# Tag Info

6

There is a whole family of examples similar to the proposition that $n^3-n$ is divisible by $6$ for each natural number $n$. Proof by induction isn’t hard, but it’s certainly unnecessarily complicated.

4

1) For all relatively prime positive integers $a$ and $m$ there is a prime number in the arithmetic progression $a$, $a+m$, $a+2m$, $a+3m,\dots$ Of course the general theorem is that there are infinitely many primes in each such progression, but the way I stated it by quantifying over all $a$ and $m$ is equivalent to the general version; for beginners I ...

3

There's the classic example that $$1 + 2 + \ldots + n = \dfrac{n(n+1)}{2},$$ which can be proved without induction using Gauss's trick, or the geometric argument involving a rectangular grid. In a similar vein, showing $${n \choose k} = {n - 1 \choose k - 1} + {n - 1 \choose k} \quad \text{for } 1 \le k \le n - 1$$ has a really straightforward ...

3

Step $2$ is an instance of Axiom $6$ of formal deduction (see Enderton's A Mathematical Introduction to Logic page $112$): $\vdash (x=y)\to(\alpha\to\alpha ')$, where $\alpha$ is atomic and $\alpha '$ is obtained from $\alpha$ by replacing $x$ in zero or more places by $y$. In your case (where $\equiv$ replaces $=$), the atomic formula is $Pb$ and, ...

3

Let $f(x) = \sin x - \frac x2$. $f(\pi/2) = 1 - \frac \pi4 > 0$ while $f(\pi) = 0 - \pi/2 < 0$ Now use the intermediate value theorem to show that there exists a zero of $f(x)$ in $(\pi/2, \pi)$, which will also be the solution you were looking for :-) Another way would be to compare their derivatives; both $\sin x$ and $\frac x2$ are $0$ on the ...

3

Hint: You can find a counterexample with $$A=\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}$$

2

There are infinite number of primes. The proof by contradiction is as such: Suppose to the contrary there exists a finite number of primes. Let's call this set of finite number of primes $$p_1\text{,...,}p_n$$ If you add 1 to this set, we have a new set and we shall denote this new set as S: $$S=\left\{p_1\text{,..., }p_n\text{\}+1}\right.$$ There ...

2

Step two could actually be $b\equiv a \rightarrow (Pb\leftrightarrow Pa)$. If two items are equal, they have the same properties.

2

Hint: Let $S\subseteq\Bbb N$ with no least element. So $0\notin S$ for otherwise it would have a least element. So, $0\in\Bbb N\setminus S$. Now use induction to show $\Bbb N=\Bbb N\setminus S$.

2

Start with the list of all $2^n$ vectors of length $n$ of $+1$s and $-1$s. Someone changes some of the entries to $0$. Show that there is always a non-empty collection of rows which sums to the zero vector. This looks like the perfect case for induction (on $n$), but the two proofs I'm aware of don't use induction; induction just doesn't seem to help here.

2

For all n, there is no m such that m < n < m+1.

2

Theorem: For all $n\in \mathbb{N}$, $$n^2 + 1 \geq 2n.$$ Easy to prove by just observing that $(x-1)^2 \geq 0$ for all $x\in \mathbb{R}$. Not only is this way quicker than writing out a proof by induction, but it works for all real numbers, not just natural numbers. So you have an easier proof of a stronger result!

2

Yes, you're on the right track. If $n=0$ the statement is obvious, because it reduces to $v\ne0$. So suppose we have proved the statement for the pairs $(w,g)$ with $g^{n-1}(w)=0$ and $g^n(w)=0$, for $n>0$. Let $\alpha_1v+\alpha_2f(v)+\dots+\alpha_nf^{n}(v)=0$. By applying $f$ to this equality and setting $w=f(v)$, we get $$... 2 One thing that come to mind is that you have to think carefully about using \in and \subset. Unless your function is bijective it is a very real possibility that f^{-1}(y) is not an element, but a set. In this event the statement f^{-1}(y) \in Y is not true. 2 here is another way which is not use derivative: first we need to know: \sqrt{\dfrac{x+y+z}{3}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} (1) to prove it, we have \sqrt{\dfrac{x+y}{2}} \ge \dfrac{\sqrt{x}+\sqrt{y}}{2} \implies \sqrt{\dfrac{x+y+z+t}{4}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{t}}{4} let t= \dfrac{x+y+z}{3} we have (1) squre ... 2 I think your confusion boils down to this: what is the definition of \overline{F}(P)? Now \overline{F} \in \Gamma_h(V) = k[X_1, \ldots, X_{n+1}]/I(V) is not a polynomial; it is a coset, i.e., \overline{F} = F + I(V). So what does it mean to evaluate a coset at a point? Well, the first idea that comes to mind is to choose a representative f \in F + ... 2 Your proof is fine, but I added some points and shortened it. You can say that f_n \rightarrow f pointwise, where f = \begin{cases} 0 & \text{if } x = 0 \\ \frac{x}{\sin(x)} & \text{if } x \in (0, \frac{\pi}{2}] \end{cases}. For x = 0, \lim_{n \rightarrow \infty}f_n(0) = 0 and For x \in (0,\frac{\pi}{2}], \lim_{n \rightarrow ... 1 Expanded HINT: For one direction, assume (a), and let A be an infinite subset of V. Since A is infinite, it contains a sequence \langle a_n:n\in\Bbb Z^+\rangle whose points are all distinct. (That is, a_m\ne a_n if m\ne n.) By (a) this sequence has a subsequence \langle a_{n_k}:k\in\Bbb Z^+\rangle that converges to some x\in V. Use the ... 1 What Daniel Fischer, Brian Tung, Michael Hardy, and myself now are referring to is that the empty sum is taken to be zero; on the other hand, the empty product is taken to be one. More concretely,$$ \sum_{i=r}^k\Omega_i=0\quad\text{when}\quad k<r\quad\text{and}\quad\prod_{i=r}^k\Omega_i=1\quad\text{when}\quad k<r. $$Now, consider Brian's comment; ... 1 As the comments have indicated, the equation is actually valid for n = 1, after a fashion. It is clearer when written as a summation:$$ \sum_{k=2}^n (5k-4) = \frac{n(5n-3)-2}{2} $$Spelled out in words, the above equation reads, "The sum of (5k-4), from k = 2 to n, equals \ldots" When n = 1, the summation on the left is empty, and an empty ... 1$$\begin{align} x\in f^{-1}(X) & \implies f(x)\in X\\ & \implies f(x)\in Y\\ & \implies x\in f^{-1}(Y) \end{align}$$1 These are the standard definitions: Let X, Y be any sets whatsover.$$ f^{-1}(X) = \{ e \in E \ \mid \ f(e) \in X \} f^{-1}(Y) = \{ e \in E \ \mid \ f(e) \in Y \}$$Your requirement is to prove that the former is a subset of the latter, given X \subseteq Y. So we prove the implication$$ a \in f^{-1}(X) \implies a \in f^{-1}(Y) $$So to ... 1 The proof looks pretty good! There are some things that I can say though. For starters, we want to show the implication x\in f^{-1}(X)\implies x\in f^{-1}(Y). Because of this, we don't really need to look for y\in Y. You made this claim, and then the corresponding claims about the preimages with a subtle implication to if y\in f^{-1}(X)\cap f^{-1}(Y) ... 1 There is a slick solution to (i) using compactness: simply observe that if X,Y are compact then so is X\times Y. Then f(x,y)=x+y is continuous, so f(X\times Y)=X+Y is also compact (as the continuous image of a compact set). You can get (ii) by using subsequential convergence of compact sets. (iii) is false, as is seen by the example X=\mathbb Z ... 1 Yes you are on the right way! Suppose that the a_n's are such that$$ a_0 v + a_1 f(v) +\dots + a_n f^n(v)=0 $$Taking the function times on both side (because of linearity) you obtain that$$ a_0 f(v) + a_1 f^2(v) +\dots + a_n f^{n+1}(v)=0 $$and since f^{n+1}(v)=0 you get$$ a_0 f(v) + a_1 f^2(v) +\dots + a_{n-1} f^{n}(v)=0. $$Now you can do ... 1 When n is odd, let n = 2k + 1, k >= 0.  \quad \ \sum\limits_{i=1}^n (-1)^{i-1} i^2 \\ = 1 - 2^2 + 3^2 - 4^2 + 5^2 - \cdots - (2k)^2 + (2k + 1)^2 \\ = 1 + 3^2 - 2^2 + 5^2 - 4^2 + \cdots + (2k + 1)^2 - (2k)^2 \\ = 1 + (3 + 2) + (5 + 4) + \cdots + (2k + 1 + 2k) \\ = 1 + 2 + 3 + 4 + 5 + \cdots + 2k + 2k + 1 \\ = (k + 1)(2k + 1) = n(n + 1) / 2\\ = ... 1 We know \log_y{x}= \frac{\log{x}}{\log{y}}. Thus, \log_r{s}.\log_s{a}= \frac{\log{s}}{\log{r}} \cdot \frac{\log{a}}{\log{s}}=\frac{\log{a}}{\log{r}}=\log_r{a}. 1 This is just the change of base formula that has been rearranged. Let:$$\log_sa=x\tag{1}\therefore a=s^x$$Now takes logs of both sides to base r to get:$$\log_ra=\log_r(s^x)=x\log_rs\tag{2}Now substitute (1) into (2) and you are done.

1

If $AB=AC$ then $A(B-C)=0$, and we can use the fact that the product of two non-zero matrices can be zero, so that both $A$ is non-zero and $B-C$ is non-zero, so that $B\neq C$.

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