Q.Prove by the method of contradiction $\sqrt{5}$ is irrational.

Assume the opposite (that $\sqrt5$ is rational) and derive a contradiction: both the numerator and denominator of the assumed lowest-terms fraction would have to share the common factor 5.

Assume the contrary

Suppose $\sqrt5$ is rational. Then it can be written as

$\sqrt5 = \frac{p}{q}$

where $p,q$ are integers, $q\ne0$, and $\gcd(p,q)=1$ (the fraction is in lowest terms).

Deriving a relation

Squaring both sides:

$5 = \frac{p^2}{q^2} \implies p^2 = 5q^2 \quad (\ast)$

p must be divisible by 5

From $(\ast)$, $5$ divides $p^2$. Since $5$ is prime, $5$ must divide $p$ itself. Write

$p = 5k \quad \text{for some integer } k$

q must also be divisible by 5

Substitute $p=5k$ into $(\ast)$:

$(5k)^2 = 5q^2 \implies 25k^2 = 5q^2 \implies q^2 = 5k^2$

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