solve this question ...↓ if p is a prime number, prove that √p is irrational ....
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Answers & Comments
kartik338
If possible,let √p be a rational number. also a and b is rational. then,√p = a/b on squaring both sides,we get,
(√p)²= a²/b² →p = a²/b² →b² = a²/p [p divides a² so,p divides a] Let a= pr for some integer r →b² = (pr)²/p →b² = p²r²/p →b² = pr² →r² = b²/p [p divides b² so, p divides b] Thus p is a common factor of a and b. But this is a contradiction, since a and b have no common factor. This contradiction arises by assuming √p a rational number. Hence,√p is irrational.
Answers & Comments
also a and b is rational.
then,√p = a/b
on squaring both sides,we get,
(√p)²= a²/b²
→p = a²/b²
→b² = a²/p [p divides a² so,p divides a]
Let a= pr for some integer r
→b² = (pr)²/p
→b² = p²r²/p
→b² = pr²
→r² = b²/p [p divides b² so, p divides b]
Thus p is a common factor of a and b.
But this is a contradiction, since a and b have no common factor.
This contradiction arises by assuming √p a rational number.
Hence,√p is irrational.
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