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9

From the perspective of an elementary calculus student (by which I mean that it can supposedly be made rigorous later on, but isn't in introductory classes), the $du$, $dx$ stuff is absolute nonsense and I will never understand why it continues to be used by so many professors. It really is the one glaring hole in most otherwise rigorous calculus courses. ...


6

Note that $n^2$ is even if and only if $n$ is even. Since $m^2+n^2$ is even, you see that either both $m,n$ are even, or both are odd. If both $m$, $n$ are even, then $mn$ is divisible by $4$. If both $m$, $n$ are odd, you have $m=2k+1$, $n=2l+1$ so $m^2+n^2=4k^2+4k+4l^2+4l+2$ is not divisible by $4$. This gives the result.


5

To put it simply, you only need to use that the sum of rationals is again a rational. If $a+b \in \Bbb Q$, you would get that $b = (a+b) - a$ would be a rational, contradiction.


4

Hint: Use induction. To get you started, suppose we have the statement $$ S(n) : \sum_{i=1}^n 1 = n $$ that you are trying to prove. Fix some $k\geq 1$ and assume $$ S(k) : \sum_{i=1}^k 1 = k $$ is true. Then we need to show that $$ S(k+1) : \sum_{i=1}^{k+1} 1 = k+1 $$ follows. Beginning with the left side of $S(k+1)$, \begin{align} \sum_{i=1}^{k+1} 1 ...


4

HINT: If $x$ is irrational, then $x$ is not an integer, so there is a unique integer $n$ that $n<x<n+1$. Clearly the distance from $x$ to the closer of $n$ and $n+1$ is at most $\frac12$. If $x$ is irrational, can that distance be $\frac12$? Added: Please take note also of Jack M’s excellent comment under the original question; I was a bit rushed when ...


4

The matter of "evenness" (which means nothing more than being a multiple of $2$) is something that was important in the original problem regarding $\sqrt 2$. In your case, you should not necessarily be concerned with evenness. Continuing from where you left off, i.e., $10b^2 = 2c^2$, we can divide both sides by $2$ to get: $$5b^2 = c^2.$$ From here you can ...


4

HINT: Assume that every set of $n$ natural numbers has a maximum, and let $A$ be a set of $n+1$ natural numbers. Let $a$ be any element of $A$; then $A\setminus\{a\}$ has $n$ elements, so it has a maximum element $m$. Now compare $a$ and $m$.


3

Obviously, if $p_1p_2\mid n$, then $p_1\mid n$ and $p_2\mid n$. So assume that $p_1\mid n$ and $p_2\mid n$. If $p_1$ and $p_2$ are distinct primes, then, using Bezout's Identity, there are $x,y\in\mathbb{Z}$ so that $$ xp_1+yp_2=1\tag{1} $$ Equation $(1)$ implies that $$ \begin{align} n &=nxp_1+nyp_2\\ ...


3

Or if this isn't correct (by contradiction) then let $f_a:\mathbb R \setminus \mathbb Q\to \mathbb Q$ with $f_a(b)=a+b$. Then $f_a$ is $1-1$ which is wrong because $\mathbb Q$ is countable.


3

Your problem is a misunderstanding of the technique of proof by contradiction. When performing a proof by contradiction, you make your usual assumptions (e.g., $a\in\mathbb Q, b\in\mathbb R\setminus \mathbb Q$) and you assume the logical negation of what you're trying to prove (e.g., $a+b\in\mathbb Q$). Per your description, it looks like you made your ...


2

$$\begin{align*} \lambda &\color{red}\leftarrow \color{red}{\textrm{integer}} \\ \beta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\ \xi &\color{red}\leftarrow \color{red}{\textrm{integer}} \\ \zeta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\ \alpha\zeta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\ ...


2

It’s not in general true that $(A_1\times A_2)\times A_3=A_1\times(A_2\times A_3)$. Suppose, for instance, that $A_1=A_2=A_3=\{0\}$; then the only element of $(A_1\times A_2)\times A_3$ is $\big\langle\langle 0,0\rangle,0\big\rangle$, while the only element of $A_1\times(A_2\times A_3)$ is $\big\langle 0,\langle 0,0\rangle\big\rangle$, and ...


2

Hint: Let $n = 2m+1$ be an arbitrary odd integer and expand the square. Then analyze the structure of $n^2-2$


2

This baffled me too as I first came along, Jack M gave a good answer what really is behind it. And for the practicing mathematician, the „symbolic approach“ via variables, differentials and so on is just like a mental shorthand, that works by clever choosen notation. Maybe bear in mind that the chain rule could be read in two directions, one if you see at ...


2

Yes your argument is correct, you can also conclude the same by using the fact that $f'$ is bounded on $[0,1]$ and hence $f$ is Lipschitz implying uniform continuity.


1

It is false that $f(1.5)=3$. As you have pointed out, $(\forall y)(1.5,y)\notin f$, so $(\forall y)y\neq f(1.5)$. In particular $3\neq f(1.5)$. The fact that $1.5$ is not in the domain of $f$ does not stop the proposition "$f(1.5)=3$" from being false.


1

Since $\lim\limits_{\vert x\vert\to \infty}f(x)=L$, there fore for every $\epsilon>0$, there exists $M>0$ such that $\vert f(x)-L\vert< \epsilon$ for all $\vert x\vert>M$. Also due to continuity of $f$, $f$ is bounded on $[-M,M]$


1

So, for the first one I don't know what I have to do to prove that it is a map, it seems clear that it is, as if $f$ is defined for every set in $P_1$ to every set in $P_2$, the inverse would be a map as well, moreover, we can take "some" (the down-set) of those elements in $P_2$, the inverse function will be defined for it. I would appreciate the input ...


1

An analytic proof is probably beyond the scope of your discrete math course. I suspect brute force is the intended approach from here.


1

You have just proved limit exists for a specific convergent sequence $x_n\to x_0$, you have to show then the limit is independent of the convergent sequence. That is if $x_n\to x_0,y_n\to x_0$, then $\lim f(x_n)=\lim f(y_n)$, which should be easy if you make use of the uniform continuity of $f$ on $X$ and pass to limit.


1

ADDED: To be clear, (1) I can see why such (perhaps sloppy) notation can be confusing and it's quite natural that many would find it confusing at first and I used to too and (2) nonetheless I believe that there is a value in this type of abuse of notation. But yes it demands explanation. I will first explain in generality (which you probably know, but for ...


1

Your statement is equivalent to $$\forall a,b,c\in\mathbb Z\left( \text{ if } a\mid b+c, \text{ then either } a\mid b\text{ and } a\mid c \text{ or } a\not\mid b \text{ and } a\not\mid c\right)$$ Assume $a\mid b+c$. Then $b+c=ak, k\in\mathbb Z$. Now check two cases, which are exhaustive: $1)$ $a\mid b$. Then $b=am, m\in\mathbb Z$ and so $b+c=am+c=ak\iff ...


1

$\Rightarrow$ part: If $p_1 p_2 \mid n$ then $n = m p_1 p_2 = (m p_1) p_2 = (m p_2)p_1$. It is obvious then $p_1 \mid n$ and $p_2 \mid n$ $\Leftarrow$ part: If $p_1 \mid n$ then $n = n_1p_1$. Now if $p_2 \mid n$ then $p_2 \mid n_1p_1$. But $p_2 \nmid p_1$ since $p_1$ and $p_2$ are different prime numbers, then $p_2 \mid n_1$ must hold, which means $n_1 = ...


1

Since $\{e_1,e_2,e_3\}$ is a basis, for every $u$ there exist unique scalars $\lambda_1$, $\lambda_2$, and $\lambda_3$ such that $$ u=\lambda_1\cdot e_1+\lambda_2\cdot e_2+\lambda_3\cdot e_3\tag{1} $$ This implies \begin{align*} \langle u,e_1\rangle &= \langle \lambda_1\cdot e_1+\lambda_2\cdot e_2+\lambda_3\cdot e_3,e_1\rangle \\ &= ...


1

Use the definition of $z$ transform $$Z\left\{n\right\} = \sum_{n=0}nz^{-n}=\frac{z} {(z-1)^2}. $$ now you can use the geometric series (see my answer) to find the desired result. Note: The $z$ transform of a sequence $a_n$ is given by $$ \sum_{n=0}^{\infty} a_n z^{-n}.$$


1

Prove that $2$ does not go into $n^2-2$ without a remainder for odd $n$ This basically means: “Prove that for odd $n\in\mathbb{N}$, $n^2-2$ is odd ($\Leftrightarrow$ leaves a remainder when filling it with $2$'s, saying it with your words). The key here is to observe that if $n$ is odd, so is $n^2$, and therefore is $n^2-2$. Let me give you a short ...


1

Looks reasonable. Your handwritten version doesn't exactly explain $$\begin{align} x\in A \lor x \in B \\ \text{ and ... }x\not\in A \land x \in B \\ \text{ ... to ...}\\ x \in B\\ \end{align}$$ but you do explain that in the text here. I think I would have kept the implication $x\in A \implies x \in B $ and gone through like this: $$\begin{align} x\in A ...


1

I would write it something like this. We want to show $A\cup B = B$. There are two parts: First, we want to show $B\subseteq A\cup B$. Assume $x\in B$. Then certainly $x\in A$ or $x\in B$, so we are done. Now we show $A\cup B\subseteq B$. Assume $x\in A\cup B$; we want to show $x\in B$. Since $x\in A\cup B$, either $x\in A$ or $x\in B$. if ...


1

$$3\mid a+b\mid (a+b)((a+b)^2-3ab)=(a+b)^3-3ab(a+b)=a^3+b^3$$ or $$3\mid a+b\mid (a+b)(a^2-ab+b^2)=a^3+b^3$$ This uses the fact that $a\mid b\mid c\implies a \mid c$, where $a\neq 0$. It can be proved by definition as follows: $\begin{cases}a\mid b\implies b=am,m\in\mathbb Z\\b\mid c\implies c=bk,k\in\mathbb Z\end{cases}\implies ...



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