Question

prove that √3 is an irrational number ​

Answer 1

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Answer 2

SHORT FORM:-

Let us assume that root 3 is rational.

Root 3 = a/b where a and b are integers and coprimes.

Root 3 * b = a

Square LHS and RHS

3b2 = a2

b2 = a2/3

Therefore 3 divides a2 and 3 divides a.

Now take ,

a = 3c

Square ,

a2 = 9c2

3b2 = 9c2

b2/3 = c2

Therefore 3 divides b2 and b.

LONG FORM:-

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