Show that
is an irrational
number
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pulakmath007
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Answer:
Let us assume that √2 is a rational number.
then, as we know a rational number should be in the form of p/q
where p and q are co- prime number.
So,
√2 = p/q { where p and q are co- prime}
√2q = p
Now, by squaring both the side
we get,
(√2q)² = p²
2q² = p² .( i )
So,
if 2 is the factor of p²
then, 2 is also a factor of p .( ii )
=> Let p = 2m { where m is any integer }
squaring both sides
p² = (2m)²
p² = 4m²
putting the value of p² in equation ( i )
2q² = p²
2q² = 4m²
q² = 2m²
So,
if 2 is factor of q²
Then 2 is also factor of q
Since
2 is factor of p & q both
So, our assumption that p & q are co- prime is wrong
Hence √2 is an irrational number
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square root of prime numbers is always irrational number.
Irrational number has non- terminating and non-recurring decimal.