Question

prove that √3 is not a rational number​

Answer 1

Step-by-step explanation:

We will prove the given condition by contradiction.

Now,

Let us assume that √3 is a rational number.

But, we know that a rational number should be in the form of p/q,

where, p and q are integers and q ≠ 0 or simply p and q are co- primes.

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 }

Again, squaring both sides, we get

=> p² = (3m)²

=> p² = 9m²

Again, putting the value of p² in equation ( i ),

We get,

=> 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.

Therefore, √3 is an irrational number.

Hence, √3 is not a rational number.

Thus, proved.

Answer 2

Answer:

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Step-by-step explanation:

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