Question

Prove, that between any rational numbers there exists an irrational number.

Answer 1

Lets prove this:

We have 2 ratios: m/n and a/b are rational if m,a∈︎ℤ , n,b∈︎ℕ\{0}.

Lets Assume both are postive rationals or the quotient of naturals where the denomiantor isn’t zero rather than integers.

2mb/2nb = m/n, 2an/2nb = a/b. But now we have common denominator and guaranteed even number in both numerator and denominator.

WLG let m/n < a/b

m/n = 2mb/2nb

< (2mb+1)/2nb ≤︎ (2an-1)/2nb <

2an/2nb = a/b

(since 2mb < 2an and both numbers are even, there is at least one odd number between them,

if exactly one odd exists, then 2mb+1 = 2an-1.)

Since the d/dx √︎(x²+1) < d/dx x for all x≥︎1,

2mb+1 > √︎((2mb)²+1) > 2mb (second part trivial)

So I assert that:

m/n < √︎(4m²b²+1)/2nb < a/b

and the middle term as the square root of a perfect square plus 1, (all divided by a rational), is clearly irrational.