Prove that √7 is irrational.
Answers
Answer:
Let us assume that
7
is rational. Then, there exist co-prime positive integers a and b such that
7
=
b
a
⟹a=b
7
Squaring on both sides, we get
a
2
=7b
2
Therefore, a
2
is divisible by 7 and hence, a is also divisible by7
so, we can write a=7p, for some integer p.
Substituting for a, we get 49p
2
=7b
2
⟹b
2
=7p
2
.
This means, b
2
is also divisible by 7 and so, b is also divisible by 7.
Therefore, a and b have at least one common factor, i.e., 7.
But, this contradicts the fact that a and b are co-prime.
Thus, our supposition is wrong.
Hence,
7
is irrational.
Answer:
let us assume to the contrary that √7 is irrational
therefore, √7=a/b (where a and b are co primes)
or √7b=a
on squaring both sides we get,
7b²=a²
therefore it is clear that a²is divisible by 7 so therefore will be also divisible by 7
now let us take, a=7c
therefore,
√7b=7c
on squaring both sides we get,
7b²=49c²
or b²=7c²
so it is clear that b²is divisible by 7 and so b is also divisible by 7
therefore a and b must contain 7as there prime factors .but it contradicts the fact that √7is irrational
this contradiction has arisen due to our wrong assumption therefore we conclude that √7 is irrational .