Problem Statement
Prove that $\sqrt{5}$ is an irrational number by contradiction using the Fundamental Theorem of Arithmetic.
Verified Solution & Marking Scheme
Assume rationality and express in co-prime form a/b
Let us assume, on the contrary, that $\sqrt{5}$ is rational.
Then there exist positive integers $a$ and $b$ such that $\sqrt{5} = \frac{a}{b}$, where $a$ and $b$ are co-prime (i.e. $\gcd(a, b) = 1$) and $b \neq 0$.
Square both sides and deduce that 5 divides a
Squaring both sides:
$5 = \frac{a^2}{b^2} \implies a^2 = 5b^2$
Therefore, $5$ divides $a^2$. By the Fundamental Theorem of Arithmetic, if a prime number $p$ divides $a^2$, then $p$ divides $a$.
Hence, $5$ divides $a$.
Express a = 5c and deduce that 5 divides b
Since $5$ divides $a$, we can write $a = 5c$ for some integer $c$.
Substituting $a = 5c$ into $a^2 = 5b^2$:
$(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2$
This means $5$ divides $b^2$, which implies that $5$ divides $b$.
Conclude the contradiction
Therefore, $5$ is a common factor of both $a$ and $b$. But this contradicts our assumption that $a$ and $b$ are co-prime with no common factors other than $1$.
This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.
Hence, $\sqrt{5}$ is irrational. (Q.E.D.)