I am kind of partial to the old $n^2 + n + 41$ chestnut, namely that the expression is prime for all $n$. It fools an awful lot of people.

Here is a mathematical scherzo. $$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3.$$

To be separable means to have a countable dense subset. Suppose that $(M, d)$ is a metric space and that $U\subseteq M$ be an uncountable subset and $r > 0$. Suppose that for all $x \neq y \in U$...

This should work. $\forall x\in A\cdots$

The humps for $x\mapsto \sin(x^2)$ go up and down. Each has an area smaller than that of the last. The areas converge to 0 as you progress down the $x$-axis. By the alternating series test, this ...

Imagine a pond with an infinite linear progression of lily pads. You have a frog who, if he hops on one pad, he is guaranteed to hop on the next one. If he hops on the first pad, he'll visit them all....

The uniform boundedness principle of Functional Analysis is a very important application of the Baire Category Theorem. Added: (t.b.) See also Sokal's A really simple elementary proof of the uniform ...

Let us remember that an empty product is always 1. Hence, 1 has the empty product as its prime factorization. This product is vacuously a unique product of primes.

In a cup of coffee, one molecule of coffee is in its original location, even though the contents are undergoing convection. This is the Brouwer fixed point theorem.

Here are two resons. We can define $n!$ to be the number of rearrangements of $n$ distinct objects in a list. The empty list has one rearrangement: itself. We can define $n!$ as the product of all ...

Let $x$ be a number. Then if $b$ is any base, $x \% b$ ($x$ mod $b$) is the last digit of $x$'s base-$b$ representation. Now integer-divide $x$ by $b$ to amputate the last digit. Repeat and this ...

Suppose you have two expressions $e_1$ and $e_2$ and you know $$e_1 = e_2.$$ Then, if you apply a function to both sides, you have $$f(e_1) = f(e_2).$$ However, this logic in general does not reverse,...

Computing income taxes in a bracketed system.

You can obtain these formulae recursively. Look at this $$n^5 = \sum_{k=1}^n \left( k^5 - (k-1)^5\right) $$ Expand the second term. Cancel the $k^5$ terms. Then apply the identities above. You ...

Look at the arithmetic operations and their actions. With + and *, these matrices form a field. And we have the isomorphism $$a + ib \mapsto \left[\matrix{a&-b\cr b &a}\right].$$

You have $${\sqrt{1/n} - \sqrt{0}\over{1/n - 0}} = {1/\sqrt{n}\over {1\over n}} = \sqrt{n}.$$ This ratio can be made as large as you like by choosing $n$ large. Therefore the square-root function ...

A field homomorphism must in particular be a ring homomorphism, so its kernel is an ideal. The only ideals of a field are the zero ideal and the field itself.

Divide this into cases. The expression is not defined if $x = 1$. If $x > 1$, you can multiply both sides by $x - 1$ to get $1 >0$ So, if $x > 1$ the inequality is satisfied. If $x < ...

Take $$a_n = \sum_{k=1}^n {1\over k}.$$

No. Enumerate the rationals in [0,1] with the sequence $\{r_n\}_{n=1}^\infty$. Now define $f_n(x)$ by $f_n(x) = 1$ if $x = r_k$ for some $1\le k \le n$ and 0 otherwise. For all $n$, we have $$\...

Suppose the integer $z$ is even. Write it as $z = 2n$, where $n\in\mathbb{Z}$. Then $z^2 = 4n^2$; $z$ is divisible by 4. Suppose the integer $z$ is odd. Write it as $z = 2n + 1$ where $n\in\mathbb{Z}$...

Suppose that $f:(0,1) \rightarrow [0,1]$ is 1-1 and continuous. By the intermediate value theorem, the image of any interval under $f$ is an interval. Since $f$ is 1-1, it is either (strictly) ...

Do this: $$a^{\ln(b)} = e^{\ln(a)\ln(b)} = b^{\ln(a)}.$$

I editorialize here, but I think it is useful for parallelism to be an equivalence relation. Hence, a line should be parallel to itself.

This does not exist. If you have a sequence $\{x_n\}$, then if $x_n \rightarrow l$, for any open interval $I$ with $l\in I$, $x_n\in I$ for all but finitely many $n$. Intuitively, any open interval ...

You can also do this. Generate three random numbers $(a,b,c)$ in $[-1,1]$; if $a^2 + b^2 + c^2\le 1$, then normalize them. Otherwise try again and pick triplets until you have a usable triplet. The ...

The map $t\mapsto e^{2\pi it}$ is a homomorphism from $\mathbb{R}$ onto the circle group. Its kernel is $\mathbb{Z}$. Invoke the first isomorphism theorem.

Here is another way to think about it. You have the word TOFFEE and six blanks - - - - - - you want to place the letters in. Begin with the E. Choose two of the six blanks and pop the Es in. This ...

The open mapping theorem and closed graph theorem of functional analysis are two vital applications.

It is fairly unlikely if $n$ is large. The probability of $n$ heads and $n$ tails is $${2n\choose n} {1\over 4^n}.$$ We have $${2n\choose n} = {(2n)!\over n!n!}.$$ Stirling's formula states that as ...