Question

4. Prove that square root of 3 is irrational no​

Answer 1

Answer:

The square root of 3 is irrational. It cannot be simplified further in its radical form and hence it is considered as a surd. Now let us take a look at the detailed discussion and prove that root 3 is irrational.

Step-by-step explanation:

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Answer 2

$\huge\mathfrak\blue{Answer}$

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

It can be expressed in the form of p/q

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

⇒ √3 = p/q

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

⇒ √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

So we have p = 3rwhere r is some integer.

So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)

So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)

So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)⇒ 3q2 = 9r2

So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)⇒ 3q2 = 9r2⇒ q2 = 3r2

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