Math, asked by swastidt2494, 10 months ago

Prove that root 7 is an irrational number so prove 3 root 5 - root 7 is also irrational

Answers

Answered by thedragonslayer668
3

Step-by-step explanation:

Let us assume that √7 is a rational no.

∴√7=p/q where p and q are integers with no common factor other than 1 and q≠0

On squaring both sides,

(√7)²=(p/q)²

=> 7= p²/q²

=>7q²= p²

=> 7 divides p²         [∵7/7q²]

=> 7 divides p

Let p=7r for some integer r

On squaring both sides,

=>p²=49r²

=>7q²=49r²              [∵p²=7q²]

=>q²=7r²

=>7 divides q²         [∵7/7q²]

=>7 divides q

So, 7 is a common factor of both p and q.

But this contradicts the fact that a and b are co- primes.

Hence, √7 is irrational

Let us assume 3√5-√7 to be rational.

3√5-√7=a/b

=>3√5=a/b+√7

=>45=a²/b²+7+2√7a/b

=>38-a²/b²=2√7a/b

=>38b²-a²/b²=2√7a/b

=>38b²-a²/2ab=√7

Since a and b are integers

So, 38b²-a²/2ab is rational and hence, √7 is rational

But this contradicts the fact that √7 is irrational

Hence, our assumption is wrong and,

3√5-√7 is irrational

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