Question

Prove that √3 is an irrational number.​

Answer 1

Answer:

assume that √3 is rational

√3=a/b

squaring on both sides

3=a^2/b^2

3b^2=a^2

therefore 3 divides a^2 and a

put a=3c

3b^2=(3c) ^2

3b^2=9c^2

b^2=3c^2

therefore 3 divides b^2 and b

so, 3 is the common factor of a and b

but we assume that 1 is the common factor of a and b

By contradiction,

√3 is irrational

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Answer 2

Answer:

Let us assume that √3 is rational.

Hence, √3 can be written in the form

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

Hence, √3 =

So, b√3 = a.

Squaring on both sides, we get

3b2 = a2

= b2

Hence, 3 divides a2.

By theorem: Let p is a prime number and p divides a2 , then p divides a, where a is a positive integer,

3 divides a also …(1)

Hence, we can say a = 3c for some integer c

Now, we know that 3b2 = a2

Putting a = 3c

3b2 = (3c)2

3b2 = 9c2

b2 = 3c2

Hence, 3 divides b2

By theorem: Let p is a prime number and p divides a2, then p divides a, where a is a positive integer,

So, 3 divides b also …(2)

By (1) and (2)

3 divides both a and b

Hence, 3 is a factor of a and b

So, a and b have a factor 3

Therefore, a and b are not co-prime.

Hence, our assumption is wrong

Therefore, by contradiction √3 is irrational.