Suppose that I have a positive integer $n$. Let $n^2$ denotes its square, i.e. $n^2$ is $n$ times $n$. For example, if $n$ is $3$ then $n^2$ denotes $9$, and if $n$ is $5$ then $n^2$ denotes $25$, etc.

I noticed that the recurrence operation $n\mapsto n^2$ has an intriguing behaviour. Namely, after the second step, we get $(n^2)^2$, which happens to be equal to $n$ multiplied with itself four times. For example, if $n=3$ then $n^2=9$ and $(n^2)^2=9^2=81$. But also $3\times 3\times 3\times 3=81$. Hence, I use the notation $n^4$ to mean $n\times n\times n\times n$. Then, what I observe is that $(n^2)^2=n^4$. That is already intriguing, but it does not stop here. At the third step we get $((n^2)^2)^2$ which also happens to be $n$ multiplied with itself six times. You will now guess that we must define $n^6$ accordingly, and then our observation is that $((n^2)^2)^2=n^6$. I checked this many times and we always find the same pattern. Is this a coincidence, or does this follow from more general considerations?

Now, I am trying to devise a theory of repeated multiplications. My goal is to show that $$(\cdots((n^{k_1})^{k_2})^{\cdots})^{k_r}=n^{k_1+k_2+\cdots+k_r},$$ where by now I think you can guess the meaning of the notation. For example it is actually true that $$((3^5)^2)^7=3^{5+2+7}.$$ I verified with my calculator: both numbers match up.

Now, given that we make the following operation: $$a+_k b=(k^{a})^b.$$ Call it multiplicative addition with base $k$. By the preceding remark, this operation does not depend on the order: $a+_kb=b+_ka$. Moreover, it does not depend how we bracket it, for several numbers: $$(a+_kb)+_kc=a+_k(b+_kc).$$ Now I am exploring the possibility of combining different bases. Is there a relation between $(a+_mb)+_nc$ and $a+_m(b+_nc)$? Let us see an example: $$(3+_52)+_47=5^5+_47=4^{5^5+7}$$ and $$3+_5(2+_47)=3+_54^9=5^{3+4^9}.$$ So in that case the number are not equal. But do you see any pattern?


closed as unclear what you're asking by Did, Shailesh, Leucippus, JonMark Perry, Claude Leibovici Sep 6 '16 at 5:43

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ $((n^2)^2)^2=n^8$ $\endgroup$ – Hagen von Eitzen Sep 5 '16 at 20:08
  • 2
    $\begingroup$ The unique pattern (widely known) here are the properties of exponents. $\endgroup$ – Crostul Sep 5 '16 at 20:14
  • 2
    $\begingroup$ The tag "groebner-basis" has absolutely nothing to do with the question. $\endgroup$ – Mariano Suárez-Álvarez Sep 5 '16 at 20:49

$(k^a)^b$ can be proved to be equal to $k^{ab}$, not $k^{a+b}$. Also with your proposed operator $+_k$, the exponentials stack up in a more complicated way than you think: it does matter where you put the brackets (in technical jargon, $+_k$ is not associative). To see this, note that if $a +_k b$ is defined to be $(k^a)^b= k^{ab}$, we have: $$ (a +_k b) +_k c = (k^{(a +_k b)})^c= (k^{((k^{ab})})^c = k^{(k^{ab}c)} $$ and $$ a +_k (b +_k c) = (k^a)^{(b +_k c)} = (k^a)^{(k^{bc})} = k^{(ak^{bc})} $$ but $k^{(k^{ab}c)} \neq k^{(ak^{bc})}$ in general. E.g., take $k = 2$, $a = b = 1$ and $c = 3$.


I think you haven't done your homework.

In more detail, there's two things missing here. First, if you think you've discovered something new in math, research it! For instance, on the wikipedia page https://en.wikipedia.org/wiki/Square_(algebra), the second full sentence links you to the page on exponentiation, where you'll find the correct versions of the patterns you're looking at. Other people have done a lot of math already; take advantage of it! This is not to dissuade original work, but it's important not to work in a vacuum.

More importantly, you've noticed a pattern, and you've guessed how it goes; and then you've made a lot of assertions without actually checking them (which I know because they are false). It's important not to jump the gun like this in mathematics.

Let's look at $((n^2)^2)^2$. By definition, this is $$[(n^2)^2]\times [(n^2)^2]=[(n^2)\times (n^2)]\times [(n^2)\times (n^2)]=n\times n\times n\times n\times n\times n\times n\times n.$$ That's eight $n$s, not six. Similarly with the other pattern you claim to have verified with your calculator (I'm really not sure what happened here). Bottom line: the pattern you've guessed at from looking at $(n^2)^2$ is incorrect. That's fine, but this is why we check things carefully. Instead, exponents combine multiplicatively: $(a^b)^c=a^{bc}$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.