Step by step
The question
Question: prove that √3 is irrational.
1 Assume the opposite
Rational means: it can be written as a fraction of whole numbers.
Irrational means it cannot.
Suppose the opposite: √3 is rational.
Then √3 = a/b.
We cancel the fraction as far as it goes: a and b have no common factor.
Careful!
Without lowest terms there is nothing to break.
Remember it: a and b have no common factor.
2 Work it out
Square both sides: √3 squared is 3, and a/b squared is a squared over b squared.
Times b squared on both sides: a² = 3b².
3 times a whole number: so a squared is a multiple of 3.
3 is a prime number.
When a prime goes into a times a, it goes into a.
So a is a multiple of 3 too: a = 3c, with c a whole number.
Put it into a squared = 3 b squared.
3c squared is 9c squared.
Divide both sides by 3: b² = 3c².
The same as before: b squared is a multiple of 3, so b is a multiple of 3.
3 Find the contradiction
But now a and b are both a multiple of 3.
3 is a common factor.
That breaks lowest terms!
The opposite led to nonsense, so it is false.
√3 is irrational.
4 Check
A calculator: √3 is about 1.73205081, and the digits go on with no repeating pattern.
Even a good fraction like 19/11 squares to about 2.983, close to 3 but not 3.
Right!
Let's look back.
First: assume the opposite.
Then: work it out.
Then: find the contradiction.
Last: check.