Proposition 1 Prove ${\sqrt{2}}$ is irrational.

Here is a proof using a traditional method (See Euclid’s Elements Book X which incorporates Theatetus work on incommensurable numbers. It includes a proof that ${\sqrt{2}}$ is irrational (Proposition 22), and ends with a proof that there are infinitely many distinct irrational numbers (Proposition 115):

Proof: (Contradiction) Suppose ${\sqrt{2}}$ is rational. Then ${\sqrt{2}=\frac{a}{b}}$, ${a,b\in\mathbb{Z}}$, ${b\neq0}$ and ${gcd(a,b)=1}$. Squaring both sides, we get ${2=\frac{a^{2}}{b^{2}}}$ or ${a^{2}=2b^{2}}$. Note, this means ${a}$ is even. If ${a}$ were odd, then ${a=2n+1}$ for ${n\in\mathbb{Z}}$,

$\displaystyle \begin{array}{rcl} a^{2} & = & (2n+1)^{2}\\ & = & (2n+1)(2n+1)\\ & = & 4n^{2}+4n+1\\ & = & 2(2n^{2}+2n)+1\end{array}$

So, an odd number squared is odd, thus ${a}$ is even. Subsitituting ${a=2n}$ into the original equation we get,

$\displaystyle \begin{array}{rcl} 2 & = & \frac{(2n)^{2}}{b^{2}}\\ 2b^{2} & = & 4n^{2}\\ b^{2} & = & 2n^{2}\end{array}$

Note, ${b^{2}}$ is even, which means ${b}$ is even. So, ${gcd(a,b)=2},$ hence a contradiction. Thus, ${\sqrt{2}}$ is irrational. $\Box$

We also can prove this by finding the zeros (i.e., roots) of integral polynomials (i.e., a polynomial with only integer coefficients). Let us consider the integral polynomial ${p(x)=3x^{3}+x^{2}-6x-2}$, and look for the rational zeros (i.e., zeros which are rational number) of this polynomial. At the first glance, there seems to be infinitely many possibilities for the rational zeros. However, there is a very useful theorem for this problem, called the Rational Zero Theorem. It helps us narrow down the possible rational zeros of an integral polynomial to a list of finite rational numbers.

Theorem 1 (The Rational Zero Theorem). Let ${p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_{1}x+a_{0}\in\mathbb{Z}[x]}$ with ${a_{n}\neq0}$. If ${s}$ and ${t}$ are nonzero integers such that ${p(\frac{s}{t})=0}$ and ${gcd(s,t)=1}$ (i.e., there is no common factor among ${s}$ and ${t}$ except ${1}$), then ${s|a_{0}}$ and ${t|a_{n}}$.

Proof: Let ${s,t\in\mathbb{Z}}$ such that ${t\neq0}$. Assume that ${p(\frac{s}{t})\neq0}$ and ${gcd(s,t)=1.}$ So,

$\displaystyle \begin{array}{rcl} a_{n}(\frac{s}{t})^{n}+\cdots+a_{1}(\frac{s}{t})+a_{0} & = & 0;\\ a_{n}\cdot\frac{s^{n}}{t^{n}}+\cdots+a_{1}\cdot\frac{s}{t}+a_{0} & = & 0;\\ a_{n}\cdot s^{n}+a_{n-1}\cdot s^{n-1}t+\cdots+a_{1}\cdot st^{n-1}+a_{0}t^{n} & = & 0.\end{array}$

So,

$\displaystyle t|(a_{n}\cdot s^{n}+a_{n-1}\cdot s^{n-1}t+\cdots+a_{1}\cdot st^{n-1}+a_{0}t^{n}).$

Thus, ${t|a_{n}\cdot s^{n}}$. Since, ${gcd(s,t)=1}$, we have ${t|a_{n}}$. Similary,

$\displaystyle s|(a_{n}\cdot s^{n}+a_{n-1}\cdot s^{n-1}t+\cdots+a_{1}\cdot st^{n-1}+a_{0}t^{n}).$

So, ${s|a_{0}t^{n}}$. Since, ${gcd(s,t)=1}$, we have ${s|a_{0}}$. $\Box$

Remark 1 If ${p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_{1}x+a_{0}\in\mathbb{Z}[x]}$ with ${a_{n}\neq0}$ and ${\frac{s}{t}\in\mathbb{Q}}$ (with ${s,t}$ nonzero integers and ${gcd(s,t)=1}$; this means that we cannot reduce the expression ${\frac{s}{t}}$ further) is a zero of ${p(x)}$, then the denominator ${t}$ must divide the leading coefficient ${a_{n}}$ of ${p(x)}$ and the numerator ${s}$ must divide the constant term ${a_{0}}$ of ${p(x)}$.

Example 1 Consider the integral polynomial ${p(x)=3x^{3}+x^{2}-6x-2}$, Since ${p(0)=-2}$, ${x=0}$ is not a zero of ${p(x)}$. We can also deduce the fact that ${0}$ is not a zero of ${p(x)}$ by noting that the constant term of ${p(x)}$ (which is ${-2}$) is not zero. If ${\frac{s}{t}\in\mathbb{Q}}$ (with ${\frac{s}{t}\neq0}$ and ${gcd(s,t)=1}$) is a zero of ${p(x)}$, then the numerator ${s}$ divides the constant term ${-2}$ (i.e., ${s|-2}$) and the denominator ${t}$ divides the leading coefficient ${2}$, (i.e., ${t|2}$). So, the possible ${s}$ are ${\pm1}$ and ${\pm2}$. Similarly, the possible ${t}$ are ${\pm1}$ and ${\pm3}$. Thus, the possible rational zeros ${\frac{s}{t}}$ are ${\pm1}$, ${\pm2}$, ${\pm\frac{1}{3}}$, and ${\pm\frac{2}{3}}$.

A special case of the Rational Zero Theorem shows us that the rational zeros of an integral polynomial with the leading coefficient 1 (i.e., a monic integral polynomial) are all integers.

Corollary 1 Let ${p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_{1}x+a_{0}\in\mathbb{Z}[x]}$. If ${a_{n}=1}$ (i.e., the leading coefficient is ${1}$) and the integral polynomial ${p(x)}$ has a rational zero, then the rational zeros of ${p(x)}$ are all integers, and they must divide the constant term ${a_{0}}$.

Proof: Let ${p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_{1}x+a_{0}\in\mathbb{Z}[x]}$. Assume that ${a_{n}=1}$ and ${p(x)}$ has a rational zero. Let ${s}$ and ${t}$ be nonzero integers such that ${gcd(s,t)=1}$ and ${\frac{s}{t}}$ is a zero of ${p(x)}$. According to the Rational Zero Theorem, the integer ${t}$ must divide the leading coefficient ${1}$. So, ${t}$ must be either ${1}$ or ${-1}$. Thus, the zero ${\frac{s}{t}}$ of ${p(x)}$ is either the integer ${s}$ or ${-s}$. Again, by the Rational Zero Theorem, we have ${s|a_{0}}$; so, ${s|a_{0}}$and ${-s|a_{0}}$. $\Box$

Now we are ready to prove ${\sqrt{2}}$ is irrational using the Rational Zero Theorem:

Proof: Let us show that ${\sqrt{2}}$ is an irrational number. Note that it is a zero of the polynomial ${p(x)=x^{2}-2}$. Now, possible rational zeros of this polynomial are ${\pm1}$ and ${\pm2}$. Now, ${p(-1)=-1}$, ${p(1)=-1}$, ${p(-2)=2}$ and ${p(2)=2}$. So, there is no rational zero for the polynomial ${p(x)=x^{2}-2}$. So, ${\sqrt{2}}$ can not be a rational number, i.e., ${\sqrt{2}}$ is an irrational number. $\Box$