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I tried to solve it by first showing that the function is monotonically increasing which shows that it is one to one. Then finding a number that lies between two consecutive values of the function.

1)$f(x+k)=(x+k)^{(x+k)} > x^{x+k} > x^x$(by binomial theorem) when $k$ is a natural number greater than or equal to one.

2) $2^2<5<3^3$

Is my solution correct?

Caleb Stanford
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Jinn
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  • If it is defined on $\mathbf Z$, its range is not contained in $\mathbf N$… – Bernard Nov 04 '15 at 19:10
  • You are right, does that make the very question obsolete ? – Jinn Nov 04 '15 at 19:12
  • @Jinn It means that the question as written does not make any sense. However, we can just assume you meant the range to be $\mathbb{Q}$ instead of $\mathbb{N}$. – Caleb Stanford Nov 04 '15 at 19:15
  • If you suppose the domain is $\mathbf N$, it's not meaningless – Bernard Nov 04 '15 at 19:15
  • @6005: You're right, but the answer is obvious then. – Bernard Nov 04 '15 at 19:16
  • It would still be undefined at $0$ – Jinn Nov 04 '15 at 19:16
  • @Bernard Ah yes, I see. – Caleb Stanford Nov 04 '15 at 19:16
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    @Jinn Not necessarily. We can say $0^0 = 1$. – Caleb Stanford Nov 04 '15 at 19:17
  • Isn't that undefined – Jinn Nov 04 '15 at 19:19
  • @Jinn More often than not in discrete contexts it is defined to be $1$. – Caleb Stanford Nov 04 '15 at 19:20
  • I would like to read about it, since it contradicts my present knowledge. Can you cite some references? – Jinn Nov 04 '15 at 19:22
  • @Jinn You can read about it in this question. As you can see from the answers, it's often important to say it is undefined (an "indeterminate form") but it is also often important to define it as $1$. It depends on the context. – Caleb Stanford Nov 04 '15 at 19:26
  • @Jinn The empty product is defined to be $1$. -- Also for cardinalities, $a^b$ is the number of maps from a set of cardinality $b$ to a set of cardinality $a$. -- Even to allow the notation $\sum_{k=0}^n a_kX^k$ for polynomials where $X$ is just a symbol makes the deifniion (anything)${}^0=1$ useful. -- It is just so that when extending $x^y$ to real numbers as far as one can, it is not possible to make the function continuous at $(0,0)$, but that is a whole differnt story than not being defined (after all the ways to defined powers start with naturals and are only extended to reals, – Hagen von Eitzen Nov 04 '15 at 19:30
  • @6005 indeterminate $\ne $ undefined – Hagen von Eitzen Nov 04 '15 at 19:31
  • What is the difference? – Jinn Nov 04 '15 at 19:33
  • @HagenvonEitzen Thanks for the correction, I wasn't aware of that distinction (and it seems to only matter in the case of $0^0$, since all the other standard indeterminate forms are actually undefined, except maybe $0 \cdot \infty$ which is sometimes defined to be $0$.) – Caleb Stanford Nov 04 '15 at 19:36
  • On second thought, $0 \cdot \infty$, $0^0$, $0^\infty$, $1^\infty$, and $\infty^0$ all have natural interpretations in set theory and combinatorics. – Caleb Stanford Nov 04 '15 at 19:38

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This is not actually a function $\mathbb{Z} \to \mathbb{N}$. If you want the function to be defined for negative integers, you need the codomain to be $\mathbb{Q}$. But then if it is defined for $0$ we would have $0^0 = 1 = 1^1$, so it wouldn't be one-to-one.

Therefore, let me assume you meant for the function to be $f: \mathbb{N} \to \mathbb{N}$, $f(x) = x^x$. Then your solution is correct! Since the function is strictly increasing ($f(x+k) > f(x)$) it is one-to-one, and since it skips $5$ it is not onto. Good work.

Caleb Stanford
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