Question

prove that root 3 is irrational​

Answer 1

Answer- The above question is from the chapter 'Real Numbers'.

Given question: Prove that √3 is an irrational number.

Solution: Let us suppose that √3 is a rational number.

√3 = p/q (where p,q are co-primes, q≠0)

Transposing q to L.H.S., we get,

√3q = p

Squaring both sides, we get,

3q² = p² --- (1)

3 is a factor of p².

Using Fundamental Theorem of Arithmetic, we get,

3 is also a factor of p.

Put p = 3c.

Put the value of p in equation 1, we get,

3q² = 9p²

⇒ q² = 3p²

3 is a factor of q².

Using Fundamental Theorem of Arithmetic, we get,

3 is also a factor of q.

⇒ √3  is a rational number.

This contradicts the statement that p and q are co-primes and √3 is a rational number.

∴ We arrive at a wrong result due to our wrong assumption that √3  is a rational number.

∴ √3 is not an irrational number.

Answer 2

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Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence, √3 is an irrational number