Question

4. Prove that √3 is an irrational
number.
​

Answer 1

Answer:

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2

Where q2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

Answer 2

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Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p²/q² (Squaring on both the sides)

⇒ 3q² = p²………………………………..(1)

It means that 3 divides p²and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p² = 9r²………………………………..(2)

from equation (1) and (2)

⇒ 3q² = 9r²

⇒ q² = 3r²

Where q² is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

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