show that 7 root 3 is irrational no..
Answers
Answered by
96
Let's assume that 7√3 is rational. Then, there exists two positive integers such that
7√3 = a
--- and HCF(a,b) = 1
b
Squaring,
(7√3)² = a²
---
b²
49 × 3b² = a²
147b² = a² →1
⇒147 / a²
⇒147 / a
∴a = 147c, for some integer c
Squaring,
a² = 21609c²
147b² = 21609c²
b² = 147c²
⇒147 / b²
⇒147 / b
∴147 / a, 147 / b
∴a and b has common factor as 147.
But our assumption is HCF(a,b) = 1.
∴Our assumption is wrong.
Hence, 7√3 is an irrational no.
7√3 = a
--- and HCF(a,b) = 1
b
Squaring,
(7√3)² = a²
---
b²
49 × 3b² = a²
147b² = a² →1
⇒147 / a²
⇒147 / a
∴a = 147c, for some integer c
Squaring,
a² = 21609c²
147b² = 21609c²
b² = 147c²
⇒147 / b²
⇒147 / b
∴147 / a, 147 / b
∴a and b has common factor as 147.
But our assumption is HCF(a,b) = 1.
∴Our assumption is wrong.
Hence, 7√3 is an irrational no.
Answered by
6
Answer:
We have to prove that 3+
7
is irrational.
Let us assume the opposite, that 3+
7
is rational.
Hence 3+
7
can be written in the form
b
a
where a and b are co-prime and b
=0
Hence 3+
7
=
b
a
⇒
7
=
b
a
−3
⇒
7
=
b
a−3b
where
7
is irrational and
b
a−3b
is rational.
Since,rational
= irrational.
This is a contradiction.
∴ Our assumption is incorrect.
Hence 3+
7
is irrational.
Hence proved.
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