Set Theory Proof of $m^n < 2^{mn}$ for $m,n \in Z^{+}$

$n^m<2^{mn}$ for $m,n \in Z^{+}$

I chanced upon a rather interesting proof of this while studying sets, relations and functions in math class. Try to see if you can find it!

To get you started, consider two non-empty sets $A$ and $B$ containing $m$ and $n$ elements respectively. I'll post the solution in a few days...

SOLUTION

Consider two non-empty sets $A$ and $B$ containing $m$ and $n$ elements respectively.
The CARTESIAN PRODUCT of $A$ and $B$ , given by $A \times B = \{(a,b):a \in A, b \in B\}$ , is the set that contains all possible ordered pairs of the form $(a,b)$ . Obviously, this set contains $mn$ elements.
A RELATION between two sets $A$ and $B$ is defined as a subset of their Cartesian product. Then, how many relations are possible between $A$ and $B$ ? This is the number of possible subsets of the Cartesian product, or, in other words, the number of elements in the POWER SET of $A \times B$ . How do we find the number of possible subsets of $A \times B$ ? To choose a subset from a set, what do we do? We either choose an element to be part of the subset or not, i.e. we make a choice between 2 alternatives. Since there are $mn$ elements in the set, and we need to make a choice for each element, the number of subsets possible is given by $\boxed{2^{mn}}$ .
A FUNCTION between two sets $A$ and $B$ is a subset of their Cartesian product too (just like a relation) but with an additional constraint: all elements of $A$ should have exactly one image in $B$ . Then, how many functions are possible between $A$ and $B$ ? Since each element in $A$ has $n$ possible choices of an image in $B$ and there are $m$ elements in $A$ , there are a total of $\boxed{n^{m}}$ possible functions from $A$ to $B$ .
The final step is just logic: if a function is a relation with a constraint, there have to be fewer possible functions than relations.

Hence, $n^m < 2^{mn}$ .

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2 \times 3

2 \times 3

2^{34}

2^{34}

a_{i-1}

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\frac{2}{3}

\frac{2}{3}

\sqrt{2}

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`italics` or `_italics_`	italics
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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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${ m }^{ n }\quad ?{ \quad 2 }^{ mn }\\ { m }^{ n }\quad ?\quad { ({ 2 }^{ m }) }^{ n }\\ m\quad <\quad { 2 }^{ m }\\ { m }^{ n }\quad <{ \quad 2 }^{ mn }$

*I use the ? to show that the relation is not yet known

Niven Achenjang - 5 years, 6 months ago

Well, of course you're right algebraically, but that's not the curious proof I saw. I'll post the solution soon, maybe in two more days... :)

Raj Magesh - 5 years, 6 months ago

Subbed for any good solutions other than simple inequality. =3=

Samuraiwarm Tsunayoshi - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Set Theory Proof of mn<2mnm^n < 2^{mn}mn<2mn for m,n∈Z+m,n \in Z^{+}m,n∈Z+

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