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