Prove that root 3 is an irratirrational
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Let us assume that √3 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, √3 = p/q { where p and q are co- prime}
√3q = p
Now, by squaring both the side
we get,
(√3q)² = p²
3q² = p² ... ( i )
So,
if 3 is the factor of p²
then, 3 is also a factor of p ... ( ii )
Let p = 3m { where m is any integer }
squaring both sides
p² = (3m)²
p² = 9m²
putting the value of p² in equation ( i )
3q² = p²
3q² = 9m²
q² = 3m²
So, if 3 is factor of q²
then, 3 is also factor of q
Since, 3 is factor of p & q both
So, our assumption that p & q are co- prime is wrong.
hence, √3 is an irrational number.
#BeBrainly
So, √3 = p/q { where p and q are co- prime}
√3q = p
Now, by squaring both the side
we get,
(√3q)² = p²
3q² = p² ... ( i )
So,
if 3 is the factor of p²
then, 3 is also a factor of p ... ( ii )
Let p = 3m { where m is any integer }
squaring both sides
p² = (3m)²
p² = 9m²
putting the value of p² in equation ( i )
3q² = p²
3q² = 9m²
q² = 3m²
So, if 3 is factor of q²
then, 3 is also factor of q
Since, 3 is factor of p & q both
So, our assumption that p & q are co- prime is wrong.
hence, √3 is an irrational number.
#BeBrainly
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