# Tag Info

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Induction is, in essence, a way of proving statements about integers. The "fact" that it works is really an axiom, and it says: If $P(n)$ is some property of natural number $n$ which satisfies the following conditions: $P(1)$ is true. $P(n)\implies P(n+1)$ is true. Then the statement $$\forall n\in\mathbb N: P(n)$$ is also true. In ...

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Hint If $A \times B =A \times C$ and $a \in A$ then $$\{ a \} \times B \subset A \times B=A \times C$$ This implies that $B \subset C$. Same way you get $C \subset B$: Details: Let $b \in B$. Then $(a,b) \in A \times B= A \times C$. Since $(a,b) \in A \times C$ we get by definition that $b \in C$. Therefore, as long as you can pick some $a \in A$ you can ...

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Since $x_i\geq 1$, we have $$x_{n+1}\le x_n\iff 2x_n+1\leq x_n^2\iff 2\le(x_n-1)^2\iff x_n\geq1+\sqrt{2}.$$ Now, show the latter by induction: $$x_{n+1}=\sqrt{2x_n+1}\ge \sqrt{3+2\sqrt{2}}= 1+\sqrt{2},$$ where the last equality follows since $$(1+\sqrt{2})^2=1+2\sqrt{2}+2=3+2\sqrt{2}.$$ Note: This also shows that $x_n$ is bounded below and hence converges by ...

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Simply use the definition of $x_{k+1}$. We have to show $x_{k+1}\geq x_{k+2}$. By definition, this is equivalent to $\sqrt{2x_k+1}\geq \sqrt{2x_{k+1}+1}$ Take the square this inequality: $2x_k+1\geq 2x_{k+1}+1$. This is true by the induction assumption.

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You may be misinterpreting the statement due to ambiguous parenthesization. It is almost certainly meant to be interpreted $$\forall a, b \in \mathbb R.\ ((\forall \epsilon > 0.\ a \le b + \epsilon) \rightarrow a \le b).$$ The important thing is that the scope of the quantification of $\epsilon$ should stop before the $\rightarrow$. If you think about ...

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Induction Step: Suppose we know that there exist positive integers $a$, $a_1,a_2,\dots a_k$ such that $a^2=a_1^2+a_2^2+\cdots+a_k^2$. For $i=1$ to $k$, let $b_i=3a_i$, and let $b_{k+1}=4a$. Let $b=5a$. Then $$b_1^2+b_2^2+\cdots+b_k^2+b_{k+1}^2=9a^2+16a^2=(5a)^2=b^2.$$

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HINT: The base case is $n=2$: you must find positive integers $a,a_1$, and $a_2$ such that $a^2=a_1^2+a_2^2$. All you need is one example of this; think about familiar right triangles. For the induction step step you’ll assume that you’ve found positive integers $a,a_1,\ldots,a_n$ such that $$a^2=a_1^2+a_2^2+\ldots+a_n^2\;,\tag{1}$$ and you’ll try to use ...

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Maybe I got it wrong, but I think that the reason for proving $$\prod_{k=1}^{n}\frac{2k-1}{2k}\leq \frac{1}{\sqrt{3n}}$$ by proving a stronger inequality, is just that the stronger inequality $$\prod_{k=1}^{n}\frac{2k-1}{2k}\leq \frac{1}{\sqrt{3n+1}}$$ is way easier to prove by induction. So it is just a technical reason. Anyway, we may also notice ...

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I'd say don't give up. This kind of thing comes with experience. One of the best advice I've seen (from this site) is to practice everyday, even a little, or you'd get rusty even on things that you are quite comfortable with. Also, when you read a math book, don't just read it, work on it too. Do a lot of exercise to "get the feeling" of a theorem or a ...

2

Dividing $a$ and $b$ by $d$, we have two relatively prime numbers. By the relatively prime numbers equation, it exist $r$ and $s$ such that $r\frac{a}{d} + s\frac{b}{d} = 1.$ Those numbers are relatively prime. If not, the left part could be factorise giving an integer factorisation of 1... Multiplying everything by $d$ again to obtain what you need

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We know that $a = d\bar{a}$ and $b = d\bar{b}$, so dividing $$ar + bs = d$$ by $d$ we get $$\bar{a}r + \bar{b}s = 1$$ which is what you call the "relatively prime equation", hence $r$ and $s$ must be coprime. To conclude without assuming this, just suppose that there is an $f > 1$ that divides both $r$ and $s$. Then the above equation implies $f \mid ... 2 A far shorter proof : Define the function :$F(x) = \int_0^x f(t) dt$By hypothesis,$\forall x \in [0,1], F(x) = 0$:$F$is the null function. But as$f$is continuous, we can use the fundamental theorem of calculus, to get that$f = F'$. So$f = 0$2 Your idea is a neater form : Let$f \not \equiv 0$a continuous function that verify the hypothesis. There exists$c \in [0,1]$such that$f(c) =M$Without loss of generality, we can take$M>0$, and by continuity of$f$,$c\in]0,1[$By continuity of$f$, there exists$\delta > 0$such that$\forall x \in [c-\delta,c+\delta], f(x) > \frac{M}{2}$. ... 2 $$\frac{10^n}{n!} = \frac{10^{10}}{10!} \cdot \frac{10}{11} \cdots \frac{10}{n} < \frac{10^{10}}{10!} \frac{10}{n}$$ As$n$approaches$\infty$, the above fraction approaches$0$. 2 If you (really already) know you have a basis, then$\{v_1,\ldots, v_n\}$are linearly independent. So if$v = \sum_{1}^k \lambda _i v_i$is in the intersection then also$ v= \sum_{k+1}^n \lambda _i v_i$and then $$0=\sum_{1}^k \lambda _i v_i - \sum_{k+1}^n \lambda _i v_i$$ which by definition of linear independence implies$\lambda_i=0$for all$i $an so ... 2 [The right side of your equality should be$bf(b)-af(a)$.] You are correct that this equality holds under the assumption that$f$is continuous and strictly increasing. No differentiability is needed, and your "area proof" is doubtless the best way to see this. The differentiability hypothesis is a mere expedient: It permits your to view the two sides of ... 2 Hint: Square both sides of the equation $$|a + b| = |a| + |b|.$$ Can you take it from here? 1 If$x \in \mathbb{R}$, then $$|f(x) - f(1)| = |x^{2}+x|x-1|-1| \leq |x-1|(|x| + |x+1|);$$ if$|x-1| < 1$, then$0 < x < 2$, so $$|x-1|(|x| + |x+1|) < 5|x-1|;$$ take$k := 5. 1 From some intermediate step to the conclusion: \begin{align*} x \in \bigcup \{A,B,C\} &\iff \left(\exists X \in \{A,B,C\}\right) x \in X \\ & \quad \ \vdots \\ &\iff \exists X\left( \left(\left( X = A \vee X = B \right) \vee X = C\right) \wedge x \in X \right) \\ &\iff \exists X\left( \left(\left( X = A \vee X = B \right) \wedge x \in X ... 1 First Proof Claim.kd\mid d$. Proof$d\mid a \land k\mid r \implies kd\mid ard\mid b \land k\mid s \implies kd\mid bs\therefore kd\mid ar+bs\implies kd\mid d\implies ??$Second Proof But I think that the proof can be done in much simple way if you just notice that$r\left(\dfrac{a}{d}\right)+s\left(\dfrac{b}{d}\right)=1$. ... 1 Since you are learning set theory, you won't need to read Chapter 1. You'll probable need to skim over it, to see what's covered, the notation etc. Judging by the edit you made to your post, I'm assuming that you're primary goal is to be able to use basic theorems and methods in general topology, as they apply to other subjects. This is as opposed to ... 1 We can apply the ratio test to the sequence$a_n = \frac{10^n}{n!}$: $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{10^{n+1}\cdot n!}{10^{n}\cdot (n+1)!}=\lim_{n\to\infty}\frac{10}{n+1}=0\color{red}{<1}\implies \lim_{n\to\infty}a_n = 0$$ 1 Proceeding with your argument: $$\alpha \beta -\varepsilon(\alpha+\beta -\varepsilon)<ab\le \alpha \beta$$ and this is true for every$\varepsilon>0$. Now let$\eta >0$be arbitrary, and put$\varepsilon = \min \left \{\frac{2\eta}{\alpha+\beta} , \frac{\alpha+\beta}{2}\right \}$. It follows that: ... 1 The product $$\prod_{p\equiv 1 (\bmod 3)}\frac{p}{p-1}=\prod_{p\equiv 1 (\bmod 3)}\left(1+\frac{1}{p-1}\right)$$ converges if and only if the sum $$\sum_{p\equiv 1 (\bmod 3)}\frac1{p-1}$$ does (just take the logarithm of the infinite product to see it). Now, the sum in question is infinite because of Dirichlet's theorem on primes in arithmetic ... 1 In case anyone was interested : $$B = B + 0 = B + ( A + (-A) ) = (B + A) + (-A) = (A + B) + (-A)$$ We know (A + B) equals (A + C) so we'll have : $$(A + B) + (-A) = (A + C) + (-A) = C$$ which implies $$B = C$$ 1 I don't believe your proof is quite correct as written. You say "let$q$be the first point at which$f = 0$," but what does this mean? It's not automatic that there is a smallest such value. (For instance, there are continuous functions vanishing on all the$1/n$.) 1 Not sure what the well ordering principle is, but here is how I would prove this:$A_{18} = \{4,7,7\}A_{19} = \{4,4,4,7\}A_{20} = \{4,4,4,4,4\}A_{21} = \{7,7,7\}A_{n} = \{4\} \cup A_{n-4}$1 Your product of prime powers with odd exponents is not square-free unless the$\alpha_i$are all$0$. Use$p_i^{2\alpha_i+1} = p_i\cdot p_i^{2\alpha_i}$to reduce it to the product of a square-free integer and a square. 1 Every (positive) real number has a simple continued fraction expansion, one which terminates if and only if the number is rational. Let's start with how one expands$x \gt 0$in a simple continued fraction, and then reflect on the special case where$x = a/b$is rational. First off, we take the integer part and fractional part of$x$:$$x = \lfloor x ... 1 The original statement uses "whenever", so implicitly it's saying something about all "times", that is, "occasions".$x$ranges over "times", "occasions" – times when the other person does that, times when you get mad. So: let$P(x)$mean "I get mad at (or a little after) time x", and let$Q(x)\$ mean "you do that at time x". Then "I get mad when you do ...

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