Step by step
The question
Question: prove that log₂ 5 is irrational.
1 Assume the opposite
Suppose the opposite: log₂ 5 is rational, log₂ 5 = m/n, with m and n whole numbers.
log₂ 5 = mn
5 is bigger than 1, so the log is positive.
m and n are positive.
log₂ 5 > 0 (5 > 1), so m, n > 0
Careful!
The log is a power, not a number times 2.
Careful: log₂ 5 is the power of 2 that gives 5
2 Work it out
A log is a power: log₂ 5 is the power of 2 that gives 5.
2mn = 5
Raise both sides to the power n: 2ᵐ = 5ⁿ.
2m = 5n
(2^(m/n))ⁿ = 2ᵐ
3 Find the contradiction
The left side is even.
The right side is odd.
An even number can never equal an odd number!
2ᵐ is even, 5ⁿ is odd ✗
The opposite led to nonsense, so log₂ 5 is irrational.
So log₂ 5 is irrational. ∎
4 Check
A calculator: log₂ 5 is about 2.321928, the digits going on with no pattern.
log₂ 5 ≈ 2.321928...
The closest pair: 2 to the 7 is 128, 5 to the 3 is 125.
Close, but never equal.
Right!
2⁷ = 128, 5³ = 125: close, not equal
Let's look back.
First: assume the opposite.
Then: work it out.
Then: find the contradiction.
Last: check.
Answer
So log₂ 5 is irrational. ∎