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7

An overkill proof. By the Lindemann--Weierstrass Theorem, $e^i$ is transcendental. I can get $e^i$ from $\sin(1) = (e^i-e^{-i})/(2i)$ by solving a quadratic equation. Thus, if $\sin(1)$ were algebraic, then also $e^i$ would be algebraic.


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

Well, I think that by "closure" your text-book means that addition and scalar multiplication are well-defined. Indeed, call $V$ your candidate vector space of all the sequences with finite support (i.e., with a finite number of non-zero terms), then you can think to $V$ as a subset of the cartesian product $\mathbb K^{\mathbb N}$ (the space of all sequences ...


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

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


2

The rules of the game are unclear. You have to tell us the underlying (abelian) group of which $A$ is a subset. E.g., if $A={\mathbb Z}_2\oplus{\mathbb Z}_2$ is the four group then $|A|=4$ and $|A+A|=|A|=4$.


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

You could start with Wilsons's theorem $(p-1)!\equiv -1 \mod p$, and then try to make $(p-3)!$ appears in this identity...


2

Your answer is the right idea, but you need a few more details. Let $c = \sup \{ t \in [a,b] | f(t) = f(a) \}$, then we have $c <b$ and $f(c) = f(a)$ by continuity. Furthermore, $f(x) > f(a)$ for all $ x > c$. Now let $d = \inf \{ t \in [c,b] | f(t) = f(b) \}$. We have $c < d$ and $f(d) = f(b)$. Furthermore, $f(x) < f(b)$ for all $ c \le x ...


2

This is known as Bernoulli's Inequality: for all $x\ge-1$ and non-negative integer $n$, $$ (1+x)^n\ge1+nx $$ This can be proven by induction: Note that the inequality above is true for $n=0$. Suppose that $x\ge-1$ and for a non-negative integer $n$, we have $$ (1+x)^n-nx\ge1 $$ Then $$ \begin{align} (1+x)^{n+1}-(n+1)x &=(1+x)^n-nx+x(1+x)^n-x\\ ...


2

Let's get there without using division. We are given that $ab=0$ If $b>0$, we have $\begin{align} (a+1)b&=ab+b \\ \text{but }\hspace{1in}ab&=0\\\implies (a+1)b&=b\\ \implies (a+1)&=1\\ \implies a&=0\\\end{align}$ and similarly $a>0 \implies b=0$ Therefore $ab=0 \implies a=0$ or $b=0$


1

I feel like your question needs some more context (what axioms you have to use, etc.), but I will try to provide an easy proof that uses minimal "sophistication": Problem: If $ab=0$, then either $a=0$ or $b=0$. Proof. Suppose $ab=0$ and, without loss of generality, that $a\neq 0$. Then $a^{-1}$ exists and $$ b=1\cdot b = (a^{-1}\cdot ...


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

Let a rectangle be defined by two points $[a,b]$ in $\mathbb{R}^{n}$ as the the set of $x$ such that $a\le x\le b$. (The inequalities must hold for every component.) If for some component $a_i>b_i$, the rectangle is void. Now the intersection of the rectangles $[a,b]$ and $[c,d]$ is the set of $x$ such that $a\le x\le b\land c\le x\le d$. In other ...


1

The trick is not to induct on $m$, just do the induction on $n$. You don't have to induct for both variables. Let $m\ge 0$ be arbitrary instead, then the base case $n=0$ is verified since $1+0=(1+m)^0$. Next assume we have proven this for some $n\ge 0$. Then $$1+mn\le (1+m)^n$$ Multiply both sides by $(1+m)$ to get $$1+mn+m+m^2n\le (1+m)^{n+1}$$ now we ...


1

This is another approach which avoids $\sup$ and $\inf$. As mentioned in my answer there is a last value of $x$ in $[a, b]$ for which $f(x) = f(a)$. This value of $x$ we denote by $c$. Clearly $a \leq c < b$. Also note that if $x > c$ then $f(x) > f(c) = f(a)$ (why?). Similarly there is a first value of $x \in [c, b]$ for which $f(x) = f(b)$. This ...


1

 am+bn=1=ck+cj=c(k+j) should be $$ am+bn=1=ckm+cjn=c(km+jn)=1$$ And your proof is valid.


1

Let’s look at the first projection: you want to show that $\operatorname{pr}_1(\alpha\setminus\beta)\supseteq\operatorname{pr}_1\alpha\setminus\operatorname{pr}_1\beta$. Suppose that $a\in\operatorname{pr}_1\alpha\setminus\operatorname{pr}_1\beta$; you need to show that $a\in\operatorname{pr}_1(\alpha\setminus\beta)$. What does it mean for $a$ to be an ...


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

Hint: If $k$ is odd then, $2^k+1=2^k-(-1)^k$, now use that $x^n-y^n $is divisible by $x-y$ to conclude. Proof that $x^n-y^n$ is divisible by $x-y$: We prove this by induction, $n=1$ follows, let $x^k-y^k$ be divisible by $x-y$. Then $x^{k+1}-y^{k+1}=x(x^k)-x(y^k)+x(y^k)-y(y^k)=x(x^k-y^k)+y^k(x-y)$, which is divisible by $x-y$. Hence by induction the ...


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 ...


1

You start with $X$ and with $f$. Now you define $B$, and you say that $b\in X$ is such that $f(b)=B$. Since $B$ is a subset of $X$ and $b$ is an element of $X$, it has to be that either $b\in B$ or $b\notin B$. Deriving $b\in X$ from $b\in B$ is meaningless, it tells us nothing new. But asking whether or not $b$ is also an element of $B$ or not, that is ...


1

(a). $X \setminus (X \setminus A) =A.$ Note that $x \notin X \setminus A \Leftrightarrow x \in A.$ Thus $x \in X \setminus (X \setminus A) \Leftrightarrow x \notin X \setminus A \Leftrightarrow x \in A.$ (b). $A\subset B \Rightarrow X\setminus B \subset X\setminus A.$ Let $x \in X\setminus B.$ Then $x \in X, x \notin B.$ Since $A \subset B, x \notin B ...


1

$(m+n)^2=m^2+n^2+2mn$, let $m^2+n^2=4k$ then $(m+n)^2=4k+2mn$ , now as RHS is divisible by $2$ LHS will also be divisible by $2$ but LHS is square of some quantity thus if $2$ divides it $4$ will also divide now consider $(m+n)^2-4k=2mn$. Now LHS is divisible by $4$ so $2mn$ is divisible by $4$ thus $mn$ is divisible by $2\implies$ atleast one of $m$ or $n$ ...


1

I am following your idea. $$A = \{a \le x < x_0 | f(x) = 0 \}$$ $$B = \{x_0 < x \le b | f(x) = 0 \}$$ Let $c = \sup A$ and let $d = \inf B$ ($\sup B$ doesn't work here, since it's always $b$). Note $c<x_0<d$ Step 1: prove $f(x) > 0$ for $x \in (c, d)$ we will use contradiction. Assume $f(x) < 0$, for some $c<x<d$, by IVT ...



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