# Tag Info

## New answers tagged proof-verification

0

The proof works up to line 5 as user21820 explained. The assumption made on your line 6 would not be easy to discharge later. Rather take the derived line 5 and continue with it. Here is a completed proof using a Fitch-style proof checker which you may use to check this and other proofs. On lines 7 to 9, I derived $\neg\neg R$ to use modus tollens on ...

0

If you want to avoid using partial fractions, then you could take this approach. We are given that $$\frac{dx}{dt}=3x(x-5)$$ which rearranges to (observing that $x\equiv 0$ and $x \equiv 5$ are solutions to the ODE but not the IVP) \frac{1}{x(x-5)}dx=3dt \implies-\frac{1}{5}\Big(\frac{(x-5)-x}{x(x-5)}\Big)dx=3dt\implies -\frac{1}{5}\Big(\frac{1}{x}-\... 1 HINT \begin{align*} \frac{1}{3x(x-5)} = \frac{1}{3}\times\frac{1}{x(x-5)} = \frac{1}{15}\times\frac{x - (x-5)}{x(x-5)} = \frac{1}{15(x-5)} - \frac{1}{15x} \end{align*} 2 Before giving you a complete solution I want to point out the mistakes in your solution. Your partial fractions are correct. As mentioned in the comments \int \frac{1}{ax} dx \ne \ln | a x |. Instead we have \int \frac{1}{ax} dx = \frac{1}{a} \int \frac{1}{x} dx = \frac{1}{a} \ln | x| $$for a \ne 0. We have e^{\ln(a) - \ln(b)} \ne a - b. This is ... 0$$ 1 = \frac{A}{3x} + \frac{B}{x-5} $$if and only if$$ (3x)(x-5) = A(x-5) + B(3x) \text{,} $$with x \not\in \{0,5\}. How do we get "15xB"? 0 There's a mistake. Let \alpha = \sup(\mathbb{N}) which exists by the reasons you mentiond. It is true that this means that for all \varepsilon>0 there exists n\in\mathbb{N} such that \alpha-\varepsilon <n\leq \alpha. From this you conclude that \alpha<n+\varepsilon which is fine. However this does not mean that \alpha\leq n. You ... 0 You may have the right idea, but for the proof to be complete and rigorous you need to more clearly justify the following crucial inference: When we square a number, we merely repeat its factors, therefore A^2 and B^2 must also not share any factors". As it stands, your justification "when we square a number, we merely repeat its factors" could be ... 1$$ \sum\limits_{n = 1}^{ + \infty } {\left( { - 1} \right)^n } \frac{1} {{\sqrt {4n + 1} }} $$is convergent by Leibniz. The same is true for$$ \sum\limits_{n = 1}^{ + \infty } {\left( { - 1} \right)^n } \frac{1} {{\sqrt {5n + 1} }} $$thus the given series is convergent because is the difference of two convergent series. 1 Your proof is fine. You showed that the sequence is decreasing and bounded below (thus convergent). You might elaborate that the equation a = (a+1/a)/2 has two solutions (a= \pm 1), but only a=1 can be the limit. Alternatively observe that$$ 0 \le x_{n+1} - 1 = \frac{(x_n-1)^2}{2x_n} \le \frac{(x_n-1)^2}{2} $$which also implies convergence x_n \... 3 I will mention one (easily corrected) logical error and one stylistic piece of advice that could make the proof more readable. But the upshot is that this is a well-argued proof by any standard, and especially impressive for a first effort. When you said that A^2 and B^2 share no factors aside from 1, that does not imply that \frac{A^2}{B^2} is not ... 1 If one doesn't count multiplicities of roots$$ f(x)=36x^4-76x^3+42x^2+1 $$has one point at which f(x)=1, two points at which f(x)=2, three points at which f(x)=3, and four points at which f(x)=4. 2 Since f(x) - 1 has only one root and all coeficients are in \mathbb{R}, we have only two cases to consider: Case 1. f(x) - 1 = a(x - \alpha)^4 In this case, the function obtained is convex (if a > 0) and concave otherwise. In particular, it does not have any horizontal line that crosses it 3 times. Case 2. f(x) - 1 = a(x - \alpha)^2 g(x) ... 1 Consider f a polynomial of degree 4, then its derivative is a polynomial of degree 3, so it has at most three critical points. There are only two cases. First case: localmin-localmax-localmin (or max-min-max). The plot you have to imagine is this one. If you count the roots from the global minimum upwards you find 1 \rightarrow 2 \rightarrow 3 \... 3 It is correct. I would have used Abel's test to justify the convergence of \sum_{n=1}^\infty\frac{a_n}n without the assumption that (\forall n\in\mathbb N):a_n\geqslant0. It allows you to deduce that, say, \sum_{n=1}^\infty\frac{na_n}{n+1} also converges. 0 By the physics rules, I have that: l=\frac{1}{2}at^2. Substituing: l=\frac{1}{2}g\cdot \sin(\alpha)\cdot t^2. Also: \sin(\alpha)=\frac{h}{l}. From this: 2l^2=ght^2 and so: t=l\sqrt{\frac{2}{hg}}. 1 I think your mistake is in deriving the constant of integration C. s measures distance along the plane, not vertical distance. So you should have C=l, not C=h. If the acceleration due to gravity is g then you should get t=l\sqrt{\frac{2}{gh}} If l is in feet and you take g to be 32 feet per second^2 then you have t=\frac{l}{4\sqrt{h}... 1 There is an error in your calculation. The acceleration 32\sin A is along the slanted plane, not vertically. Thus, the object travels a distance of l, not h. So, the correct equation after your derivation is instead,$$-16t^2\sin A + l = 0,$$which leads to the correct answer. 2 You know that \alpha: X \to P satisfies$$\forall j: \beta_j \circ \alpha = \text{pr}_j\tag{1}$$and \beta: P \to X satisfies$$\forall j : \text{pr}_j \circ \beta= \beta_j \tag{2}$$now using (1) and (2) we get that for any j:$$\text{pr}_j \circ (\beta \circ \alpha) = (\text{pr}_j \circ \beta) \circ \alpha = \beta_j \circ \alpha = \text{pr}_j\...

0

Anothet approach is to use the theorem that A is connected iff for all continuous f:A -> {0,1} with the discrete topology, f is constant. To prove your problem and the generalization: connected A and A subset B subset $\bar A$ implies B is connected, one can use the above theorem and the fact that for continuous f, f($\bar A$) subset $\overline {f(A)}$.

1

1) Depending on your definition of monotonically increasing, being $p_{n} > p_{n-1}$ or $p_{n} ≥ p_{n-1}$. The first case is easily disprovable, just pick $a_{n} = 2$, and $p_{1}$ and $p_{2}$ are both 2. For the second case, since $p_{n}|a_{n} \rightarrow p_{n}|a_{n} + p_{n} \rightarrow p_{n}|a_{n+1}$, so $p_{n+1}$ is at least $p_{n}$. So depending on ...

3

It's better to start with an element $n\in N$ then show that $n$ commutes with any $g\in G$, like so. Since $N$ is normal and $g$ is arbitrary in $G$, we have $gng^{-1}\in N$. Since $N$ is a subgroup of $G$, we have $$h:=\underbrace{gng^{-1}}_{\in N}n^{-1}\in N.$$ But $h$ is the commutator $[g, n]$ of $g$ with $n$, so $h\in N\cap G'=\{e_G\}$, whence $h=gng^{... 2 Well, this is a consequence of the famous Steinitz exchange lemma: If$\{v_{1},\dots ,v_{m}\}$is a set of$m$linearly independent vectors in a vector space V, and$\{w_{1},\dots ,w_{n}\}$spans$V$, then$m\leq n$and after reordering$\{v_{1},\dots ,v_{m},w_{m+1},\dots ,w_{n}\}$spans$V$. Here you take$\{v_{1},\dots ,v_{m}\}$as a basis of$W_1$and$\...

0

you can just say: Let $\{w_1, \ldots, w_n\}$ be a base for $W_1$, and let $\{w_1, \ldots, w_m \}$ be a base for $W_2$. Because $W_2 \subseteq W_1 \to Sp(W_2) \subseteq Sp(W_1) \to \{w_1, \ldots, w_m \} \subseteq \{w_1, \ldots, w_n \}$, hence $m \le n$ (if $W_1 \subseteq W_2 \to m=n$, and $W_1 = W_2$).

0

If the function is of exponential order, then by definition it follows that we can find constants $a$ and $M>0$ such that $$|f(t-t_0)|\le Me^{a(t-t_0)}$$ whenever $t>t_0.$ Since this function is continuous it follows that as $t_0\to 0,$ we must have $f(t-t_0)\to f(t).$ Therefore, in this limit, we have $$|f(t)|\le Me^{at},$$ since the exponential ...

5

There is no such thing as "more correct" when talking about proofs. A proof can only be one of two things: correct or incorrect. A half correct proof is incorrect. A 99.999999% correct proof is incorrect. Both your proofs are correct, and there is nothing better or worse about either of them. Personally, I think the second is a little easier to understand, ...

3

The mistake is in thinking that $x \in E_n$ iff $\frac 4 {10^{n}} \leq x \leq \frac 5 {10^{n}}$. In fact $E_n$ is a union of $10^{n-1}$ disjoint intervals of length $\frac 1 {10^{n}}$ each, so $m(E_n)=\frac 1 {10}$. For each $n$. Using Borel - Cantelli Lemmas we can show that $m(E)=1$.

-1

Disregarding the validity of its mathematical correctness, here is my thought on how you can structure it in the form of a mathematical statement and proof. Club the $key$ and $formula$ section together as $Keys$, starting with the formula section, ending with your key section. Then put $X=E$ as the $Formula$ and the remaining part of the proof as its ...

5

If $F\colon \mathcal{C}\to \mathbf{Set}$ is any functor and $f$ is an iso in $\mathcal{C}$, then $Ff$ has inverse $F(f^{-1})$. This is an inverse in $\mathbf{Set}$, as you've noticed, and the isos in that category are precisely the bijections. However, what I suspect you want to ask is if in every concrete category, an arrow is an iso iff its underlying ...

2

If different linear combinations give the same vector $v$, we have $v=a_1v_1+\dots+a_nv_n=b_1v_1+\dots +b_nv_n$ and then subtracting, we get $(a_1-b_1)v_1+\dots+(a_n-b_n)v_n=0$, without all $a_i-b_i=0$. This contradicts linear independence of the $v_i$.

1

By contradiction assume otherwise, so there exists scalars $a_i$ and $b_i$ such that $a_j\neq b_j$ for some $j$ and $\sum a_i v_i=\sum b_i v_i$, then $$\sum (a_i-b_i)v_i=0$$ can you take it from here?

1

Here is a proof of the result by cases using a proof checker: Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

2

$$\frac{dx}{dt}=3x(x-5)\implies\frac{dx}{x(x-5)}=3dt\implies$$ $$\int\frac{dx}{x(x-5)}=\int3dt\implies\int\frac{1}{5}\left(\frac{1}{x-5}-\frac{1}{x}\right)dx=3t+c\implies$$ $$\frac{1}{5}\ln|x-5|-\frac{1}{5}\ln|x|=\frac{1}{5}\ln\left|\frac{x-5}{x}\right|=3t+c$$ Taking the exponential of both sides, $$\left(\frac{x-5}{x}\right)^{\frac{1}{5}}=e^{3t+c}=e^{3t}e^... 0 Put more formally, prove that, for L, K, n \in \mathbb{Z}, n \ge 1,$$ \tag{1}\label{proposition} (L < K) \wedge (L < m \le K) \wedge (K/n \text{ is upper bound}) \wedge (L/n \text{ is no upper bound}) \Rightarrow (\exists m) (m/n \text{ is upper bound}) \wedge ((m-1)/n \text{ is no upper bound}) $$Proof is by induction over K - L. Base case: ... 0 Yes, just a bit more is needed. Assume x\to y as a premise. Assume y\to z as a premise. Assume x. Derive y from 1 and 3 via modus ponens. Derive z from 2 and 4 via modus ponens. Deduce x\to z from subproof 3-5. \blacksquare~x\to z is derivable from x\to y, y\to z 2$$y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$$So$$y^{'}(x)=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$import these power series to main differential equation y^{'}(x)=1+xy(x)to receive to$$\sum_{n=0}^{\infty}na_{n}x^{n-1}=1+\sum_{n=0}^{\infty}a_{n}x^{n+1}$$by simplifying this equation:$$a_{1}=1 , a_{3}=\frac{1}{3}, a_{5}=\frac{1}{15}, \cdot\cdot\cdot,a_{2n-1}=...

1

If the conditions of the Picard–Lindelöf theorem are satisfied then the function sequence $(y_n)$ defined iteratively by $$y_0(x) = 0 \, , \\ y_{n+1}(x) = Ty(x) = x + \int_0^x t y_n(t) \, dt$$ converge to a solution of the initial value problem. This is called Picard-iteration. The first iterates are \begin{align} y_0(x) &= 0 \\ y_1(x) &= x + ... 2 (\Rightarrow) Your proof is correct, but you start with p, then you start calling it n and in the end it's p again. Don't do that. (\Leftarrow) There is the same problem here. Besides, when you wrote 1<a<n, you should have written 1\leqslant a<b. Finally, you should explain how you passed from a\nmid p to the assertion that \mathbb ... 1 I have a possibly more rigorous proof for the \Leftarrow proof: Assume p is prime. Now, consider ab \equiv 0 \pmod p. This means, ab=np for some n \in \Bbb{Z}. Thus, p \mid ab, so by the definition of a prime element, either p \mid a or p \mid b. Therefore, either a \equiv 0 \pmod p or b \equiv 0 \pmod p. This proves that \Bbb{Z}_p ... 1 Your partial fraction set up is \begin{align*} \frac{1-v^2}{v(1+v^2)}&=\frac{A}{v}+\frac{B\color{red}{v}+C}{1+v^2} \end{align*} Then you got the values as A=1, B=-2 and C=0. This means \begin{align*} \frac{1-v^2}{v(1+v^2)}&=\frac{1}{v}-\frac{2\color{red}{v}}{1+v^2} \end{align*} Now when you integrate you get \begin{align*} \int \frac{1-v^2}{v(1+... 0 Let d=\gcd(a,b) with d>1. Because of the proprieties of \gcd(a,b), I have: (d\mid a) \land (d\mid b). Squaring, I obtain: (d^2\mid a^2) \land (d^2\mid b^2). d is the \gcd(a,b) so also d^2 is the \gcd(a^2,b^2). Thus d^2>1 and \gcd(a^2,b^2)>1. 0 You don't need a proof by contradiction for such a statement: if d=\gcd(a,b), d divides a, hence it trivially dids a^2. Similarly, it divides b^2, hence, it divides \gcd(a^2, b^2), and it is greater than 1… 2 All that you have proved was that if your limit exists, then it cannot be greater than 0. You can prove that it is 0 using the fact that\left\lvert(x^2-y^2)\sin\left(\frac1{x^2+y^2}\right)\right\rvert\leqslant x^2+y^2.$$0 Here are three ways to see that if X\to Y and Y\to Z then X\to Z. Consider how this might look in a diagram. If X is a sufficient condition for something else, say, Y, then whenever we have X we have Y. That could be represented as one circle containing everything in X being fully contained in a circle containing everything in Y. Here is ... 0 There is an easier way: 4^n+6n-1\iff (3+1)^n+6n-1=(9M+3n+1)+6n-1=9(M+n) 1 And now for something totally different: your condition implies that G has unique Sylow subgroups and hence is nilpotent. This reduces the problem to p-groups. Now maximal subgroups of p-groups all have index p, hence have the same order. Again, the condition now implies G to have a unique maximal subgroup M. Pick a g \in G with g \notin M, ... 0 Schur's inequality:$$a^3+b^3+c^3+3abc \ge a^2(b+c)+b^2(c+a)+c^2(a+b) \iff \\ (a+b+c)^3+9abc\ge 4(a+b+c)(ab+bc+ca) \Rightarrow \\ 1+9abc\ge 4(ab+bc+ca) \quad (1)$$Rearrangement:$$a+b+c=1 \Rightarrow a^2+b^2+c^2=1-2(ab+bc+ca)\ge ab+bc+ca \Rightarrow \\ 1\ge 3(ab+bc+ca) \quad (2)$$Now add (1) and (2). 1 What you have written is actually how most people would prove this in their heads. But if you are looking for a valid mathematical proof, you have to go backwards. What I mean by going backwards is something like this: Proof. Let M \gt 0 be arbitrary. Let N = M. Then for all n \gt N we have:$$a_n = n+1 \gt n \gt N = M.$$Thus we have found the ... 0 Yes, your proof is correct. Without returning to the definition, you could also say that$$ a_n = n+1 \rightarrow + \infty $$when n \rightarrow + \infty. 0 A proof by SOS:$$2+9abc-7(ab+ac+bc)=2(a+b+c)^3+9abc-7(a+b+c)(ab+ac+bc)==\sum_{cyc}(2a^3+6a^2b+6a^2c+4abc+3abc-7a^2b-7a^2c-7abc)==\sum_{cyc}(a^3-a^2b-ab^2+b^3)=\sum_{cyc}(a-b)^2(a+b)\geq0. Also, $uvw$ kills it immediately.

2

Your proof looks good. Here's an unsophisticated alternative proof, using only elementary algebra . . . We don't even need $a,b,c$ to be nonnegative. As shown below, if $a,b,c\;$are real numbers such that $a+b+c=1$, and if at least one of $a,b,c\;$is between $-1$ and ${\large{\frac{7}{9}}}$ inclusive, then the inequality holds. Without loss of generality,...

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