Question

सिद्ध कीजिए की 7 अंडर रूट 2 एक अपरिमेय संख्या है ।​

Answer 1

Given : 7√2

To Find : Prove that   7√2 is an irrational number

Solution:

Assume that  7√2 is a rational number  

Hence 7√2 = p/q

where p and q are coprime integers

Dividing both sides by 7

=> √2 = p/7q

p/q is rational number hence p/7q is also rational number

but √2 is irrational number

LHS is an irrational number

while RHS is  Rational number

which is not possible Hence our assumption was wrong

so 7√2 is  not a rational number  

Hence 7√2  is an irrational number

QED

Hence Proved

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Answer 2

SOLUTION

TO PROVE

7√2 is an irrational number

PROOF

First we prove that √2 is an irrational number

If possible let √2 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,

√2 = p/q ( where p and q are co- prime )

⇒ √2q = p

Now, by squaring both the side

we get,

(√2q)² = p²

⇒ 2q² = p² - - - - - - - (1)

If 2 is the factor of p²

then, 2 is also a factor of p - - - - - (2)

⇒ Let p = 2m ( where m is any integer )

Squaring both sides we get

p² = (2m)²

⇒ p² = 4m²

Putting the value of p² in equation (1) we get

2q² = p²

⇒ 2q² = 4m²

⇒ q² = 2m²

If 2 is factor of q²

then, 2 is also factor of q

Since 2 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

Hence √2 is an irrational number

Now 7 is rational number and √2 is an irrational number

Since product of a rational number and an irrational number is irrational

Hence 7√2 is an irrational number

Hence the proof follows

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