Question

prove that √3 irratuonal​

Answer 1

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step-by-step explanation :

Let us assume that √3 is a rational number.

then,

as we know that,

a rational number should be in the form of p/q

where p and q are co- prime number and q ≠0.

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 bot

It is contradiction.

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

hence,

√3 is an irrational number

Answer 2

Assume that √3 is a rational number.

As we know that,

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

√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

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 bot

It is contradiction.

Our assumption that p & q are co- prime is wrong

hence,

√3 is an irrational number