# Tag Info

9

Here is an easy proof. Suppose it has a rational root say $a/b$ and $\gcd (a,b)=1$. Then putting $x=a/b$, we obtain $$a^3+ab^2+b^3=0.$$ Note that $|a|>1$ or $|b|>1$, since $\pm 1$ are not roots. Let $p$ be a prime such that $p\mid a$ (assume $|a|>1)$. Then $p\mid a^3$ and $p\mid ab^2$ and hence $p\mid a^3+ab^2$. Now, from above equation we have ...

9

If it were rational it would be $m/n$, where $m$ and $n$ are integers. Then we would have $$\left(\frac m n \right)^3 + \frac m n + 1 =0.$$ From that we get $$m^3 + mn^2 + n^3 = 0.$$ If the fraction $m/n$ is in lowest terms, then $m$ and $n$ have no prime factors in common. But $$n^3 = -m(m^2+n^2) = (-m\cdot\text{something})$$ so $n^3$ is divisible ...

9

From the perspective of an elementary calculus student (by which I mean that it can supposedly be made rigorous later on, but isn't in introductory classes), the $du$, $dx$ stuff is absolute nonsense and I will never understand why it continues to be used by so many professors. It really is the one glaring hole in most otherwise rigorous calculus courses. ...

8

Hint: $P(x)$ is a cubic, so it must have at least one real root. There exists an $x_0$ such that $e^{P(x_0)}=1$. This means $$e^{P(x_0)}\geq \sin x_0$$ The leading coefficient of $P(x)$ is $1$, so $$\lim_{x\to-\infty} P(x)=-\infty$$ This means that $$\lim_{x\to-\infty} e^{P(x)}=0$$ Now, why must there be a solution to $e^{P(x)}=\sin x$ in the interval ...

8

Consider an $11$-sided regular polygon inscribed in the unit circle, with one vertex at $(-1,0)$. The centre of gravity of the eleven vertices is the origin. Looking at the $x$-coordinate, this shows that $$0 = -1 + \sum_{k = 0}^4 \cos\frac{(2k+1)\pi}{11} + \sum_{k = 0}^4 \cos\frac{-(2k+1)\pi}{11}.$$ But in view of the fact that the cosine function is even, ...

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.

5

We have that $\forall \epsilon>0. \; \exists N \in \mathbb N. \; \forall n > N. \; |a_n+b_n-0|=|a_n+b_n|<\epsilon$. Because $a_n \ge0$, $b_n\ge0$, $|a_n+b_n|\ge|a_n|$, $|a_n+b_n|\ge|b_n|$. Therefore, $|a_n-0|<\epsilon$, $|b_n-0|<\epsilon$ when $n>N$. Hence $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=0$.

5

Hint:- $6\equiv1 \pmod 5\implies 6^n\equiv1\pmod 5\tag{1}$ $-5(n-1)\equiv 0\pmod 5\tag{2}$ Solution:-

5

Directly: $$6^{n}-5n+4=\left(6-1\right)\left(6^{n-1}+6^{n-2}+\cdots+6^{1}+6^{0}\right)-5n+5=$$$$5\left(6^{n-1}+6^{n-2}+\cdots+6^{1}+6^{0}-n+1\right)$$

4

Like you, we give a semantic (model-theoretic) argument. There are also syntactic arguments, where the details depend very much on the setup (axioms, rules of inference). Let $L$ be our formal language. We will show that the sentence $\exists x(P(x)\to \forall y P(y))$ is true in any $L$-structure $M$. Suppose first that $\forall yP(y)$ is true in ...

4

Yes. You need to cite them. You are using someone else's idea, and you need to give them credit for it. Anything else is plagiarism.

4

Hint: Apply the Residue theorem for $$f(z)=\frac{\pi\cot \pi z}{1+z^2}$$ using the circle at $0$ of radius $n+\frac12$ and take the limit.

4

Using Fourier Series, you can show that: $$\cot(\pi z) = \frac 1\pi \left( \frac 1z - \sum_{k=1}^\infty \frac{2z}{k^2 - z^2} \right)$$ Let $z = i$ to essentially get your sum, modulo some small simplifications which I'll leave to you.

4

If $3|(a+b)$ then $3|((a+b)^3-3a^2b-3ab^2)$ so $3|a^3+b^3$.

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

HINT: If $x$ is irrational, then $x$ is not an integer, so there is a unique integer $n$ that $n<x<n+1$. Clearly the distance from $x$ to the closer of $n$ and $n+1$ is at most $\frac12$. If $x$ is irrational, can that distance be $\frac12$? Added: Please take note also of Jack M’s excellent comment under the original question; I was a bit rushed when ...

4

HINT: Assume that every set of $n$ natural numbers has a maximum, and let $A$ be a set of $n+1$ natural numbers. Let $a$ be any element of $A$; then $A\setminus\{a\}$ has $n$ elements, so it has a maximum element $m$. Now compare $a$ and $m$.

4

Your teacher is probably pretty smart, but that's not why he can prove the theorem and you can't. The difference between you two lies somewhere else: You only saw the proof once, and you also probably only saw somewhere between several ten and a hundred proofs in your life. Your professor proved the theorem in question once every year for several years, and ...

4

The matter of "evenness" (which means nothing more than being a multiple of $2$) is something that was important in the original problem regarding $\sqrt 2$. In your case, you should not necessarily be concerned with evenness. Continuing from where you left off, i.e., $10b^2 = 2c^2$, we can divide both sides by $2$ to get: $$5b^2 = c^2.$$ From here you can ...

4

You are very nearly there. You just need to show that this function satisfies that definition. So suppose that $f(x) = f(y)$ for some $x, y \in \mathbb{R}$. Then can we have $x \neq y$? Why/why not? Use the property that $x \neq y \implies f(x) \neq f(y)$ for this function!

4

I’ll give a fairly detailed commentary on your argument. $[\Rightarrow]$ Assume that int(cl($\{ x \})) = \varnothing$. Thus, $X \neq \{ x \}$ and the result trivially holds. It’s true that if $\operatorname{int}\operatorname{cl}\{x\}=\varnothing$, then $X\ne\{x\}$, but this by no means implies that $x$ is not an isolated point. For a simple ...

3

Obviously, if $p_1p_2\mid n$, then $p_1\mid n$ and $p_2\mid n$. So assume that $p_1\mid n$ and $p_2\mid n$. If $p_1$ and $p_2$ are distinct primes, then, using Bezout's Identity, there are $x,y\in\mathbb{Z}$ so that $$xp_1+yp_2=1\tag{1}$$ Equation $(1)$ implies that \begin{align} n &=nxp_1+nyp_2\\ ... 3 HINT: Suppose that \langle x,y\rangle\in S. Then either \langle x,y\rangle\in R, or x=y; why? Can you finish it from here? Remember, you want to conclude that \langle y,x\rangle\in S. 3 Hint: by the binomial theorem, (1+2+\cdots +n +(n+1))^2=(1+2+\cdots +n)^2+(n+1)^2+2(n+1)(1+2+ \cdots n). Now, you can show by induction (or by Gauss' trick) that 1+2+\cdots +n= \dfrac{n(n+1)}{2}, and hus reach the answer. 3 This is the drinker's paradox: "In every (populated) bar there is a person such that, if that person is drinking, then everyone is drinking". It takes advantage of the 2 cases of vacuous implication: (1) "False implies anything" (2) "Anything implies true" So divide the theorem into 2 cases: Case (1): Someone is not drinking. Then that person is an ... 3 First Question For another approach without using the Rational Root Theorem is to solve the cubic. The roots of a cubic equation of the form x^3+px+q=0 are given by, \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\tag{1} ...

3

If we start at $r=1$, then the result is true. Using $1+x\le e^x$, we get \begin{align} \prod_{r=1}^\infty\left(1+\frac1{2^r}\right) &\le\frac32\prod_{r=2}^\infty e^{1/2^r}\\ &=\frac32e^{1/2}\\ &\lt\frac52 \end{align} since $e\lt\frac{25}9$. Another Approach Lemma: if $0\le a_k\le1$, then $$... 3 I suppose x is an integer and you series is$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{n}}$$For n=1, our inequity is obviously true. Suppose it is true for n=k. Then$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{k}}<2\sqrt{k}$$... 3 I think your try has a little problem. Notice that for f=\sin and [a,b]=[0,4\pi], we have A=[0,\pi]\cup[2\pi,3\pi], but then \sup A=3\pi and \inf A=0; f is not always positive on (\beta,\alpha). Focus on the existence of x_0 such that f(x_0)>0, and let$$c=\sup\{a\le x<x_0\mid f(x)=0\}$$and$$d=\inf\{x_0<x\le b\mid f(x)=0\}. ...

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