7

I would assume that the teacher considered it an invalid answer, because your son assumed that the equality is true before showing that it is, if that even makes any sense. This is obviously a guess, but in the mind of the teacher, what he did was not "showing" that the equality is true, but merely "verifying" that it is. In other words, ...


5

When you write $|x-2| < \varepsilon/3-1$ this means $$ 0\le|x-2| < \dfrac{\varepsilon-3}{3}\\ \varepsilon \ge 3 $$ For this inequality to hold true But in the limit definition we require $ \forall \epsilon > 0$ EDIT To prove that a sequence don't have a limit or in other terms it diverges we need to show that there doesn't exist any real number $l$ ...


5

All of these methods of proof are valid, and should be graded as such (unless the question specifically said to use a certain method). Method $1$ - Start with one side, simplify to the other. $$\frac{\csc(\theta)-1}{\cot(\theta)}=\frac{(\csc(\theta)-1)(\csc(\theta)+1)}{\cot(\theta)(\csc(\theta)+1)}=\frac{\csc^{2}(\theta)-1}{\cot(\theta)( \csc(\theta)+1)}=\...


5

I would at least have skipped the first line and started with $\dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta.$ One reason why an instructor might have qualms about this argument is the frequency with which students do this in the opposite direction, i.e. they write $$ \require{cancel} \xcancel{ \begin{align} \frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{...


5

There are multiple motivations for writing proofs: perhaps to pass an exam (from a regular student's viewpoint), or perhaps to test understanding (from a teacher's perspective). Here, I'd like to highlight another important motivation: to enjoy the process of discovery. Here's a geometric proof of the trigonometric identity. It is not practical to use in an ...


4

Since $2|a^3$, then $2|a$ because $2$ is prime (Euclid's Lemma). Later on when you say $4|b^3$, you can argue that this implies $2|b^3$. By the same argument as before, $2|b$ (as mentioned in the comments). This contradicts that $\text{gcd}(a,b)=1$ (you will need to state this as well in your proof because the fraction $\dfrac{a}{b}$ must be reduced to ...


4

The issue is that in going from $x$ to $x+1$ you assume the hypothesis is true for both $x$ and $x-1$. But if $x=1$ then $x-1$ is out of range, so the step from $1$ to $2$ doesn't work.


4

With a little bit of rigor, you can do this of course. You need to define what means $B$ approaches $A$. For example, you can take the angle $\theta=\angle DEB$ and say $\theta\to\theta_{\tau}$, where $\theta_\tau$ is the angle between $CD$ and the tangent from $E$ to the circle (let's call the point $A_\tau=B_\tau=T$). Next, you want to show that indeed $...


4

The answer is negative: if $A$ is finite and $B$ is countably infinite then $B^A$ is either finite or countably infinite, and hence is certainly not uncountably infinite. You can prove this by induction on the size $|A|$ of $A$. If $|A| = 0$, then $A$ is empty and $B^A$ comprises the single function $\emptyset$. If $|A|= k + 1$, pick $a_0$ in $A$ and observe ...


3

Hint $x$ can not be further away from $0$ than both $a$ and $b$. Consider that $x$ must have the same sign as at least one of $a,b$.


3

Of course it's fine to show directly it's a finite set if that is what you're trying to prove! Your proof is fine. If we write it out very explicitly: you have shown shown for an arbitrary element $(x,y) \in A$, we have $xy \leq 4$. Since we know that $x$ and $y$ are natural numbers, this then implies that ($x=1 \text{ and } y=4$) or ($x=2 \text{ and } y=2$) ...


3

Because then what we would have would be: For any number $a$, one, and only one, of the following holds: (i) $a=0$, (ii) $a<0$, (iii) $0<a$. It is not obvious that you can jump from this to: For any number $a$, one, and only one, of the following holds: (i) $a=0$, (ii) $0<-a$, (iii) $0<a$. You will need some connection between $<$ and the ...


3

We have $ \int_0^t f(s)dB_s \sim N(0;\int_0^t f^2(s)ds)$. Let's denote $a =\int_0^t f^2(s)ds$, then it suffices to calculate $E(e^{aX^2})$ with $X \sim N(0,1)$. And $E(e^{aX^2})$ exists if and only if $a<\frac{1}{2}$. If this condition is satisfied, then $E(e^{aX^2}) = \frac{1}{\sqrt{1-2a}}$ Conclusion: if $\int_0^t f^2(s)ds < \frac{1}{2}$ then $$E\...


3

It's often better to prove it in one line: $2(k+1) = 2k + 2 < k!+2 < k!\cdot (k+1) = (k+1)!$, where the first inequality is by induction hypothesis and the second inequality holds for $k\geq 4$ anyway (for completeness, it could also be proved by induction).


3

Fixe that $2k(k+1) > 2(k+1)$ and not $2k(k+1) > 2k$ and your proof is fine. Alternatively it might be simpler to go left to right rather than right to left $2k < k!$ $2k + 2 < k! + 2 < k! + 2\cdot 3 < k! + 2\cdot 3\cdot 4 < ...... < k! + k! =$ $2k! < (k+1)k! = (k+1)!$. .... You know.... as far as inequalities go... this aint even ...


3

Per OP's request This is not an answer. However, the work is too long-winded to be placed as a series of comments. $a>0$ and $b>0$ $$2+\log_{2}{a} =3+\log_{3}{b}=\log_{6}{(a+b)}$$ Compute $$ \log_{ab}{\left(\frac{1}{a}+\frac{1}{b}\right)}$$ Let $x = \log_2 a, ~y = \log_3 b, ~z = \log_6(a+b) \implies$ $2^x = a, 3^y = b,~$ and $~6^z = (a+b).$ Also ...


3

A probably awful solution. Rewrite the sustem as $$2+\log_{2}{a} =3+\log_{3}{b}=\log_{6}{(a+b)}\color{red}{=k}$$ Switching to natural logarithms $$2+\frac{\log (a)}{\log (2)}=3+\frac{\log (b)}{\log (3)}=\frac{\log (a+b)}{\log (6)}=k$$ So $$a=2^{k-2} \qquad \text{and} \qquad b=3^{k-3}$$ and we face the problem of solving now $$2^{k-2}+3^{k-3}=6^k \qquad\text{...


3

Induction step: Suppose given statement holds for $k$, an even number. To show: the given statement holds for $k+2$. $$\begin{align}2^{k+2} - 1 &= 4\cdot2^k - 1\\&=3\cdot2^k + (2^k - 1)\\ &=3\cdot2^k + 3a \textrm{, since } 3|(2^k-1) \\&=3\cdot(2^k +a) \end{align}$$


3

Remember that, $\sqrt{x+2}≥0$ and multiplying each side by $(-1)$ then you get, $$-5\sqrt{x+2}-2=0 \Longleftrightarrow 5\sqrt{x+2}+2=0$$ which is impossible. The sum of non-negative and positive numbers can not equal to zero. EDIT: Your argument is not generally correct. Because, $$a=b \not \Longleftrightarrow a^2=b^2.$$ For example, $$-1=1 \\ (-1)^2=1^2 ...


3

Your analysis of $\delta_B'$ is accurate. Seperately, measures of the sort you seek do not really exist. Suppose $C$ is a countable set with $\mu(C)>0$. Suppose moreover that, for each $x\in C$, the set $\{x\}$ is measurable. Then, by countable additivity, we must have $\mu=\sum_{x\in C}{\mu(\{x\})\delta_x}$ (as measures on $C$). So any countably-...


3

It's disappointing to see that the teacher believes there is only one way to prove trigonometric identities. Actually, there are a number of approaches, all equally valid and useful in different circumstances. Sometimes it is more elegant to 'start with one side and prove that it is equal to the other', but a proof is a proof. Your son is actually using ...


2

Hint: If $x^2=y^2$, then $(x-y)(x+y)=0$ and so $x=y$ or $x=-y$.


2

Yes, the path you used in your attempt (as given in the main comments) works. Below I outline a strategy for discovering such a path . . . To show that $g$ is unbounded in a punctured neighborhood of $(0,0)$, it suffices to find a path terminating at $(0,0)$ such that $g(x,y)$ approaches infinity as $(x,y)$ approaches $(0,0)$ along that path. Consider a ...


2

Here is a coordinate-free proof. Consider the inclusion $i : N \to M$. As $N$ is a submanifold, $i$ is an embedding, and for each $p \in N$, its differential $\mathrm{d}i_p : T_p N \to T_pM$ is injective, and identifies $T_pN$ as a linear subspace $\mathrm{d}i_p(T_pN) \subset T_pM$.


2

From the inductive step we have $(1+x)^{l-1}=\frac1l\sum_{k=1}^lk\binom lkx^{k-1}$. Thus$$\begin{align*}(l+1)(1+x)^l&=(1+x)\left[(l+1)(1+x)^{l-1}\right]\\&=(1+x)\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom lkx^{k-1}\right]\\&=\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom lkx^{k-1}\right]+x\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom ...


2

For every nonnegative integer $k$, and every integer $m\geq k$, you have $\frac{d^k}{dx^k}(1+x)^m=\frac{m!}{(m-k)!}(1+x)^{m-k}$ and \begin{eqnarray*} \frac{d^k}{dx^k}(1+x)^m&=& \frac{d^k}{dx^k}\sum_{s=0}^m\binom{m}{s}x^s\\ &=&\sum_{s=0}^m\binom{m}{s}\frac{d^k}{dx^k}x^s\\ &=&\sum_{s=k}^m\binom{m}{s}\frac{s!}{(s-k)!}x^{s-k}\\ &=&...


2

Your proof is correct but you can make a shorter proof. If $x$ is odd and $\sqrt{2x}$ is an integer, then $2x$ is even, which implies $\sqrt{2x}$ is also even. So, you have $$\sqrt{2x}=2m, ~ m\in \mathbb Z^{+}$$ $$\implies 2x=4m^2$$ $$\implies x=2m^2$$ which gives a contradiction.


2

So, you wrote the error formula $$ |E_2(x)| \leq |(x-x_0)(x-x_2)(x-x_2)| \frac{M}{3!}. $$ Now you can simply maximize the 3rd degree polynomial... The maximum will be attained at a stationary point, that you can compute explicitly. The stationary points are $x_0 + (1 \pm \frac{\sqrt{3}}{3})h$, and the corresponding maximum value will be $\frac{2h^3}{3 \sqrt{...


2

The present problem may be resolved if we can show that for primes $p$ with $p \mid a^n$, $n \ge 1$, then $p \mid a$. We note that if $p \not \mid a \tag 0$ then $\gcd(p, a) = 1; \tag 1$ for the only divisors of $p$ are $p$ and $1$; and if $p \not \mid a$, we are left with $1$ as the sole common divisor of $p$ and $a$; now in light of (1) we have, via Bezout'...


2

" a=k+1 and let x′ and y′ be two positive integers such that max(x′,y′)=k+1. Now we have max(x′−1,y′−1)=k,"-1$ But $x'-1$ or $y'-1$ need not be a positive integer.


Only top voted, non community-wiki answers of a minimum length are eligible