Question

prove that 6+√2 is irrational

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Answer 1

First of all we will prove that √2 is irrational.

Let us, on the contrary, assume that √2 is a rational number of the form of p/q where p and q are integers, q ≠ 0 and HCF(p, q) = 1.

Then, we have

√2 = p/q

Squaring both sides,

2 = p²/q²

→ p² = 2q²

This means that

p² is divisible by 2

Hence, p is also divisible by 2

So, we can write p as 2a for some integer a.

p = 2a

→ p² = (2a)²

→ p² = 4a²

Also, p² = 2q²

→ 4a² = 2q²

→ q² = 2a²

This means that q² is also divisible by 2, which further means that q is divisible by 2.

Now, we had taken that p and q are co primes, that is, their HCF = 1. But since p and q both are divisible by 2, their HCF would not equal 1. This contradiction has occurred because we assumed √2 to be a rational number. Hence, our assumption is wrong and √2 is an irrational number.

Now, again, let 6 + √2 be a rational number of the form of p/q where both p and q are integers and q ≠ 0

→ 6 + √2 = p/q

→ √2 = p/q - 6

Taking LCM,

√2 = (p - 6q)/q

Now, since p and q are integers, (p - 6q)/q would give a rational value. But √2 is irrational. Hence, those two can not be equal. This contradiction has now occurred because we took 6 + √2 to be a rational number of the form of p/q. Hence, 6 + √2 is not a rational number. It is therefore an irrational number.

Answer 2

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❚ QuEstiOn ❚

# Prove that $(6+\sqrt{2})$ is irrational

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❚ ANsWeR ❚

✺ Given :

• $(6+\sqrt{2})$

✺ To ProVe :

• $(6+\sqrt{2})$ is irrational .

✺ Explanation :

Let , we assume to the contrary that , $(6+\sqrt{2})$ is rational .

Again let , $(6+\sqrt{2})=\dfrac{a}{b}$

( where a and b are Co-prime numbers )

✏ $\ \ {(6+\sqrt{2})=\dfrac{a}{b}}$

✏ $\ \ {\sqrt{2}=\dfrac{a}{b}-6}$

✏ $\ \ {\sqrt{2}=\dfrac{a-6b}{b}}$

Now , as a and b are Integers ,

So , $\ \ {\dfrac{a-6b}{b}}$ is rational ;

Hence , $\ \ {\sqrt{2}}$ is rational ;

But , $\ \ {\sqrt{2}}$ is irrational .

✺ This contradiction has arisen because of our wrong assumption that $(6+\sqrt{2})$ is rational .

✺ Therefore :

$(6+\sqrt{2})$ is irrational . (proved)

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