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If a statement is true, prove it. If a statement is false, disprove it. : If p and q are prime numbers for which p<q, then 2p+q^2 is odd.

Solution

If p and q are prime numbers for which p<q, then 2p+q^2 is odd.

This statement is true.

Proof:-

Prime numbers = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 …)

2 is the only even prime number, all other prime numbers are odd. After 2, the primes are an irregularly spaced set of odd numbers that continues to infinity.

Consider the following cases:-

Case 1:-

If p = 2 and as p < q, take q = 11

So, 2p+q^2 = (2 * 2) + (11)^2 = 4 + 121 = 125 which is odd.

As p < q, for p = 2, q will always be odd.

As 2p = 4 will be even and q^2 will be always odd as square of an odd number is odd,

then 2p+q^2 is always odd for p = 2. (even + odd = odd)

Case 2:-

Take p other than 2.

If p = 7 and as p < q, take q = 11

So, 2p+q^2 = (7 * 2) + (11)^2 = 14 + 121 = 135 which is odd.

As 2p = 14 will be even (any number multiplied by 2 is always even) and q^2 will be always odd as square of an odd number is odd,

then 2p+q^2 is always odd for p = prime number other than 2. (even + odd = odd)

Therefore, from above cases, it is clear that

as p < q, q can never be 2 ( as 2 is the smallest prime number), and so 2p+q^2 is odd ( As 2p is always even and q^2 is always odd, even + odd = odd)