Prove that 3√2+2√3 is an irrational number.
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Answer:
Rational numbers are expressed in the form a/b,
where a and b are co- prime and b≠0
\implies 3\sqrt{2}-3=\dfrac{a}{b}⟹3
2
−3=
b
a
\implies 3\sqrt{2} = \dfrac{a}{b} +3⟹3
2
=
b
a
+3
\implies 3\sqrt{2} = \dfrac{a+3b}{b}⟹3
2
=
b
a+3b
\implies \sqrt{2} = \dfrac{a+3b}{3b}⟹
2
=
3b
a+3b
The RHS is a rational number
=> LHS is also a rational number
=> √2 is also a rational number
But, this contradicts to the fact that it is an irrational number.
Hence, our assumption is wrong.
\boxed{\boxed{\bold{Therefore, \ 3\sqrt{2} -3 \ is \ an \ irrational \ number}}}}}}
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