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Two irrational numbers between √2 and √3 are-

A)$2^{\frac{1}{2}}$ and $6^{\frac{1}{4}}$

B)$3^{\frac{1}{4}}$ and $3^{\frac{1}{6}}$

C)$6^{\frac{1}{8}}$ and $3^{\frac{1}{4}}$

D)None

By calculator the answer seems to be none. But can u suggest me some more formal way?

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    $\sqrt 2 < x < \sqrt 3$ if and only if $2 < x^2 < 3$ – 2017-02-24
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    For positive real number $x$ you have $\sqrt{2}\leq x\leq \sqrt{3}$ iff $2\leq x^2\leq 3$ iff $4\leq x^4\leq 9$ iff ... – 2017-02-24
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    @stevengregory: I'm afraid your statement as written is wrong: $\sqrt{2}\not<-5/2$, but $2<(-5/2)^2<3$. – 2017-02-24
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    It's pretty clear that $x>0$ in this context, so his statement holds – 2017-02-24
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    @celtschk - of course you are right, but the context of the problem suggests that we are talking about positive integers. but if you'd prefer, $\forall x \in \mathbb R, \sqrt 2 < |x| < \sqrt 3 \iff 2 < x^2 < 3$. – 2017-02-24

2 Answers 2

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We have:

$\sqrt{2} < x<\sqrt{3} \Leftrightarrow 2

From the wording of the question, it is sufficient to prove that one of the numbers in each part doesn't lie between $\sqrt{2}$ and $\sqrt{3}$ to disprove that part.

For part (A):

let $6^{\frac{1}{4}} = x \Rightarrow 4<6<9$ hence this works

But clearly $x=\sqrt{2}$ fails

For part (B):

we try $3^{\frac{1}{4}} = x \Rightarrow$ we require $4<3<9$ which is a contradiction, so B is false.

Finally (C):

From B we can see that C fails, therefore the answer is D.

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We have that exponentiation is monotone. That is $a

For example candidate A:

$$(2^{1/2})^4 = 4$$ $$(6^{1/4})^4 = 6$$

compare these with $\sqrt{2}^4=4$ and $\sqrt3^4 = 9$$. These are not strictly between (but if you accept equality they work).

The same method can be used for the rest of the candidates, but the power will have to differ. For candidates B you raise to the power of $12$ (multiple of both $4$ and $6$), C you raise to the power of $8$. Of course as mentioned if you raise to the power of $24$ it will work for all sets of candidates.