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16

Write $0\le (x-y)^2=x^2-2xy+y^2$, hence $1<2xy\le x^2+y^2$.


12

Suppose $x^2=4y+2$. The RHS is divisible by $2$ but not by $4$. But if the LHS is divisible by $2$, it must be divisible by $4$.


6

A square can only be 0 or 1 mod 4. Looking at above mod 4, you get $x^2 = 2 \mod 4$. Hence no solution.


4

Since $x^2=4y+2=2(2y+1)$ Suppose that $x=2k$, then $x^2=4k^2$. It is obvious that $4k^2 \not= 2(2y+1)$.


4

A different approach $$S=1+3+\cdots+2n-1$$ $$S=2n-1+2n-3+\cdots +1$$ adding both of them yields $$2S=2n+2n+\cdots+2n$$ implying $$2S=(2n)\cdot n $$ thus $$S=n^2$$


2

To supplement the answers already given, there is also a geometric proof of this result. For example: $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $$ 1 + 3 + 5 + 7 + 9 = 5^2 $$


2

If $t>a$ then we have $$\frac{f(a)-f(t)}{t-a} =f'(\xi)$$ where $a<\xi < t$ by Lagrange Theorem. Now letting $t\to a$ we obtain $$f'_+ (a) =f'(a+).$$


2

Hint You're on the right track. You can show by exhaustion that $n=0$ and $n=1$ are solutions, but $n\in \{2,3,4\}$ are not. Next we test our base case $n=5$ and it is a solution. Suppose for some $n=i$, it is the case that $n^2<2^n$. Hence $i^2<2^i$. We will show that $(i+1)^2<2^{i+1}$, which would complete the proof. Note: $$ \begin{align} ...


2

Nice write-up; looks great! Perhaps a slightly slicker way of doing the induction step (and probably what the problem intended you to do) is to observe that: \begin{align*} 3^k + 3^{k-1} + 3^{k-2} &< 3^k + (3)3^{k-1} + (3^2)3^{k-2} \\ &= 3^k + 3^k + 3^k \\ &= 3(3^k) \\ &= 3^{k+1} \end{align*}


2

Not sure if this is "calculusy" enough for you: $$a = \dfrac{dv}{dt} = v \cdot \dfrac{dv}{dt} \cdot \dfrac{1}{v} = v \cdot \dfrac{dv}{dt} \cdot \dfrac{dt}{dx} = v\cdot \dfrac{dv}{dx}$$


2

For (a), it boils to comparing $1$ and $$ \frac{100!}{10^{100}}=\frac{1}{10}\cdot\frac{2}{10}\cdot\cdots\frac{100}{10}.\tag{*} $$ Only the fractions $\frac{1}{10},\ldots,\frac{9}{10}$ are less than $1$ but they are more than compensated for by $\frac{100}{10},\frac{90}{10},\ldots,\frac{20}{10}$. Other fractions are greater or equal to $1$. So it should be ...


2

See the definition of rank function in Dirk van Dalen, Logic and Structure (5th ed - 2013), page 12. For a non-atomic formula $\varphi$, $rank(\varphi)$ is at least $+1$ with respect to its "immediate" subformulas. Thus, if $\psi$ is a proper subformula of $\varphi$ : $rank(\varphi) \ge rank(\psi) +1 > rank(\psi)$.


2

Since $a-b-1\gt 0$, there exists a real number $\alpha\gt 0$ such that $a=b+1+\alpha$. Then, since there exists an integer $m$ such that $m-1\le b\lt m$, we have $$b\lt m\lt m+\alpha\le b+1+\alpha =a,$$ which implies that $m$ satisfies $b\lt m\lt a$.


2

The paragraph is saying that there are only finitely many elements that do not satisfy the inequality $|x_{m(k)} - x| < \varepsilon$, namely, all elements with $k < K$. This then implies that the inequality is satisfied for infinitely many $k$ (for $k \ge K$).


2

Since $x,y\not=0\Rightarrow x^2,y^2\gt 0$, by AM-GM inequality, we have $$x^2+y^2\ge 2\sqrt{x^2y^2}=2|xy|=2xy\gt 2\cdot \frac 12=1.$$


2

Basically you did it all right but you missed the last step.Namely you have the first step of induction P(1) done(usually called the base case).Now you have the hypothesis that P(k) is true which implies that $13| 3^{k+2} + 4^{2k +1}$ from which we can conclude follwing for some natural number $q$: $$13q = 3^{k+2} +4^{2k+1} \implies 13q-4^{2k+1}=3^{k+2}$$ ...


2

If $x$ divides $a$ and $a+2$, it divides their difference as well.


2

Contour integration is a bit less painful. For first, it is better to write our integral as: $$ I = \frac{1}{4}\int_{0}^{2\pi}\frac{\cos(3x)+\cos(5x)}{(2-\cos x)^2}\,dx, $$ then, since $\cos x = \frac{e^{ix}+e^{-ix}}{2}$, by setting $z=e^{ix}$ we get: $$ I = -\frac{i}{4}\left(\oint\frac{2(z^6+1)}{z^2(z^2-4z+1)^2}\,dz + ...


2

By the triangle inequality, $d(x,y)\leq d(x,x_k)+d(x_k,y)$. So substracting $ d(x,x_k)$ on both sides, we get $d(x,y)-d(x,x_k)\leq d(x_k,y)$.


1

Before starting, I believe there is something to say about differentiability; the formulae in the OP can be proven if the partial derivatives of $f$ are continuous at all point $(x,y)$ in the compact $[a,b]\times [c,d]$. Therefore, in what follows "differentiable" means "belonging to class $C^1([a,b]\times [c,d])$". On the second formula; we begin by ...


1

Your proof is not correct. You have proven that $P(n)$ implies $$13 | (4\cdot 3^{n+1} + 3\cdot 4^{2n + 2})$$ which is not the same as $P(n+1)$. You need to prove $P(n+1)$, which is equal to $$13| (3^{(n+1)+2} + 4^{2(n+1) + 1})$$


1

I would divide by $\sqrt5 - 2$ and then subtract 2. Or divide by $\sqrt5+2$ and then add 2.


1

If $P(n):13|(3^{n+2}+4^{2n+1})$ holds true for $n=m$ $13|(3^{m+2}+4^{2m+1})$ Now,$3^{(m+1)+2}+4^{2(m+1)+1}-3(3^{m+2}+4^{2m+1})=4^{2m+1}(4^2-3)$ which is divisible by $13$ So, $P(n)$ will hold true for $n=m+1$ iff $P(n)$ will hold true for $n=m$ Now you have already established the base case. If induction is not mandatory, ...


1

I'm probably going to be a little grammatically picky, but I personally think proper grammar is good in a proof, (and an analysis lecturer drilled it into our class to use 'proper English sentences' where the maths would read as part of the sentence). I would remove the comma after the first line of mathematical notation say something like 'we have/get', ...


1

The only problem is the "which means". $2xy>1$ doesn't imply $x^2+y^2≥2xy$ in any way. $x^2+y^2≥2xy$ is true because $x^2-2xy+y^2 = (x-y)^2 ≥ 0$, which means $x^2+y^2≥2xy$.


1

OK, here's a guess as to what is meant: among the numbers $1,2,3,4,\ldots,(p_1\cdots p_n)$, where $p_1<\cdots<p_n$ are the $n$ smallest primes, only $p_1\cdots p_n - (p_1-1)\cdots(p_n-1)$ of them can be divisible by at least one of the primes $p_1,\ldots,p_n$; therefore the others must be divisible by some other primes; hence there are more primes than ...


1

The approach seems to be: Consider the $n$-th prime $p$. Between $p+1$ and $p^2$ every number is either prime or divisible by one of the first $n$. TAMO. Therefore, there is at least one prime between $p+1$ and $p^2$, hence there are infinitely many primes. I don't know what the miraculous step in the middle is supposed to be. I mean, you can't use the ...


1

You tried to use Weierstrass substitution but $\cdots$ you have not be patient. Doing it, I arrived to $$\int\frac{\cos x \cos 4x}{(2-\cos x)^2}dx=\int\frac{2 \left(1-t^2\right) \left(t^8-28 t^6+70 t^4-28 t^2+1\right)}{\left(t^2+1\right)^4 \left(3 t^2+1\right)^2}dt$$ Using partial fraction decomposition, the last integrand is $$-\frac{2882}{3 \left(3 ...


1

For the sake of contradiction, assume that $A \neq \emptyset$, along with $A \cap B = \emptyset$ and $A \cup B = B$. This implies that there exists an element in $A$, i.e. $\exists x \in A$. Now, $A \cap B = \emptyset$ implies that $x \notin B$. Why does this contradict $A \cup B = B$?


1

When you don't need to use the induction hypothesis, it means that you don't need to use induction. You can just do a direct proof.



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