Real Numbers - Irrational Numbers Proof

Explanatory Answer

i. \\frac{1}{\sqrt2})

Let us assume the contrary that \\frac{1}{\sqrt2}) is rational.
∴ we can write \\frac{1}{\sqrt2}) = \\frac{a}{b})
Where a and b are coprime integers such that a, b ≠ 0. In this case a ≠ 0. Else the value of \\frac{1}{\sqrt2}) = 0, which it is not.
So, √2 = \\frac{b}{a})
Because 'b' and 'a' are integers, \\frac{b}{a}) is rational implying that √2 is rational.
But this contradicts with the fact that √2 is irrational.
So, the assumption that \\frac{1}{\sqrt2}) is rational is incorrect.
∴ we can conclude \\frac{1}{\sqrt2}) is irrational.

ii. 7√5

Let us assume the contrary that 7√5 is rational.
∴ we can write 7√5 = \\frac{a}{b}) where a, b are co-prime integers such that b ≠ 0
√5 = \\frac{a}{7b})
Because a, b and 7 are integers, \\frac{a}{7b}) is rational implying that √5 is rational.
But this contradicts with the fact that √5 is irrational.
So, the assumption that 7√5 is rational is incorrect.
∴ 7√5 is irrational.

iii. 6 + √2

Let us assume the contrary that 6 + √2 is rational.
So, we can write 6 + √2 = \\frac{a}{b}), where a, b are co-prime integers such that b ≠ 0
6 + √2 = \\frac{a}{b})
√2 = \\frac{a}{b}) – 6
Or √2 = \\frac{a - 6b}{b})
Because a, b and 6 are integers \\frac{a - 6b}{b}) is rational implying that √2 is rational.
But this contradicts with the fact that √2 is irrational.
So, the assumption that 6 + √2 is rational is incorrect.
∴ 6 + √2 is irrational.