prove root 3 and 3root 5 are irrational nos.
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AyushJotwani:
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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 ...✌✌✌✌✌...
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 ...✌✌✌✌✌...
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