6 min readPathway sheetA 2-mark or 3-mark proof

Proving numbers like 3 + 2√5 are irrational

If 3 + 2√5 were rational, then √5 would be rational too. Since √5 is irrational, 3 + 2√5 must be irrational. Isolate the root, then contradict.

Rational + irrationalAlways irrational, e.g. 2 + √3

Non-zero rational × irrationalAlways irrational, e.g. 5√2

Irrational + irrationalCan be rational: √2 + (−√2) = 0

Usually given"Given that √5 is irrational": you may use it without proof

Results to know

Combination	Result	Example
rational + irrational	irrational	3 + √2
rational − irrational	irrational	5 − √3
non-zero rational × irrational	irrational	2√5, (1/3)√7
non-zero rational ÷ irrational	irrational	1/√2, which equals √2/2
irrational + irrational	can be either	√2 + √2 = 2√2 (irrational); (5 + √2) + (5 − √2) = 10 (rational)
irrational × irrational	can be either	√2 × √3 = √6 (irrational); √2 × √2 = 2 (rational)

Method: prove that a + b√p is irrational

Here a and b are rational (b ≠ 0) and √p is known to be irrational.

1
Assume the number is rational. Write it as x/y, with x and y co-prime integers and y ≠ 0.
2
Move everything except the root to the other side.
3
Write √p as one fraction made only of integers.
4
Say: this fraction is rational, so √p would be rational.
5★
This contradicts the fact that √p is irrational. So the number is irrational.

★ key step

Example 1: prove that 4 − 3√2 is irrational

Problem

Given that √2 is irrational, prove that 4 − 3√2 is irrational.

Working

Assume, to the contrary, that 4 − 3√2 is rational.
Then 4 − 3√2 = a/b, where a and b are co-prime integers and b ≠ 0.
Rearrange: 3√2 = 4 − a/b = (4b − a)/b
So √2 = (4b − a)/(3b)
a and b are integers, so (4b − a)/(3b) is rational. Then √2 would be rational.
This contradicts the fact that √2 is irrational.

Answer

So our assumption is wrong. 4 − 3√2 is irrational.

Example 2: prove that 3√7 is irrational

Problem

Given that √7 is irrational, prove that 3√7 is irrational.

Working

Assume, to the contrary, that 3√7 is rational.
Then 3√7 = a/b, where a and b are co-prime integers and b ≠ 0.
So √7 = a/(3b)
a/(3b) is a ratio of integers, so it is rational. Then √7 would be rational.
This contradicts the fact that √7 is irrational.

Answer

So 3√7 is irrational.

In real life

A ramp or staircase may need a length like (2 + √3) m. Since this is irrational, a builder marks it with an approximate decimal, about 3.73 m (using √3 ≈ 1.732).

Exam angle

1 marks: “Is (2 + √3)(2 − √3) rational or irrational?” (2 + √3)(2 − √3) = 4 − 3 = 1, so it is rational. A product of two irrationals can be rational.

2 marks: “Given that √3 is irrational, prove that 2 + 5√3 is irrational.” Assume 2 + 5√3 = a/b (co-prime, b ≠ 0). Then √3 = (a − 2b)/(5b), which is rational. This contradicts √3 being irrational, so 2 + 5√3 is irrational.

3 marks: “Prove that √2 is irrational. Hence show that 5 + 3√2 is irrational.” Prove √2 irrational by contradiction (2 divides both a and b). Then assume 5 + 3√2 = a/b, get √2 = (a − 5b)/(3b), which is rational: contradiction.

Remember it

Rational plus irrational stays irrational: one 'impure' part spoils the whole.