Irrational Numbers and Proof of Irrationality

Statement:

Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Proof:

Step 1: Let the prime factorisation of a be
a = p₁ p₂…p_n (where p₁,p₂,…,p_n are prime numbers)

Step 2: Squaring both sides,
\[a^2=(p_1p_2\ldots p_n)^2=p_1^2p_2^2\ldots p_n^2\]​

Step 3: p divides a²
So, p must be one of the prime factors of a².

Step 4: By the uniqueness of prime factorisation, the prime factors of a² are exactly
p₁,p₂,…,p_n.

Step 5: Hence, p is one of p₁,p₂,…,p_n.
Therefore, p divides a.

\[\sqrt{2}\] is irrational.

Step 1: Assume \[\sqrt{2}\] is rational.

\[\sqrt{2}\] = \[\frac{a}{b}\]

where a and b are integers and b ≠ 0

Step 2: Square both sides.

\[2=\frac{a^2}{b^2}\Rightarrow a^2=2b^2\]

Step 3: 2 divides a².

Since 2 is prime, by the divisibility property of primes,

2 divides a.

So, let a = 2c.

Step 4: Substituting,

\[(2c)^2=2b^2\Rightarrow4c^2=2b^2\Rightarrow b^2=2c^2\]

This means that 2 divides b², and so 2 divides b.

Therefore, both a and b are divisible by 2, which contradicts the fact that a and b are coprime.

Conclusion:
The contradiction arises from the assumption that \[\sqrt{2}\]​ is rational.