Rational and irrational numbers
The diagonal of a 1 unit square cannot be written as a fraction of whole numbers. It is √2, an irrational number.
A rational number can be written p/q with q ≠ 0; its decimal terminates or repeats. An irrational number cannot; its decimal never ends and never repeats. √p is irrational when p is not a perfect square.
In the figure: √2 and π lie on the number line but are not fractions.
Rational and irrational numbers
In the app you can move this figure with your finger.
Open in the appRule 1: Sort them — Check if the decimal ends or repeats.
- Easy: 0.75 = 3/4: rational.
- Medium: 0.333… = 1/3: rational.
- Exam-type: √16 = 4 is rational but √15 is irrational.
Rule 2: Famous irrationals — 22/7 is only an approximation of π.
- Easy: √2, √3, √5, π are irrational.
- Medium: 0.1010010001… never repeats: irrational.
- Exam-type: √(25/4) = 5/2 is rational.
Every integer is rational (n = n/1).
Calling a long decimal irrational without checking for a repeating block.
"Ends or repeats: rational. Never: irrational."
Practice questions
- 1How many irrational numbers lie between any two different rational numbers?
- 2Which irrational number lies between 7 and 9?
- 3How many of the numbers √23, π, 6 + √28, 7√2, √41 are irrational?
- 4Which irrational number lies between 8 and 9?
- 5How many of the numbers π, √169, 3 + √41, 7 + √39, 1.75 are irrational?
- 6Which irrational number lies between 8 and 10?
- 7Which irrational number lies between 5 and 6?
- 8Which irrational number lies between 6 and 8?
Show answers (try them yourself first)
- 1Infinitely many
- 2√58
- 35
- 4√72
- 53
- 6√65
- 7√31
- 8√56
Operations with irrational numbers
Add a whole number to √2 and the endless decimal stays endless. But √2 × √2 brings back a plain 2.
The sum, difference, product or quotient of a non-zero rational and an irrational is irrational. Two irrationals can give a rational result, such as √2 × √2 = 2.
In the figure: Rational ± irrational stays irrational; some irrational pairs cancel to a rational.
Rule 1: Always irrational — Non-zero rational with an irrational.
- Easy: 2 + √3 is irrational.
- Medium: 5√2 is irrational.
- Exam-type: 1/√5 = √5/5 is irrational.
Rule 2: Can become rational — Two irrationals may cancel or multiply to a whole number.
- Easy: √3 − √3 = 0, rational.
- Medium: √2 × √8 = √16 = 4, rational.
- Exam-type: (3 + √2)(3 − √2) = 9 − 2 = 7, rational.
Always simplify the result before deciding whether it is rational.
Saying √2 + √3 must be rational (or irrational) without checking: only specific pairs cancel.
"Rational with irrational stays irrational."
Practice questions
- 1The sum of a rational number and an irrational number is always ___.
- 2Find the value of √108 × √3.
- 3Find the value of √24 ÷ √6.
- 4Find the value of √3 × √12.
- 5Find the value of √98 ÷ √2.
- 6Find the value of √75 × √3.
- 7Find the value of (5 + √17)(5 − √17).
- 8Simplify (3 + √3)² + (3 − √3)².
Show answers (try them yourself first)
- 1irrational
- 218
- 32
- 46
- 57
- 615
- 78
- 824
Proof that √p is irrational
To show a lock cannot be opened, pretend it can and watch the pretence break. Proof by contradiction works the same way.
Assume √p = a/b with a, b co-prime. Squaring gives pb² = a², so p divides a² and hence p divides a (p prime). Put a = pc: then b² = pc², so p divides b too. Now a and b share the factor p, which contradicts co-primeness. So √p is irrational.
In the figure: The proof of √2: both a and b turn out to be even, which breaks "co-prime".
Rule 1: The steps — Assume, square, use the prime fact, reach a contradiction.
- Easy: Prime fact: if p divides a², then p divides a.
- Medium: √3: 3b² = a² ⇒ 3 divides a ⇒ 3 divides b: contradiction.
- Exam-type: Same steps work for √5 and √7.
Rule 2: Related numbers — Reduce to a root already shown irrational.
- Easy: √2 is irrational so 3√2 is irrational.
- Medium: √6 is irrational (6 is not a perfect square).
- Exam-type: √4 = 2 is rational; the proof needs p prime.
The proof only works if p is prime, so the "p divides a² ⇒ p divides a" step is valid.
Forgetting to say a and b are co-prime at the start, so there is no contradiction at the end.
"Assume, square, p divides, clash."
Practice questions
- 1The method used to prove that √2 is irrational is called proof by ___.
- 2In the proof that √23 is irrational we get a² = 23b² and then write a = 23c. Substituting, what does b² equal?
- 3If √13 = a/b, then squaring both sides gives which equation?
- 4If a² = 23b² for integers a and b, what can we conclude?
- 5While proving that √7 is irrational, we find that 7 divides both a and b. This contradicts the assumption that ___.
- 6In the proof that √5 is irrational we get a² = 5b² and then write a = 5c. Substituting, what does b² equal?
- 7If √17 = a/b, then squaring both sides gives which equation?
- 8If √11 = a/b, then squaring both sides gives which equation?
Show answers (try them yourself first)
- 1contradiction
- 223c²
- 3a² = 13b²
- 423 divides a
- 5a and b are co-prime
- 65c²
- 7a² = 17b²
- 8a² = 11b²
Proving a + b√p is irrational
Once you know √2 is irrational, you can show many neighbours like 5 − √2 and 3√2 are irrational too.
To prove a number like a + b√p is irrational, assume it is rational (= r), isolate the root: √p = (r − a)/b. The right side is rational but √p is not, a contradiction.
Formula: √p = (r − a) / b
In the figure: Assume 3 + √2 = r; then √2 = r − 3 would be rational, which is false.
Rule 1: Isolate the root — Move the rational parts to the other side.
- Easy: 5 − √3: assume r ⇒ √3 = 5 − r rational: false.
- Medium: 3√2: assume r ⇒ √2 = r/3 rational: false.
- Exam-type: 2 + 3√5: √5 = (r − 2)/3 rational: false.
Rule 2: Conclude — Say which known irrational the result contradicts.
- Easy: So 5 − √3 is irrational.
- Medium: So 3√2 is irrational.
- Exam-type: So 2 + 3√5 is irrational.
Make sure b ≠ 0 before dividing by b.
Assuming the number is irrational instead of rational at the start.
"Assume rational, isolate the root, clash."
Practice questions
- 1In the proof for 3 + √3, we get √3 = r − 3, which is rational. This contradicts the known fact that ___.
- 2To prove that 3√3 is irrational, we assume 3√3 = r where r is rational. Then √3 = ?
- 3In the proof for 3 + √5, we get √5 = r − 3, which is rational. This contradicts the known fact that ___.
- 4To prove that 2 + 3√11 is irrational, we assume 2 + 3√11 = r where r is rational. Then √11 = ?
- 5To prove that 3 − √3 is irrational, we assume 3 − √3 = r where r is rational. Then √3 = ?
- 6To prove that (6 + √7)/2 is irrational, we assume (6 + √7)/2 = r with r rational (and non-zero where needed). Then √7 = ?
- 7To prove that 1/√5 is irrational, we assume 1/√5 = r with r rational (and non-zero where needed). Then √5 = ?
- 8To prove that 5 − 7√2 is irrational, we assume 5 − 7√2 = r with r rational (and non-zero where needed). Then √2 = ?
Show answers (try them yourself first)
- 1√3 is irrational
- 2r/3
- 3√5 is irrational
- 4(r − 2)/3
- 53 − r
- 62r − 6
- 71/r
- 8(5 − r)/7