Show that square root of 3 is an irrational number

[docnet]

Homework Statement

Show that $\sqrt{3}$ is irrational

Relevant Equations

The fundamental theorem of arithmetic

$\sqrt{3}$ is irrational. The negation of the statement is that $\sqrt{3}$ is rational.

$\sqrt{3}$ is rational if there exist nonzero integers $a$ and $b$ such that $\frac{a}{b}=\sqrt 3$. The fundamental theorem of arithmetic states that every integer is representable uniquely as a product of prime numbers, up to the order of the factors. So $a$ and $b$ are products of prime numbers.
$$\frac{a}{b}=\sqrt{3} \Rightarrow a^2=3b^2$$
Squaring $a$ and $b$ leads to the prime factors $a^2$ and $b^2$ existing in pairs. But, the right hand side is multiplied by $3$, which means there is an odd number of $3$ in the right hand side. This leads to a contradiction because there is an even number of $3$s on the left hand side.

So the negated statement is false, and hence the original statement is true.

 

[ergospherical]

Looks good.