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Key Idea: We will see a trend in all the proofs presented here; in particular, every proof is by contradiction. It’s usually very difficult to prove that something is impossible, and in our case, it’s very difficult to show it’s impossible to find integers whose ratio equals $\sqrt{2}$, i.e., that it’s impossible to find $a,b\in \Z,$ such that $\frac{a}{b} =\sqrt{2}.$ It’s far easier to assume there are two integers $a,b\in \Z$ such that $\frac{a}{b} =\sqrt{2},$ and then deduce a contradiction. This will be our strategy.

Beginning of proof: Assume for the hope of a contradiction that $\sqrt{2}$ is a rational number. It follows that we can find two integers $a,b\in \Z$ such that $\frac{a}{b} =\sqrt{2}.$ Note that we can safely assume that $a$ and $b$ are both positive. Since working with square roots and fractions can be annoying, let’s square and rearrange everything to deduce,

Intuitively, we know that if $a^2$ is even, then $a$ must also be even. Rigorously, this is a consequence of Euclid’s lemma, or Euler’s Lemma (see [10] page 90, or 82). (Or, you can do a mini proof by contrapositive: if $a$ is not even (odd), then $a^2$ is not even (odd).) In any case, we conclude that $a$ is even, and therefore $a = 2a_0,$ for some $a_0\in \Z.$ Let’s plug this into our equation relating $a$ and $b,$

Canceling a factor of 2 gives,

Similarly, we see that $b^2$ is even, and hence $b$ is even. Therefore, $b = 2b_0$ for some $b_0\in \Z.$ It follows,

Again, canceling a factor of 2 gives,

Observe that the above equation implies that $\sqrt{2}=\frac{a_0}{b_0} .$ This means that we have found a new way to express the square root of 2 as a fraction of positive integers:

Note that $0<a_0<a$ and $0<b_0<b$ follow from $a = 2a_0$ and $b = 2b_0.$ Interesting…

Taking a look back at $a_0^2 = 2b_0^2,$ you might notice that we are in the exact same situation that we started, except we are using the letters $a_0$ and $b_0$ instead of $a$ and $b.$ Thus, by going through the same argument, we can find two more positive integers $a_1$ and $b_1$ such that: (i) $a_1^2 = 2b_1^2;$(ii) $\sqrt{2}=\frac{a_1}{b_1};$ and (iii) $a_0 = 2a_1$ and $b_0 = 2b_1.$ In particular, (ii) and (iii) imply that we have found yet another way to express $\sqrt{2}$ as a fraction of smaller positive integers: $0<a_1<a_0<a$ and $0<b_1<b_0<b.$ And by starting with (i), we can go through the same reasoning yet again… and then again, and again indefinitely, always finding smaller and smaller numbers $a_n,b_n\in \N$ to express $\sqrt{2}$ as a fraction of integers.

But, this process must stop at some point since there are only a finite number of positive integers that $a_n$ and $b_n$ could be. In particular, the numerator $a_n$ can only be one of the numbers $1,\.2,\.3,\cdots, \,a-1$ and the denominator $b_n$ can only be one of the numbers $1,\,2,\,3,\,\cdots, \.b-1.$ This is our contradiction and hence $\sqrt{2}\notin \mathbb{Q}.$

Remark:  Fermat’s infinite descent is the particular type of contradiction that we reached. That is, assuming $\sqrt{2}$ is rational, let us find an infinite sequence of fractions to represent $\sqrt{2},$ where each new representation is constructed from the previous one and uses smaller positive integers. However, there is not an infinite number of positive integers less than $a$ or $b.$ Thus, a contradiction.

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Key Idea: Factorials grow super-duper fast. So fast that the series that defines the number $e$ converges quickly. So quick that, in some sense, $e$ cannot be a rational number. Let’s make this idea more precise.

Beginning of proof: Assume for the hope of a contradiction that $e$ were rational. It follows that we can find two integers $a,b\in \Z$ such that $\frac{a}{b} =e.$ First, note that we can assume that $a$ and $b$ are both positive. Second, note that $b\geq2$ by observing that $2<e< 3$ and there is no way to express a number greater than 2 and less than 3 as a fraction of positive integers with the denominator equal to 1. So, we have $\frac{a}{b} =e$ for $a,b\in \N$ and $b\geq2 .$

Now, consider $e \cdot (b!).$

Observe that the denominator is less than the numerator for the first $b+1.$ Consequently, the first $b+1$ terms are integers, and hence the sum of the first $b+1$ terms is an integer, which we denote by $N.$ To highlight this observation, let’s group the integer terms,

After some rearrangement, we have

Moreover, $e\cdot b!$ is an integer as well. Indeed, since $e = \frac{a}{b}$ it follows $e\cdot b! = a\cdot(b-1)!\in \Z.$ Thus, the left-hand side is an integer, and we conclude

However, we claim that the right-hand side is not an integer, and once we show this, we will have found our contradiction. Observe that,

So that the tail of the sum is greater than 0 but less than 1. We have found our contradiction! We conclude that $e$ is irrational.

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Key Idea: The idea behind this proof is to find a quantity that we can show to be an integer under the assumption that π is rational, which must be greater than 0 but less than 1. The difficulty and brilliance come from actually finding that quantity.

Beginning of proof: Assume for the hope of a contradiction that $\pi$ is a rational number. It follows that we can find two integers $a,b\in \Z$ such that $\frac{a}{b} =\pi.$ We can safely assume that $a$ and $b$ are both positive. Now consider the polynomials,

and

Our goal is to find some $n\in \N$ that leads to a contradiction. With this goal in mind, we make some observations regarding $f(x)$ and $F(x).$

8 Key Observations regarding $f(x)$ and $F(x)$:

$\bigstar$Observation (1): $f(x) \cdot n!$ is a polynomial with integer coefficients, i.e., $f(x) \cdot n! \in \Z[x].$

$\bigstar$Observation (2): $f(x) \gt 0$ for all $x\in (0,\pi).$ Indeed, since $x<\pi=\frac{a}{b}$ implies $a>bx.$

$\bigstar$Observation (3): $f(x) \lt \frac{(\pi a)^n}{n!}$ for all $x\in (0,\pi).$ This is seen by noting that $x \lt \pi$ for $x\in (0,\pi)$ and $a-bx \lt a$ for $x\in (0,\pi).$

$\bigstar$Observation (4): the lowest power of $x$ that shows up in $f(x)$ is $x^n.$

$\bigstar$Observation (5): $f(x) =b^n f(\pi -x).$ Indeed, since $\frac{a}{b} =\pi$ we have,

A corollary of this result is the following: $f^{(k)}(x) =b^n f^{(k)}(\pi -x).$ Which we will make use of.

$\bigstar$Observation (6): $f^{(k)}(0) =f^{(k)}(\pi) =0$ for all $0\leq k\lt n.$ This follows from observations (4) and (5). Since the lowest power of $x$ in $f(x)$ is $x^n,$ the $k^{th}$ derivative for $0\leq k\lt n$ will not yield a constant term, that is, it leaves an $x$ in every term in $f^{(k)}(x) .$ Hence, $f^{(k)}(0) =0.$ Now, using (5) we conclude that $f^{(k)}(0) =b^nf^{(k)}(\pi) =0.$

$\bigstar$Observation (7): $f^{(k)}(0) =b^nf^{(k)}(\pi) \in \Z$ for all $n\leq k\leq 2n.$ This observation requires more justification.

Note that when we expand the product in $f(x) ,$ we get

where $c_{\ell}\in \Z.$ We are interested in determining $f^{(k)}(0) .$ This means we can focus on the constant term in $f^{(k)}(0)$ because all other terms will be 0. Observe that the constant term in $f^{(k)}(0)$ will be the term with $x^{k}$ in $f(x) .$ Then, by taking the $k^{th}$ derivative, we bring down a factor of $k!$ and we deduce that the constant term in $f^{(k)}(0)$ is equal to $\frac{k!\cdot c_{k-n}}{n!}.$ And, since $n\leq k\leq 2n$ we see that $\frac{k!\cdot c_{k-n}}{n!}\in \Z.$

Now, using observation (5) we deduce $f^{(k)}(0) =b^nf^{(k)}(\pi) \in \Z.$

$\bigstar$Observation (8): $F(0)$ and $F(\pi)$ are integers. This follows from the previous items (6) and (7). In particular, the sum $F(0) + F(\pi)$ is an integer.

Now, let’s set up a differential equation in $F(x) ,$

Therefore,

We will use the fact that the integral above is an integer to find a contradiction. Our strategy is to use our observations about $f(x)$ to bound the integrand, then use this bound to find an $n\in \N$ such that the integral cannot be an integer.

Observe that the integrand, $f(x) \sin(x),$ is strictly bounded by,

for $0\lt x\lt \pi.$ These follow from Observation (2) and Observation (3). It follows that the integral, $\int_0^{\pi}f(x) \,\sin(x)\;dx ,$ is positive and bounded by

for any choice of $n.$ However, we can see from the upper bound above that the integral can be made to be strictly less than by choosing a sufficiently large $n.$

This is our contradiction! The integral cannot both be an integer and be in the interval $(0,1).$

Thus, $\pi$ is irrational!

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