Question

prove that √3 is irrational​

Answer 1

Step-by-step explanation:

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

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Answer 2

Answer:

Let √3 be a rational number

√3 = a/b (a and b are integers and co-primes and b ≠ 0)

On squaring both the sides, 3 = a²/b²

⟹ 3b² = a²

⟹ a² is divisible by 3

⟹ a is divisible by 3

We can write a = 3c for some integer c.

⟹ a² = 9c²

⟹ 3b² = 9c²

⟹ b² = 3c²

⟹ b² is divisible by 3

⟹ b is divisible by 3

From (i) and (ii), we get 3 as a factor of ‘a’ and ‘b’ which is contradicting the fact that a and b are co-primes.

Hence our assumption that√3 is an rational number is false.

∴ √3 is an irrational number.