5
$\begingroup$

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.

$\endgroup$
  • 7
    $\begingroup$ "Suppose, for contradiction that $\sqrt{3}$ is rational." $\endgroup$ – John Dec 20 '18 at 18:34
  • 2
    $\begingroup$ What a silly error. Thanks. $\endgroup$ – user503154 Dec 20 '18 at 18:37
  • 1
    $\begingroup$ Pretty slick. As a corollary (or a lemma toward), no square is of the form $4k+3$. Rather cute. $\endgroup$ – fleablood Dec 20 '18 at 19:29
3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Thank you. That is a better method. $\endgroup$ – user503154 Dec 20 '18 at 18:53
0
$\begingroup$

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 $\sqrt{3}\notin \Bbb Q$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Bill Dubuque Dec 20 '18 at 19:12
  • 1
    $\begingroup$ 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)\ \ $ $\endgroup$ – Bill Dubuque Dec 20 '18 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy