# The number √2 is a irrational.

**Proof:** Suppose, to the contrary, that √2 is a rational number, say, √2 = a/b, where a and b are both integers with gcd(a, b) = 1. Squaring, we get a^{2} = 2b^{2}, so that b\a^{2}. If b > 1, then the Fundamental Theorem of Arithmetic guarantees the existence of a prime p such that p\b. It follows that p\a^{2} and by theorem *“ if p is a prime and p\ab, then p\a or, p\b.”*, that is p\a; hence, gcd(a, b) ≥p. We therefore arrive at a contradiction, unless b = 1. But if this happens, then a^{2} = 2, which is impossible. Our supposition that √2 is a rational number is untenable, and so √2 must be irrational. **(Proved)**

## The number √5 is a irrational.

**Proof:** Suppose, to the contrary, that √5 is a rational number, say, √5 = a/b, where a and b are both integers with gcd(a, b) = 1. Squaring, we get a^{2} = 5b^{2}, so that b\a^{2}. If b > 1, then the Fundamental Theorem of Arithmetic guarantees the existence of a prime p such that p\b. It follows that p\a^{2} and by theorem “ *if p is a prime and p\ab, then p\a or, p\b.”*, that is p\a; hence, gcd(a, b) ≥p. We therefore arrive at a contradiction, unless b = 1. But if this happens, then a^{2} = 5, which is impossible. Our supposition that √5 is a rational number is untenable, and so √2 must be irrational. **(Proved)**

### Problem: The only prime of the form n^{3}-1 is 7.

**Proof**: Given that n^{3} – 1 = n^{3} – 1^{3} = (n -1) (n^{2} + n + 1)

We know if n is an integer, n-1 is an integer and n^{2} + n + 1 is also an integer.

Thus (n -1) (n^{2} + n + 1) = ab for integers a and b.

If ab is prime, either a or b must be 1(or else we would have more than one factors.)

Thus either n – 1 = 1 or (n^{2} + n + 1) = 1.

If the first is true then n = 2.

∴ n^{3} – 1 = 2^{3} – 1 = 8 – 1 = 7

If the second is true then n = 0 and n = -1. So, we test for n = 0, n = -1 and n = 2

When n = 0, then n^{3} – 1 = – 1 , is not prime

When n = -1, then n^{3} – 1 = (-1)^{3} – 1 = -1 – 1 = -2, is not prime.

When n = 2, then n^{3} – 1 = 2^{3} – 1 = 8 – 1 = 7

So, the only prime of the form n^{3}-1 is 7. **(Proved)**