# How to explain this concept in tetration?

We know that:

$$\! 2^{2^{2^2}} \ne \left[{\left(2^2\right)}^2\right]^2$$

But for some students, this might be confusing. Is there a good way to explain this?

• I might not use twos to broach this since you have $2^4 = 4^2$; instead, I might ask students to think about "the very big number" $3^3^3$: Is this the same whether it is $(3^3)^3$ or $3^{(3^3)}$? Then split the class and have them compute each by hand, and see which group finishes more quickly. Well, $27^3$ is not so tough; but you'll need some patience for $3^{27}$ (!). To settle the matter, I happen to like H Towsner's "smaller font first" mnemonic. Nov 28 '16 at 0:10
• @Benjamin Dickman, they would know it's different, perhaps, but how do I explain the "why" part? Nov 28 '16 at 16:32
• Consider another operation: $3 \div 3 \div 3$. Observe that $(3 \div 3) \div 3 = 1 \div 3 = 1/3$, which is different from $3 \div (3 \div 3) = 3 \div 1 = 3$. I suspect this is what H Towsner meant by "they've seen the same in other contexts." Nov 28 '16 at 22:06
• You could also approach this by using a horizontal notation for exponentiation, and then it becomes clear that exponentiation is not an associative operation. For example, (3^3)^3 is not equal to 3^(3^3). If you think they won't relate to ^, then maybe use "E": (3 E 3) E 3 is not equal to 3 E (3 E 3). Nov 29 '16 at 17:48

The main point here is that $[(2^2)^2]^2\neq 2^{[2^{(2^2)}]}$ --- in other words, order of operations matters. That shouldn't be too unfamiliar, because they've seen the same in other contexts (and can do the calculations themselves to verify it). The important thing to realize is that the expression is ambiguous, and we need to do something to resolve the ambiguity.