in the proof that there is no rational number whose square is 2:
Suppose x is a rational number whose square is 2.
Then x can be written in lowest terms as $\frac{a}{b}$ , where a is an integer and b is a positive integer.
Since $x^2 = 2$, $\left (\frac{a}{b} \right )^2$, so $\frac{a^2}{b^2}$
But then a is even, so $a = 2n$ for some integer n.
Then $(2n)^2$ = $2b^2$, so $4n^2 = 2b^2$.
Then $2n^2 = b^2$ , so $b^2$ is even, and thus b is even.
Then a and b both have 2 as a common factor, so $\frac{a}{b}$ cannot be in lowest terms, a contradiction.
Thus x cannot be rational.
The part I am struggling with is why we can assume that x can be written in lowest terms. Or in other proofs, they say that the gcd(a,b) must be 1, but I do not understand why this is important to the proof. Any help is appreciated
Best Answer
You can assume that $\dfrac{a}{b}$ is in lowest terms because any fraction can be written in lowest terms, so we just take $\dfrac{a}{b}$ to be that lowest terms representation to make the proof easier. This is exactly the same as assuming $\gcd(a,b)=1$. This assumption is crucial, as it allows us to go through the proof, and eventually conclude that $2\mid a$ and $2\mid b$, which gives a contradiction because of our construction that $\gcd(a,b)=1$. Since $\gcd(a,b)=1$ there cannot be any common factors of $a$ and $b$ other than $\pm 1$. In other words, if $\dfrac{a}{b}$ is in lowest terms, the numerator and denominator cannot both be even, because then they would both be divisible by two, and thus the fraction is not in lowest terms.