Question

Prove √3 is an Irrational Number.
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Answer 1

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 & q≠0

So,

√3 = p/q. { where p & q are prime}

√3q = p

Now, by squaring both the sides

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 prime is wrong & q are co

hence,. √3 is an irrational number

Answer 2

Let, us assume that √3 is a rational number.

So,

√3 can be written in the form of p/q.

⟹ √3 = p/q.

⟹ q√3 = p.

Squaring on both side we get ,

(q√3)² = (p)²

⟹ 3q² = p²

⟹ q² = p²/3

It means p² is divisible by 3

So, p is also divisible by 3. - - - - - (1).

Let us assume that,

p/3 = r.

Where r is an integer.

⟹ p = 3r.

⟹ 3q² = p².

Putting value of p in eq (1)

⟹3q² = (3r)².

⟹ 3q² = 9r².

⟹ q² = 3r².

⟹ q²/3 = r².

It means q² is divisible by 3.

So, q is also divisible by 3. - - - - - (2)

From equation (1) and (2),

It can easily be determined that :

3 is a common factor of p and q.

But, p and q are not Co-prime.

This contradiction arise due to our wrong as assumption that √3 is rational

∴ √3 is an irrational number.

Hence Proved!