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Question:

When we want to add three numbers, say $a + b + c$, we don’t bother inserting parentheses because $(a + b) + c = a + (b + c)$. But with powers, this is not true -

${(a^b)}^c$ need not be equal to $a^{(b^c)}$ - so we must be careful.

Show that this really is a problem, by finding positive integers $a,b,c$ such that

${(a^b)}^c < a^{(b^c)}$

and positive integers $d,e,f$ such that

${(d^e)}^f > d^{(e^f)}$.

Do I just have to show one actual example of each as an answer or should I write a proof without examples?

If I'm supposed to write a proof, where do I start?

And what would the full proof be?

Thank you so much!

J.-E. Pin
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user30200
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    The example would be a full proof. It shows that you cannot assume that you don't need to insert parentheses. – Calvin Khor Sep 07 '20 at 07:03
  • Noting that $a^{b^c}=a^{bc}$ simplifies things a bit, you just take an arbitrary $a>1$ and say $(b,c)=(3,2)$ or $(1,2)$. – Alexey Burdin Sep 07 '20 at 07:09

3 Answers3

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There are definitions behind. The addition of three numbers is left-associative, i.e., $a+b+c := (a+b)+c$. One can show that it is also right-associative, i.e. it is associative: $(a+b)+c = a+(b+c)$.

The exponentiation of three number is right-associative, i.e., $a^{b^c} =a^{(b^c)}$. As you recognized, it is not left-associative.

Wuestenfux
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You don't even need two examples. You only need one example of values $a,b,c$ for which $$a^{(b^c)}\neq \left(a^b\right)^c$$ to prove your point.

Finding any three such numbers is enough to prove that the statement

$$\forall a,b,c: a^{(b^c)}=\left(a^b\right)^c$$ is false, and thus justify that parentheses in this case are necessary.

5xum
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  • But doesn't the question say find a,b,c, that... AND d,e,f that... so don't we have to find both? But how do you find one for the second - d,e,f? I can't seem to find an example. How can (d^e)^f > d^(e^f) ? – user30200 Sep 07 '20 at 07:16
  • @user30200 Oh, I didn't know you were quoting an existing question. Well... then I don't know what you are confused about. The question specifically says you should find some positive integers that satisfy the two inequalities. So when you say "And what would the full proof be?" what do you mean by that. A full proof of what, exactly? – 5xum Sep 07 '20 at 07:19
  • I get that you just need an example but how do you get the example for (d^e)^f > d^(e^f) ? – user30200 Sep 07 '20 at 07:28
  • @user30200 Here's a hint: look for an example where $0<d<1$. – 5xum Sep 07 '20 at 07:32
  • Because for d the larger the power the smaller. Thank you. – user30200 Sep 07 '20 at 07:38
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The question is not only to prove that ${(a^b)}^c$ need not be equal to $a^{(b^c)}$, but that no inequality holds in general between these two expressions. In logical terms, the question is to show that neither of the two following formulas is true \begin{align} \text{for all } (a, b, c) \in {\Bbb Z}^3 \quad &{(a^b)}^c \leqslant a^{(b^c)}\\ \text{for all } (a, b, c) \in {\Bbb Z}^3 \quad &{(a^b)}^c \geqslant a^{(b^c)} \end{align} To prove this, you need to exhibit a counterexample for each of them, that is\begin{align} \text{there exists } (a, b, c) \in {\Bbb Z}^3 \quad &{(a^b)}^c > a^{(b^c)}\\ \text{there exists } (a, b, c) \in {\Bbb Z}^3 \quad &{(a^b)}^c < a^{(b^c)} \end{align} which is exactly what you are asked to do.

J.-E. Pin
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