# Tag Info

8

The question is a reasonable one. I think part of the issue is that there are secretly a few types of math texts and typically it's not clear which book is of which type. Some are written for people who "already know the material" but need to either fill in minor gaps or have a reference for some proofs. These will typically be written concisely ...

6

Your presentation is good, in fact almost perfect, however you should simply omit the initial word "If" which serves no purpose. The point of your presentation is that you are starting from the statement to be proved, then applying laws of algebra to find a sequence of equivalent statements, and then observing that the final statement of this ...

4

If $c$ is bigger than the local maximum, $y$ value about $3.723601932658154682760729230 ,$ then there are, at most, three real solutions to $x(x^2-1)(x^2-10) = c.$ Just draw a horizontal line at $y=c$ for some $c$ value of interest. Then, when $|c| \leq 3.723601932658154682760729230,$ we do get five real roots, but these are not all integers as the ...

4

The point of a proof is to convince the reader that your argument is sound. To that end, streamlining should not be your goal. Verbose is OK if it's instructive. Do avoid monotony. There is no algorithm for writing a good proof. Mathematical exposition is an art. Absent the details about $A$, $B$, $C$ and $D$ in your example I might suggest something like: ...

3

The second method is what everyone writes down on a bit of scrap paper. Students often write it down as a proof as well, and often forget that a double-arrow is needed, which leads to proofs like this: To prove $2 = 1$: $2 = 1$ hence $1 = 2$ as well. Add these to get $3 = 3$, which is true. so we're done and it's true! That pattern has led to a general ...

3

Doing the proof in reverse is perfectly fine, although certainly something I have never seen in a written proof. You can even write your proof in the following way $$\dfrac{x + y}2 ≥ \sqrt xy ~~\Longleftarrow~~ x + y ≥ 2\sqrt{xy} ~~\Longleftarrow~~ x + y - 2\sqrt{xy} ≥ 0 ~~\Longleftarrow~~ (\sqrt x - \sqrt y)^2 ≥ 0$$ where every "$\Longleftarrow$&...

3

Given that physics is all about converting physical problems into mathematical problems and the latter's solutions back into physical solutions, what you propose isn't in concert with physicists' true motive for introducing such intuitions. It's not about augmenting the opportunities for understanding that proofs provide. It's about several factors (which ...

3

Just introduce a "fresh" variable (to avoid overloading) and be explicit about what you're doing. For example, suppose I've proved Lemma: $\exists x\varphi(x)$. Next I have some argument. If along the way I want to introduce an object with property $\varphi$, I'll just write "by the Lemma, let $y$ be such that $\varphi(y)$" where $y$ is ...

3

For $n=1$, the inequality is $$2= {2 \choose 1} > \frac{2}{\sqrt{\pi}}$$ which is true since $\sqrt{\pi} > 1$. Now let's suppose that $${2n \choose n} > \frac{2^n}{\sqrt{\pi n}}$$ holds for a certain integer $n$, and let's prove the same inequality for $n+1$. One has {2(n+1) \choose n+1} = \frac{(2n+2)!}{(n+1)!^2} = \frac{(2n+2)(2n+1)}{(n+1)^2}{2n ... 3 \begin{align*} (L)\int_{0}^{1}\dfrac{1}{\sqrt{x}}dx&=\lim_{n\rightarrow\infty}(L)\int_{1/n}^{1}\dfrac{1}{\sqrt{x}}dx\\ &=\lim_{n\rightarrow\infty}(R)\int_{1/n}^{1}\dfrac{1}{\sqrt{x}}dx\\ &=\cdots, \end{align*} here (L) and (R) denote the Lebesgue and Riemann integrals repsectively, the first equality comes from Monotone Convergence Theorem. 2 Assume that  f'(0)  exists. thenf'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{-x\to 0}\frac{f(-x)-f(0)}{-x-0}=-\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=-f'(0)$$thus$$f'(0)=0$$remark If  f  is differentiable around zero then the differentiation of$$f(x)=f(-x)$$gives$$f'(x)=-f'(-x)$$and$$f'(0)=-f'(0)$$2 Hint:$$3125\cdot5^{5k+1}+1024\cdot4^{5k+2}+243\cdot3^{5k}=243\left(5^{5k+1}+4^{5k+2}+3^{5k}\right)+2882\cdot5^{5k+1}+781\cdot4^{5k+2},$$and 2882 and 781 are divisible by 11. 2 Your set-up can’t work (even after you fix the issue of the domain of your functions). To see this, consider the case where X=\mathbb{N}, and we have the chain (f_n,\{0,\ldots,n-1\}) with f_n defined on the nonempty subsets of n=\{0,\ldots,n-1\} and the restriction of f_n to the nonempty subsets of n-1=\{0,\ldots,n-2\} is f_{n-1}. Proceeding as ... 2 Your first sentence is not correct, though I understand what you mean. I think you should say something like: "Let \varepsilon > 0 be arbitrary. By making \delta smaller if necessary, we may assume that for all x \in \mathbb{R},$$0 < |x - c| < \delta \implies |h(x) - L| < \epsilon \text{ and } |g(x) - L| < \varepsilon."$$The rest ... 2 Your approach seems valid, but, rather than checking 18 cases, you could note that a^{2^7}=a^{128}=(a^{18})^7a^2\equiv a^2\bmod19 when \gcd(a,19)=1, so a^{2^8}=(a^{2^7})^2\equiv a^{2^2}, etc., so you would have to check only n=1,2,3,4,5,6. 1 Suppose that \mathcal{H} is any lower bound of \{\mathcal{F},\mathcal{G}\}. I'll denote the refinement partial order by \le. Let H \in \mathcal{H}. It is the case that \mathcal{H} \le \mathcal{F} so there exists some F_1 \in \mathcal{F} such that H \subseteq F_1. Also \mathcal{H} \le \mathcal{G} so there is a G_1 \in \mathcal{G} such that ... 1 Really we need to have \frac{1}{n^2} \lt \varepsilon. Solving it gives n \gt \frac{1}{\sqrt{\varepsilon}} . And as we need natural solutions we take N=\left \lceil \frac{1}{\sqrt{\epsilon}} \right \rceil. Now n \gt N solve initial inequality. 1 When working with the (\varepsilon,\delta) definition of a limit, a suitable N can often be found by starting with$$|a_n-L|<\varepsilonand manipulating the inequality until you end up with a statement of the form n>K. a_n=1/n^2 is no exception. \begin{align*} \left|\frac{1}{n^2}-0\right|<\varepsilon &\iff \frac{1}{n^2}<\varepsilon\... 1 Just by definition: \begin{align*} \lim_{x\rightarrow 0}\dfrac{f(x)g(x)-f(0)g(0)}{x-0}&=\lim_{x\rightarrow 0}\dfrac{f(x)g(x)}{x}\\ &=\lim_{x\rightarrow 0}\dfrac{f(x)}{x}\cdot\lim_{x\rightarrow 0}g(x)\\ &=\lim_{x\rightarrow 0}\dfrac{f(x)-f(0)}{x-0}\cdot g(0)\\ &=f'(0)\cdot g(0). \end{align*} 1 If f'(0) is undefined, then we are done. Let us assume that f'(0) is defined. For every \epsilon>0, there exists some \delta>0 such that \begin{align*} \left|\dfrac{f(h)-f(0)}{h}-f'(0)\right|<\epsilon,~~~~0<|h|<\delta. \end{align*} Then \begin{align*} 2|f'(0)|&=\left|\dfrac{f(\delta/2)-f(0)}{\delta/2}-f'(0)-\dfrac{f(\delta/2)-f(0)}{... 1 Let f(n)=5^{5n+1}+4^{5n+2}+3^{5n}f(m+1)-3^5\cdot f(m)=5^{5(m+1)+1}+4^{5(m+1)+2}-3^5(5^{5m+1}+4^{5m+2})=5^{5m+1}(5^5-3^5)+4^{5m+2}(4^5-3^5)$$which is divisible by 11 as 3^5=9(22+5)\equiv1\pmod{11}, 5^5=5^2\cdot5^3\equiv3\cdot4\equiv1\pmod{11} Similarly, 4^5\equiv1\pmod{11}$$\implies11|f(m)\iff11\mid f(m+1)$$1 Hint: Let f(n)=5^{5n+1}+4^{5n+2}+3^{5n}. Then$$ f(n+1)-f(n) =(5^5-1) \cdot5^{5 n + 1} + (4^5-1) \cdot 4^{5 n + 2} + (3^5-1) \cdot 3^{5n} = 11(284\cdot5^{5 n + 1}+ 93 \cdot 4^{5 n + 2} +22\cdot 3^{5n})  The crucial point is that $5^5-1,4^5-1,3^5-1$ are all divisible by $11$.

1

The second argument is (almost) OK because you carefully used "if and only if" ($\iff$) rather than just "implies" ($\implies$). You don't want to start that argument with "if". I would recommend the usual straightforward implication even if the argument the other way is how you discovered your proof. The other way is prone to ...

1

First method is flawless. But the second you correctly that if $b \ge a> 0$ and $y \le x < 0$ then $ay \le by$ and $ax \le bx$ But when we says $1- 2$ implies..... you have no rule that if $M < N$ and $J < K$ that $M -J < N-K$. As $J < K\implies -J > -K$ we have no way of comparing $M +(-J)$ to $N + (-K)$ when $M< N$ but $-J > -K$....

1

This proof looks just about fine. Here is one comment I have: By definition, for $0 < |x-c| < \delta$, $\delta > 0 \implies |h(x)-L| < \epsilon$, $|g(x)-L| < \epsilon$ for any $\epsilon > 0$. It is unclear whether or not you are taking $\delta$ to be the minimum of the $\delta$'s coming from $h$ and $g$ for any $\varepsilon>0$. ...

1

All I can think of is to assume $B$, and then assume $A\supset C$, and somehow maybe get to ${\sim}(A\supset C)$. Well, as you have found, you do not need to assume $A\supset C$ when you have derived it. Also that "somehow maybe get" is "modus ponens" or "conditional elimination", by which $\neg(A\rightarrow C)$ is inferred ...

1

The proof is going in the wrong direction. It starts with what it wants to prove and arrives at a truth. As all the steps are reversible, that is not a problem this time. It is bad practice because you have to commit to every step being reversible and people who check the proof have to be more careful. That is not the root of the problem, which you do ...

1

A derivation of a statement from axioms does not in itself establish an "absolute" truth of the statement. Instead, the derivation shows that the axioms entail the statement, i.e., the statement is true inasmuch as the rules of inference and the axioms are (under some interpretation of what the axioms mean). The certainty arises from the fact that ...

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