Can √2 be written as a rational number pq?
Step-by-step solution
Idea: Proof by contradiction: assume √2 is rational, follow the algebra, and reach something impossible. The key fact: if a square is even, the number itself is even (an odd number squared is odd).
- Assume √2 is rational: √2 = pq, where p, q are integers, q ≠ 0, and p, q have no common factor other than 1.½ mark
- Squaring: 2 = p2q2, so p2 = 2q2. Hence p2 is even, so p is even (if p were odd, p2 would be odd). Write p = 2k.1 mark
- Then 2q2 = (2k)2 = 4k2, so q2 = 2k2. Hence q2 is even, so q is even.1 mark
- Now p and q are both even, so they have the common factor 2. This contradicts our assumption that pq is in lowest terms.1 mark
- So the assumption is false: √2 cannot be written as pq. It is irrational.½ mark
Answer to write in the exam
Suppose √2 = pq, p, q integers, q ≠ 0, HCF(p, q) = 1
2 = p2q2 ⇒ p2 = 2q2
p2 even ⇒ p even; let p = 2k
2q2 = 4k2 ⇒ q2 = 2k2 ⇒ q even
2 is a common factor of p and q, contradicting HCF(p, q) = 1
∴ √2 cannot be written as pq; √2 is irrational.
Common mistakes that cost marks
- Forgetting to say that pq is in lowest terms at the start. Without it there is no contradiction at the end.
- Writing ‘√2 = 1.414, which is a fraction 14141000‘. 1.414 is only an approximation; (1.414)2 = 1.999396.
- Saying ‘p2 is even, so p2 = 2k‘ and then substituting the wrong thing. Write p = 2k, not p2 = 2k.
How this can come in the exam
Assertion (A): If p2 is an even integer, then p is even.
Reason (R): The square of an odd integer is odd.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
(2m + 1)2 = 4m2 + 4m + 1 is odd, so an even square cannot come from an odd number. R explains A.
In the proof that √2 is irrational we reach p2 = 2q2. The contradiction comes from
- q = 0
- p and q both being even
- p being odd
- p2 being negative
Show answer
(B) p and q both being even
Both being even gives the common factor 2, against the lowest-terms assumption.
Try one yourself
Use √2 being irrational to show that 3 + √2 is irrational.
Show answer
If 3 + √2 = r were rational, then √2 = r − 3 would be rational (difference of rationals), which is false. So 3 + √2 is irrational.
More questions like this
- Try to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7, or √10?
- We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?
- Try to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass. Generalise this method to construct a line segment of any length of the form √n, where n is a positive integer.
- We know what it means to add 2, 100, or even a lakh terms. What does it mean to add an infinite number of terms?
- It terminates: The division eventually leaves a remainder of 0. The decimal stops. 38 = 0.375. (Can you tell for which rational numbers the decimal will be terminating?)