the acceleration of the ball and apple? Someone already answered this question for us.

We use force and mass.

Mass

Every object in this universe has mass, except photons (the particle that carries light). Mass is the total amount of matter in an object. In an atom, protons, neutrons, and electrons account for its total mass.

A common misconception is that mass equals weight. This is not true. Mass is more fundamental, a total count of an object's matter expressed in the form of kilograms. Weight is a derivative of mass. You can calculate weight by multiplying the mass by gravity. An object's weight is different on other planetary bodies where the gravity varies. But, the object will always retain the same mass.

Force

Force is the push and/or pull acting on an object. A force acting on an object makes the object move, accelerate, stop, slow, or change direction. Force has magnitude and direction. We measure a force's magnitude in Newtons or Kgm/s2 (Kilogram meter per second squared), named after the father of physics, Isaac Newton. A force's direction is measured in degree or radian.

Acceleration

Acceleration is an object's change in velocity. A ball rolling downhill experiences an acceleration because it goes faster and faster as the ball rolls. A ball rolling on a level field is a deceleration because the ball gets slower and slower as the ground applies friction to the ball, which is also amplified by gravity. We measure acceleration in meter per second squared (m/s2)

Calculating Acceleration

Newton's Second Law

Mathematician and physicist Sir Isaac Newton the groundwork for the basic principles of physics. He developed three laws of motion in his book, "Principia Mathematica Philosophiae Naturalis."

His second law, in particular, is what makes our calculation work. In its pure form, the law states, "The alteration of motion is ever proportional to the motive force impress'd, and is made in the direction of the right line in which that force is impress'd."

What this means is that the vector sum of the forces (F = newtons) applied on an object is equal to its mass (m = kilograms) times its acceleration (a = meter/second).

In the real world, many different forces are acting on an object, even if this force is standing still. Wind direction and strength, gravity, and ground friction are some of the many factors. The vector sum means the total of the force's direction after it's all added up.

In an academic sense, we often gloss over this variable as a constant. Determining the source and magnitude of forces acting on an object can be much more challenging than executing the equation.

The Equation

The second law of Newton leads us to a beautifully simple equation:

F = m x a

Now, because we're solving for acceleration (a), we can rearrange that equation to:

a = F / m

Yes! It's that simple.

Example

Rolling Ball

We're going to use the question we asked earlier as an example.

A ball with a mass of 10 kg is barreling downhill with the total force of 50 N. Assuming the direction of the total force is perpendicular to the hill's slope, what's the ball's acceleration?

From that question, we can conclude that:

F = 50N

m = 10kg

Now, we're going to input that data into the equation.

a = F / m

a = 50 / 10

a = 5 m/s2

There we have it! The acceleration of the ball is 5 m/s2.

Fighter Jet

An F-35 fighter jet is summoned from the USS Eisenhower to inspect a certain event in the Atlantic Ocean. In mid-flight, the fighter jet has a massive thrust power of 200.000 N propelling the jet forward. The F-35 mass is 16.000 kg. Assuming the vector sum of forces from aerodynamic drag, gravity, and the fighter jet's thrust is indeed 200.000 N, calculate the acceleration of the fighter jet.

From that question, we can conclude that:

F = 200.000 N

m = 16.000 kg

Now, we're going to input that data to the equation:

a = F / m

a = 200.000 / 16.000

a = 12.5 m/s2

Just as easy as the first example! The acceleration of the fighter jet is 12.5 m/s2.

Conclusion

Mass is the total amount of matter in an object, measured in kilograms (kg). Force is the push or pull on an object, measured in newton (N). Acceleration is the rate of change in velocity experienced by an object, measured in meter per second squared (m/s2). The second law of Newton pioneered the equation to calculate acceleration. However, because determining the vector sum in real life is too complex, teachers make the vector sum in academic problems a constant, so students have an easier time solving it. The equation is F = m x a -- ​don't forget it!

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Kathleen Knowles