Give an explicit example of a bijective function $f:\mathbb Q \to \mathbb Q$ such that $f(x)>x^3$ for each x.
My approach :=
Consider $f(x)=x^3+1$.
Then $f$ is one-one but not onto.
Since 3 has no pre-image.
Now I am unable to construct.
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1Interesting problem. Where is it coming from? – Jun 27 '23 at 12:46
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2I'd start by defining $f(n)=n^3+1$ for integers. Thus, $f(0)=1, f(1)=2$ for example. Now write down the linear bijection $[0,1]\to [1,2]$ and so on. – lulu Jun 27 '23 at 12:47
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This problem is taken from the book Steven G. Krantz. – SUJAN DAS Jun 27 '23 at 12:57
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Nice function @lulu – SUJAN DAS Jun 27 '23 at 13:02
1 Answers
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Hint : Define $f : \mathbb{Q} \rightarrow \mathbb{Q}$ by $$f(x)=\Big((\lfloor x \rfloor+1)^3-\lfloor x \rfloor^3 \Big)\lbrace x \rbrace + \lfloor x \rfloor^3+1$$
where $\lfloor x \rfloor$ denotes the greatest integer part of $x$, and $\lbrace x \rbrace $ its fractionnal part.
TheSilverDoe
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2It is basically the same idea as lulu's comment, in an explicit form. – TheSilverDoe Jun 27 '23 at 13:02
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