Question

Prove that √3 is an irrational number.​

Answer 1

Step-by-step explanation:

Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N. We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational number

Answer 2

Step-by-step explanation:

Let us assume the opposite i.e √3 is rational.

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

where a and b(b is not equal to 0) are co prime.

Hence,√3 =a/b

√3b=a

squaring both sides:-

(√3b)^2 =a^2

3b^2=a^2

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