# Tag Info

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

If $3|(a+b)$ then $3|((a+b)^3-3a^2b-3ab^2)$ so $3|a^3+b^3$.

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 ... 13\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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