Prove that (√3-√8) is an irrational number.
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let ( √3-√8) be a rational
so,it can be written in the form of p/q
So √3-√8=p/q
= √3=p/q +√8
squaring both side
we get,(√3)^2= (p/q + √8)^2
= 3=p^2/q^2 +8 +2√8p/q
= -2√8p/q= p^2/q^2+8-3
= -√8 =(p^2/q^2+5)/(2p/q)
RHS is purely a rational and LHS isirrational so it is wrong so our countradiction is wrong
hence,it will be a irrational
so,it can be written in the form of p/q
So √3-√8=p/q
= √3=p/q +√8
squaring both side
we get,(√3)^2= (p/q + √8)^2
= 3=p^2/q^2 +8 +2√8p/q
= -2√8p/q= p^2/q^2+8-3
= -√8 =(p^2/q^2+5)/(2p/q)
RHS is purely a rational and LHS isirrational so it is wrong so our countradiction is wrong
hence,it will be a irrational
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