# Can any improvements be made to my proof that “$\sqrt{3}$ is irrational”?

Suppose, for contradiction that $$\sqrt{3}$$ is rational. Then there exists $$a,b \in \mathbb{Z}$$ such that $$\frac{a}{b}= \sqrt{3},$$

where $$a/b$$ is in its simplest form. Then the above equation implies $$a^2=3b^2.$$ If $$b$$ is even, then $$a$$ is even, which is a contradiction since $$a/b$$ is therefore not in its simplest form.

Now, consider $$b$$ to be odd, then a is odd. Then for $$m,n \in \mathbb{Z}$$, we have $$(2m+1)^2=3(2n+1)^2\\ 4m^2+4m+1=12n^2+12n+3\\ 2(2m^2+2m)=2(6n^2+6n+1)\\2(m^2+m)=2(3n^2+3n)+1.$$

The LHS is even since $$m^2+m \in \mathbb{Z}$$ and the RHS is odd since $$3n^2+3n\in \mathbb{Z}$$. This is a contradiction, and we therefore conclude that $$\sqrt{3}$$ is irrational.

• "Suppose, for contradiction that $\sqrt{3}$ is rational." – John Dec 20 '18 at 18:34
• What a silly error. Thanks. – user503154 Dec 20 '18 at 18:37
• Pretty slick. As a corollary (or a lemma toward), no square is of the form $4k+3$. Rather cute. – fleablood Dec 20 '18 at 19:29

Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $$a^2 = 3b^2$$ modulo $$4$$. If $$a$$ and $$b$$ are both odd, then $$a^2 \equiv 1 \mod 4,$$ and \begin{align*} b^2 &\equiv 1\mod 4\\ 3b^2 &\equiv 3\mod 4, \end{align*} contradiction.

Another approach is to note that in the prime factorization of $$a^2 = 3b^2$$, the power of $$3$$ dividing the left hand side is even, while the power of $$3$$ dividing the right hand side is odd.

• Thank you. That is a better method. – user503154 Dec 20 '18 at 18:53

Other proof

we know that $$1<\sqrt{3}<2$$

assume $$a=b\sqrt{3}\in \Bbb N$$

with $$b>1$$.

Let $$A=\{c\in \Bbb N, c>1 : c\sqrt{3}\in \Bbb N\}$$

$$A\neq \emptyset$$ let $$m=\min A$$.

then

$$\alpha=m(\sqrt{3}-1)\in A \text{ and } \alpha<\min A$$ which is a contradiction, thus $$\sqrt{3}\notin \Bbb Q$$.

• More conceptually: the denominator set $A$ contains coprimes $b$ and $a$ (by $a^2/b^2=3\,\Rightarrow\,\,a/b = 3b/a)\,$ thus since $A$ is closed under subtractiion it is closed under gcd, so it contains $1 = \gcd(a,b),\,$ i.e. $\,\sqrt 3\,$ can be written with denominator $1$, i.e. $\,\sqrt 3\in \Bbb Z,\,$ contradiction. See my posts on denominator descent for more on this viewpoint. – Bill Dubuque Dec 20 '18 at 19:12
• Alternatively by $\,a^2/b = 3b\,\Rightarrow\, b\mid a^2\,$ so $\,b=1,\,$ by $\,a,b\,$ coprime and Euclid's lemma (generally. the least denominator of a fraction divides every denominator, a special case of the fact that ideals are principal in $\,\Bbb Z)\ \$ – Bill Dubuque Dec 20 '18 at 19:23