Question

prove root 3 and 3root 5 are irrational nos.

Answer 1

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Arya @

Answer 2

Hey....

To prove:-√3 is an irrational no.

Proof:- Let √3be a rational no.

√3=a/b {a and b be co-prime no.}

√3b=a

{squaring on both sides}

(√3b)^2=a^2

3b^2=a^2 --------(I)

In eq. (I) 3 is a factor of 3b^2 and it is equal to a^2 so.. 3 will be a factor of a^2...

Now ..let a=3c

Put value of a in eq. (I)

3b^2 = (3c)^2

3b^2 = 9c^2

b^2 = 3c^2-------(ii)

Here , 3 is factor of 3c^2...and it is equal to b^2 so.. 3 is a common factor of b and c...

And ,A/Q to our contradiction a and b is co-prime bt here 3 is a common factor of a and b ...
so our contradict was wrong and ..

√3 is an irrational no....

NOW ...

To prove :-3√5 is an irrational no...

let 3√5 be a rational no..

3√5 ×1/3=√5-----(I)

{rational×rational =rational}

A/Q to eq. (I) √5 is a rational no bt we know that √5 is an irrational no..
so .. our contradict was wrong and...

3√5 is an irrational no.....

Here is ur answer in these figures .....

Hope this will help u ...✌✌✌✌✌...