Math, asked by Harshyadav007, 1 year ago

Give that root2 is irrational number prove the (5+3root2) is an irrational number

Answers

Answered by Simrankhan0506
6
Hope it help you.....
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Answered by AkshithaZayn
3
Hey there!

prove \: \sqrt{2 } \: is \: irrational

let \: us \: assume \: \sqrt{2} \: is \: irrational

 \sqrt{2} = \frac{a}{b}

Where b is not equal to 0
a and b are co-primes.

2 = \frac{a {}^{2} }{b {}^{2} } \:

(by squaring two sides)

a {}^{2} = 2b {}^{2}

4c {}^{2} = 2b {}^{2}

2c {}^{2} = b {}^{2}

So, b is a factor of 2

Since a and b are having 2 as a common factor, this contradict our assumption that a and b are co-primes

so \: \sqrt{2} \: is \: not \: rational

hence \: \sqrt{2} \: is \: irrational


Prove
5 + 3 \sqrt{2} \: is \: irrational

5 + 3 \sqrt{2} = \frac{a}{b}

Where a and b are integers and b is not equal to 0

3 \sqrt{2} = \frac{a}{b} - 5

 \sqrt{2} = \frac{a - 5b}{3b}

 \frac{a - 5b}{3b} is \: rational

but \: \sqrt{2} \: is \: irrational

Irrational is not equal to rational

So, this contradicts the assumption that
5 + 3 \sqrt{2} \: is \: \: rational

hence \: 5 + 3 \sqrt{2} \: is \: irrational

Hope helped
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