An irrational number cannot be written as a fraction of whole numbers. Trying fractions one by one can never prove this. Proof by contradiction does: assume the opposite and derive something impossible.
A number is rational if it can be written as a/b, where a and b are integers and b ≠ 0. Every terminating or recurring decimal is rational.
A number is irrational if it cannot be written in that form. As the previous session showed, every fraction gives a terminating or recurring decimal. So the decimal of an irrational number neither terminates nor recurs.
In Class 9, √2, √3 and √5 were stated to be irrational. This session proves it.
There are infinitely many fractions. Showing that a thousand of them are not equal to √2 leaves infinitely many unchecked. No amount of checking finishes the job.
Proof by contradiction avoids this. Assume the opposite of the statement. Reason correctly from that assumption until you reach a statement that is impossible. Since every step was valid, the assumption must be false, so the original statement is true.
Give an everyday example of reasoning: “if that were true, something impossible would follow, so it is not true.”
“If I had locked the door, my key would have been needed to open it. I opened it without a key. So I did not lock it.”
The assumption (the door was locked) led to a consequence (a key was needed) that contradicts an observation. So the assumption is rejected. A proof by contradiction uses the same logic, with every step justified.
If a prime p divides a², then p divides a.
This follows from the Fundamental Theorem of Arithmetic. The prime factorisation of a² is the factorisation of a with every prime appearing twice as often. So the primes in a² are exactly the primes in a. If p is one of the primes of a², it is one of the primes of a, so p divides a.
Reveal each step in order.
Step 1 — Assume the opposite.
Suppose √2 is rational. Then √2 = a/b for integers a and b, b ≠ 0. Any fraction can be reduced, so choose a/b in lowest terms: a and b have no common factor other than 1.
Step 2 — Remove the square root.
Multiply both sides by b: b√2 = a. Square both sides: 2b² = a². So a² is 2 times an integer, which means 2 divides a².
Step 3 — Apply the fact about primes.
2 is prime and 2 divides a², so 2 divides a. Write a = 2c for some integer c.
Step 4 — Substitute.
Substitute a = 2c into 2b² = a²: 2b² = 4c², so b² = 2c². Then 2 divides b², and by the same fact, 2 divides b.
Step 5 — The contradiction.
2 divides a and 2 divides b, so 2 is a common factor of a and b. But in Step 1, a and b were chosen with no common factor other than 1. Both statements cannot be true.
Every step from Step 2 onwards was valid. So the error is in the assumption of Step 1. √2 is not rational: √2 is irrational.
The structure: assume √2 = a/b in lowest terms; show that 2 divides both a and b; this contradicts lowest terms.
The same proof works for √3 with one change. What changes?
The prime 2 is replaced by 3 throughout.
Assume √3 = a/b in lowest terms. Squaring gives 3b² = a², so 3 divides a², so 3 divides a. Write a = 3c. Then 3b² = 9c², so b² = 3c², so 3 divides b. Now 3 is a common factor of a and b, which contradicts lowest terms. So √3 is irrational.
The same template proves √p is irrational for every prime p.
Two facts follow from the same method:
Proof for 5 − √3: suppose 5 − √3 = r, where r is rational. Then √3 = 5 − r. The difference of two rational numbers is rational, so √3 would be rational. That contradicts the result above. So 5 − √3 is irrational.
1A number is irrational if…
2What is the first step in a proof by contradiction?
3Why is a/b chosen in lowest terms in the proof for √2?
4From 2b² = a², why does 2 divide a?
5To prove √3 irrational, the proof for √2 is repeated with the prime ____ in place of 2.
6Why is 5 − √3 irrational?
7Which statement is true?
That completes the unit on real numbers.
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