Question

Topic = Rational Numbers

Prove That √3 is an rational number.

Correct Answer will be marked As brainliest
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Answer 1

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♠ Correct Question :-

Prove That √3 is an irrational number.

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♣To Prove :-

√3 is an irrational number.

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♠ Proof :-

Let, us assume that √3 is a rational number.

So,

√3 can be  written in the form of p/q.

⟹ √3 = p/q.

⟹ q√3 = p.

Squaring on both side  we get ,

    (q√3)² = (p)²

⟹ 3q² = p²

⟹ q² = p²/3

It means  p² is divisible by 3

So,  p is also divisible by 3. - - - - - (1).

Let us  assume that,

   

p/3 = r.

Where r is an integer.

⟹ p = 3r.

⟹ 3q² = p².

Putting  value of p  in  eq (1)

⟹3q² = (3r)².

⟹ 3q² = 9r².

⟹ q² = 3r².

⟹ q²/3 = r².

It means  q² is divisible by 3.

So,  q is also divisible by 3. - - - - - (2)

From equation (1) and (2),

It can easily be determined that :

3 is a common  factor of p and q.

But, p and q are not Co-prime.

This contradiction arise due to our wrong as assumption that √3 is rational

∴ √3 is an irrational number.

♠ Hence  Proved! ♠

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Answer 2

Correct Question :-

Prove That √3 is an irrational number.

To Prove :-

√3 is an irrational number

Solution :-

Let us assume that √3 is an rational number and a and b are two positive number which do not common factor except 1

√3 = a/b

Squaring both side

(√3)² = (a/b)²

3 = a²/b²

3 × b² = a²

3b² = a² (1)

Now, 3 will divide a² and b² both

Let other integer be c

3 × c = a

3c = a

Squaring both sides

(3c)² = (a)²

9c² = a²

Using 1

9c² = 3b²

9/3 c² = b²

3c² = b²

Again, 3 will divide both c² and b²

From above statement we observe that both have atleast 1 and 3 as HCF. But a and b both will be a prime number. Therefore, √3 is an irrational number