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7

You wrote I can't imagine many circumstances where mathematical notation and a formal proof system wouldn't suffice to convey the author's pertinent thoughts. The case is, mathematical proofs contain lot of hand-waving argument such as referring to symmetry or to consistent change of variables. Editors are eager to accept these arguments but theorem ...


6

I can’t think of any likely expansion of what you’ve written that makes it true. What is true is that $$\begin{align*}\binom{40}5&=\binom{17}5+\binom{17}4\binom{23}1+\binom{17}3\binom{23}2+\\&+\binom{17}2\binom{23}3+\binom{17}1\binom{23}4+\binom{23}5\;, \end{align*}$$ and it has a very straightforward combinatorial proof. HINT: You’re trying to ...


4

If you don't mind I'm gonna do some magic! $$2^{3n}=(2^3)^n=8^n=(7+1)^n$$ Now use binomial expansion to get $$\begin{align} (7+1)^n&=\binom{n}{0}7^n+\binom{n}{1}7^{n-1}+\cdots+\binom{n}{n-1}7+\binom{n}{n}7^0\\ &=\binom{n}{0}7^n+\binom{n}{1}7^{n-1}+\cdots+\binom{n}{n-1}7+1\\ (7+1)^n&=7k+1\\ \end{align}$$ So We have ...


4

Why? Because this has been the most efficient way to communicate proofs, as shown by the answers of Brian M Scott and Pew. But that may not always be so. Mathematical theorems in journals now often reach tens or hundreds of pages, for example, the famous proof of Wiles's Theorem, or the Classification of All Finite Simple Groups (itself an ongoing project ...


4

You are of course right. The issue is context, what are you expected to do? If it has already been proved in your course that all geometric series with "$|r|$" less than $1$ converge, and if a formula has been derived for the sum, then what you did is fine. But it is a peculiarly simple question to ask if one has that machinery in hand. So I am inclined to ...


4

$$ \bigcap_{n>0}\left(0,\frac 1n\right) $$is a subset of $(0,1)$. For each $x\in (0,1)$ there is some $n>0$ ($n>1/x$), such as $$ \frac 1n < x $$ and for such $n$, $$ x\notin \left(0,\frac 1n\right) $$so $$ \bigcap_{n>0}\left(0,\frac 1n\right) = \emptyset $$


3

$f$ is not surjective. For a function $f: A \to B$ to be surjective means that for every $y \in B$, there is some $x \in A$ such that $f(x) = y$. So, for example, try solving $f(x) = 2$. What happens? Is $x \in [0, \infty)$? Is is, however, injective. To see this, we put $$f(x) = f(y) \iff \frac{1}{x+1} = \frac{1}{y+1} \iff x+1 = y+1 \iff x = y.$$ As for ...


3

Hint: $2^{3n} - 1 = (2^3)^n - 1^n = (2^3-1)((2^3)^{n-1} + ... + 1)$


3

You have almost proved that $f(x)$ attains a maximum on $(0,\infty)$. Since the limit of $f(x)$ as $x$ approaches $0$ from the right is $0$, there is a $\delta$ such that $|f(x)|\lt 1/10$ if $0\lt x\lt \delta$. Similarly, there is a $B$ such that $|f(x)|\lt 1/10$ if $x\gt B$. Our function attains a maximum in the interval $[\delta,B]$. Since $f(\pi/2)\gt ...


3

$\Leftarrow\qquad$ If $\gcd(m^i,n^r)=1$ then by the Bezout's theorem there's $u,v\in\Bbb Z$ such that $$um^i+vn^r=1\iff (um^{i-1})m+(vn^{r-1})n=1\iff \gcd(m,n)=1$$ $\Rightarrow\qquad$ By the contrapositive if $1\ne p=\gcd(m,n)$ then $p$ divides $m^i$ and $n^r$ so $p$ divides $\gcd(m^i,n^r)$ so $\gcd(m^i,n^r)\ne1$.


2

Here's a neat trick: There is a continuous function $g$ on the interval $[0, \infty]$ such that $g(x) = f(x)$ for all $x \in (0, \infty)$. $g$ is more convenient to analyze, because $[0,\infty]$ is compact.


2

You have what you need to show right, but that's not the same as an injective function. With an injective function, the implication runs if $f(x)=f(y)$ then it must be that $x=y$. What we're trying to show here is that if $x$ and $y$ are, in fact, the same thing, then $f(x)$ and f(y) produce the same result regardless of whether we choose to call it $x$ ...


2

Supposing $m\leq k\leq n$, these are two different expressions for the trinomial coefficient $\binom n{m,k-m,n-k}$, the number of ways to colour $n$ points so $m$ are red, $k-m$ are white, and the remaining $n-k$ are blue. (1) Select the blue points, then among the remaining points the red points, respectively (2) select the red points, then among the ...


2

We have $n$ different doughnuts, and want to choose $k$ of them to eat today, of which $m$ are to be eaten this morning. We count the number of ways to do it. Here are two ways to do the counting. Way 1: We choose the $k$ doughnuts for today, then choose the ones for this morning from the $k$ selected. The $k$ doughnuts for today can be chosen in ...


2

I'm not familiar with "rainbow numbers", and I'm afraid I can't follow every step of your proof. But if you're just looking for a very elementary proof of this, here's the easiest one I can think of: Write the sum forwards: $S_n = 1 + 2 + 3 + ... + n$ and then backwards: $S_n = n + (n-1) + (n-2) + ... + 1$ and then sum term by term to get: $2S_n = ...


2

If $k=\gcd(c,a)$ then $k|a$ and $k|c$ and so $k|(a+b)$, meaning $k|b$ and $k|\gcd(a,b)=1$, so $\gcd(c,a)=k=1$. Similarly $\gcd(c,b)=1$.


2

You need to find two elements of the order which are incomparable. That what it means that the order is not total. In the case of a power set ordered by inclusion, this means $A$ and $B$ such that $A\nsubseteq B$ and $B\nsubseteq A$. Do note that this requires that $X$ has at least two elements.


2

First let me say that I probably know far less about this topic than you. However, I would still like to post an observation I have about this problem. Let's use $$\log a - \log b - \log \log \log c =\log\left(\frac{a}{b \log \log c}\right)$$ Now let's suppose there are natural numbers $a,b,c$ where $$\log\left(\frac{a}{b \log \log c}\right) = \gamma = ...


2

Just prove that the function is an involution, that is $f\circ f = Id_{\wp(X)}$. It will follow that it's injective (otherwise it would not be invertible) and surjective (otherwise it wouldn't be the inverse of another function).


2

Hint: Since the panels are square and you have to cover everything, you want the largest integer that divides 280, 336, and 168 simultaneously, i.e. you want the greatest common divisor.


2

There surely is another way to prove it, and they might be getting at quadratic reciprocity here, but Fermat's little theorem provides the simplest way to prove this particular question. If $p$ is a prime number, of whatever form, and $a$ is a number coprime to $p$, then $a^{p - 1} \equiv 1 \pmod p$. (I won't give a proof of Fermat's little theorem here, but ...


1

If $f:A \to B$ is a bijection and $g : B \to C$ is a bijection, then $h = g \circ f$ is a bijection from $A$ to $C$. Now that I've told you what the bijection is, can you prove the two necessary properties? I'll get you started: Injectivity. To show $h$ injective, we'll assume $h(x) = h(y)$ and will conclude that $x = y$. Assume $h(x) = h(y)$ for $x, ...


1

Hint: Complement of any independent set is a vertex cover. Complement of any vertex cover is an independent set. I hope this helps $\ddot\smile$


1

We have $$(x+a)^n=\sum_{k=0}^{n}\binom{n}{k}x^ka^{n-k}$$ Now by putting $x=3$ and $a=1$ we get $$\begin{align}(3+1)^n&=\sum_{k=0}^{n}\binom{n}{k}(3)^k(1)^{n-k}\\ &=\binom{n}{0}3^n+\binom{n}{1}3^{n-1}+\dotsc+ \binom{n}{n-1}3^{1}+\binom{n}{n}\\ \end{align}$$ And by putting $x=5$ and $a=-1$ we get ...


1

The result is true for a bounded sequence $(y_k)$ i.e. $$\exists M>0,\quad \forall k\;\; |y_k|\le M$$ and fails if $(y_k)$ is just bounded below or bounded above.


1

Normally by justify it means that at each step of the way you would put a short reason of why you made that move. So for an example : 1) $a(b+c)$ 2) $ab+ac$ (Distributive property) Just something like that.


1

Your proof hasn't actually proved $P(n+1)$. To do that, you have to show that if $A$ has $n+1$ elements then if $s\notin A$ then $A\cup\{s\}$ has $n+2$ elements. I don't think induction is necessary for this proof. If $A$ has $n$ elements, then there is a bijection $f:A\rightarrow\{1,2,...,n\}$. For $s\notin A$, you can construct a bijection ...


1

Big Hint: you could find a function $B \to C$ such that only a small part of $B$ already covers whole $C$, and make sure that $A \to B$ maps onto that part. Then the composite will be surjective, but you can make $f$ not to be. Don't try to find non-surjective $g$, since the statement $gf$ surjective implies $g$ surjective.


1

Yes, you can. It's best to give an explicit example, too; you should describe two particular numbers $a\neq b$ such that $g(a)=g(b)$.



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