Question

show that √3 is irrational​

Answer 1

Answer:

Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1. Here 3 is the prime number that divides p2, then 3 divides p and thus 3 is a factor of p. ... So, √3 is not a rational number. Therefore, the root of 3 is irrational.

Step-by-step explanation:

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Answer 2

We have to prove √3 is irrational

Let us assume the opposite,

i.e., √3 is rational

Hence, √3 can be written in the form a/b

where a and b (b# 0) are co-prime (no common factor other than 1)

• Hence, √3 = a/b

√(3b)=a

Squaring both sides

(√3b)² = a²

3b² = a²

a²/3 = b²

Hence, 3 divides a²

By theorem: If p is a prime number, and p divides a², then

p divides a, where a is a positive number

So, 3 shall divide a also. ...(1)

• Hence, we can say

a/3 = c where c is some integer

So, a = 3c

• Now we know that

3b² = a²

Putting a = 3c

3b² = (3c)²

3b² =9c²

b² = 1/3 ×9c²

b² = 3c²

b²/a = c²

Hence 3 divides b²

By theorem: If p is a prime number, and p divides a², then

p divides a, where a is a positive number

So, 3 divides b also. ...(2)

By (1) and (2)

3 divides both a & b

Hence 3 is a factor of a and b

So, a & b have a factor 3

Therefore, a & b are not co-prime.

Hence, our assumption is wrong

:: By contradiction,

√3 is irrational

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