prove that√7 is an irrational number
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Answered by
1
Answer:
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.
Answered by
0
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.
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.
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