prove that root 3 minus root 2 is an irrational number
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since rational no. can be written in the form of p/q then
_/3-_/2=p/q.
now on squaring both side we will get
= (_/3-_/2)^2=(p/q)^2
=(_/3)^2+(_/2)^2-2(_/3)(_/2)=(p/q)^2
=3+2-2_/6=p^2/q^2
=5-2_/6=p^2/q^2
=2_/6=5-p^/q^2
=2_/6=(5q^2-p^2)/q^2
=_/6=(5q^2-p^2)/2q^2
since,p and q are integers then (5q^2-p^2)/2q^2 is rational no.
then _/6 is also a rational number.
But this contradicts because _/6 is an irrational no.
so,our supposition is false.
therefore _/3-_/2 is an irrational no.
hence prooved.
Thanq. :)
_/3-_/2=p/q.
now on squaring both side we will get
= (_/3-_/2)^2=(p/q)^2
=(_/3)^2+(_/2)^2-2(_/3)(_/2)=(p/q)^2
=3+2-2_/6=p^2/q^2
=5-2_/6=p^2/q^2
=2_/6=5-p^/q^2
=2_/6=(5q^2-p^2)/q^2
=_/6=(5q^2-p^2)/2q^2
since,p and q are integers then (5q^2-p^2)/2q^2 is rational no.
then _/6 is also a rational number.
But this contradicts because _/6 is an irrational no.
so,our supposition is false.
therefore _/3-_/2 is an irrational no.
hence prooved.
Thanq. :)
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