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everyone. My question for you is why a power, such as m^n^o is calculated as m^(n^o), and not (m^n)^o. Let me explain why I don't fully understand this:

As we know, we are supposed to perform calculations from left to right, starting with brackets, then exponents, then multiplication and division, and then finally addition and subtraction.

If all the operations in a single calculation are the same, then it should(but not for exponents, for some reason) be calculated from left to right. For example: 1*2*3 = 6, when calculated from left to right.

I think the real issue is that addition and multiplication are associative, while exponentiation is not. It is strange that while multiplication is efficient adding, and while exponentiation is fast multiplication, exponential expressions are neither commutative nor associative. For example, (2^2)^3= 64, while 2^(2^3) = 256. (Try typing 2^2^3 in google. You won't get (2^2)^3, but instead 2^(2^3).) Therefore, the order in which the exponents are calculated in a single exponential expression is relevant, compared to addition and multiplication.

My theory on why it is calculated as m^(n^o), is because if we calculated as (m^n)^o, then we can simply calculate it as m^(no). Therefore, (m^n)^o must be useless.

If my theory is correct, please tell me, but if it isn't, then explain to me to whole reason for why m^n^o is calculated as m^(n^o).

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    $\begingroup$ You're probably right. The usual way of interpreting it is more useful. By the way, subtraction isn't commutative or associative either. $\endgroup$ – Matt Samuel Apr 1 '16 at 3:13
  • $\begingroup$ True, neither is division. $\endgroup$ – Eliseo Palit-ang Apr 1 '16 at 3:56
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Let us compare $a^{(b^c)}$ and $(a^b)^c$.

We see from exponent rules that $(a^b)^c=a^{bc}$.

On the other hand, $a^{(b^c)}$ is not reducible.

So the question is, what actually is $a^{b^c}$, when no parenthesis is used.

I would think that it should be evaluated as $a^{(b^c)}$ because if the parenthesis were put the other way, then it may as well be $a^{bc}$, where there would be no point in making it look like a power tower.

So I'd say your theory is right, because such notation is more or less useless if written like so.

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  • $\begingroup$ Thank you very much. Your answer is well appreciated and explained thoroughly. I like how you explained that a^(b^c) is not reducible. Perhaps that is the whole reason why people calculate exponents in such a way. $\endgroup$ – Eliseo Palit-ang Apr 3 '16 at 23:46

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