Prove √2 is irrational no. But not by long division.
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We assume that √2 is rational
√2=a/b
where, and b are integers and b≠0. Also, there is no common factor of a and b.
2=a²/b² (squaring both sides)
a²=2b² ...............(1)
2b²=a² ⇒2 divides a²
⇒2 divides a
Let a=2m ⇒a²=4m²
Putting the value of a² in eqaution (1), we have
4m²=2b²
2m²=b²
2 divides b² ⇒2 divides b
Thus 2 divides a and b
It means that 2 is a common factor of a and b. This contradicts the fact that a and b have no common factor.
THUS, √2 is not rational.
HENCE, √2 is irrational
√2=a/b
where, and b are integers and b≠0. Also, there is no common factor of a and b.
2=a²/b² (squaring both sides)
a²=2b² ...............(1)
2b²=a² ⇒2 divides a²
⇒2 divides a
Let a=2m ⇒a²=4m²
Putting the value of a² in eqaution (1), we have
4m²=2b²
2m²=b²
2 divides b² ⇒2 divides b
Thus 2 divides a and b
It means that 2 is a common factor of a and b. This contradicts the fact that a and b have no common factor.
THUS, √2 is not rational.
HENCE, √2 is irrational
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