Prove that: Let p be a prime number. If p divides a^2, then p divides

1.	Prove that: Let p be a prime number. If p divides a^2, then p divides a, where a is a positive integer.
Answer» Let the prime factorisation of a be as follows : a = p₁p₂ . . . p_n, where p₁,p₂, . . ., p_n are primes, not necessarily distinct. Therefore, a² = ( p₁p₂ . . . p_n)( p₁p₂ . . . p_n) = p²₁p₂² . . . p_n². Now, we are given that p divides a². Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a². However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a² are p₁, p₂, . . ., p_n. So p is one of p₁, p₂, . . ., p_n. Now, since a = p₁ p₂ . . . p_n, p divides a. We are now ready to give a proof that √2 is irrational. The proof is based on a technique called ‘proof by contradiction’. (This technique is discussed in some detail in Appendix 1).

Prove that: Let p be a prime number. If p divides a^2, then p divides a, where a is a positive integer.

Answer»

Let the prime factorisation of a be as follows :

a = p₁p₂ . . . p_n, where p₁,p₂, . . ., p_n are primes, not necessarily distinct.

Therefore, a² = ( p₁p₂ . . . p_n)( p₁p₂ . . . p_n) = p²₁p₂² . . . p_n².

Now, we are given that p divides a². Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a². However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a² are p₁, p₂, . . ., p_n. So p is one of p₁, p₂, . . ., p_n.

Now, since a = p₁ p₂ . . . p_n, p divides a.

We are now ready to give a proof that √2 is irrational.

The proof is based on a technique called ‘proof by contradiction’. (This technique is discussed in some detail in Appendix 1).

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