prove that
is irrational
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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.
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