Question

prove that √3 is an irrational number
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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

Step-by-step explanation:

We will prove this by contradiction.

Let's assume that $\sqrt{3}$ is a rational number.

.°. As we know a rational number should be in the form of $\dfrac{p}{q}$

where, p and q are co- prime numbers.

$=> \sqrt{3} = \dfrac{p}{q}$ [ where p and q are co- primes]

$= > p = \sqrt{3} q$

Now, by squaring both the side, we get,

$= > {p}^{2} = {( \sqrt{3} q)}^{2} \\ \\ = > {p}^{2} = 3 {q }^{2} \: \: \: \: ................(1)$

.°. If 3 is the factor of ${p}^{2}$

Then, 3 is also a factor of p ..... ( 2)

Now, let p = 3m { where m is any integer }

Again, squaring both sides, we get

$= > {p}^{2} = {(3m)}^{2} \\ \\ = > {p}^{2} = 9 {m}^{2}$

So, putting the value of p² in equation (1)

$= > {p}^{2} = 3 {q}^{2} \\ \\ = > 9 {m}^{2} = 3 {q}^{2} \\ \\ = > {q}^{2} = 3 {m}^{2}$

.°. If 3 is factor of q²

Then, 3 is also factor of q

°.° 3 is factor of p & q both

.°. Our assumption that p & q are co- prime is wrong

Hence, $\sqrt{3}$ is an irrational number.

Thus, Proved