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$f : A \to A$ and $n \in n$, Let $f^n$ be defined by $f^1 = f$ and $$f^n = f \circ f^{n-1}$$ for $n \gt 1$.

Let $n$ and $k$ be natural numbers with $k \lt n$. Prove $$f^n = f^k \circ f^{n-k}$$

Induction: $n=2$

$f^2 = f \circ f^{2-1} \implies f^2 = f^1 \circ f^1$

Hence, the base case holds true.

I.H: Suppose its true for $n=m$. We have to prove that it also holds true for $n=m+1$, $$f^{m+1} = f \circ f^m$$

How do I show that it will be equal to $$f^n = f^k \circ f^{n-k}$$

Dando18
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TheGamer
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  • What are $A$ and $\mathbb A$? – lulu Jul 31 '17 at 19:29
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    @lulu i have no idea, my teacher uses this alot – TheGamer Jul 31 '17 at 19:30
  • Well, unless $\mathbb A$ is a subset of $A$ I have no idea what composition might mean. But I suggest you clarify with your teacher. – lulu Jul 31 '17 at 19:31
  • Hint: the key point here is that composition of functions is associative (assuming you have a situation in which composition makes sense). here is a relevant question. – lulu Jul 31 '17 at 19:35
  • @lulu i think its just a function( a function from A to a function from A) – TheGamer Jul 31 '17 at 19:39
  • In your first version you wrote $f: A\to \mathbb A$ . That made it look as though $f$ were a function from one set to another set (in which case composition of $f$ with itself would not appear to be defined). Now you have changed it to $f:A \to A$ and it makes sense to compose in this context. – lulu Jul 31 '17 at 19:41
  • @lulu sorry i thought that was the format for making the function thing – TheGamer Jul 31 '17 at 19:41
  • Well, as it is written now it appears to mean "$f$ is a function from a set $A$ to itself". That seems sensible, at least with that definition we are able to define $f\circ f$. But of course it's possible that you mean something else. – lulu Jul 31 '17 at 19:43

2 Answers2

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Hint: fix $m$ and prove by induction on $n$ that $f^{n+m} = f^n \circ f^m$.

Adayah
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Let $k<m+1$. Then $k-1<m$

Using definition and inductive hypothesis,

$$f^{m+1}:=f\circ f^m=f\circ (f^{k-1}\circ f^{m-k+1})=(f\circ f^{k-1})\circ f^{m+1-k}:=f^k\circ f^{m-k+1}$$

Third equality comes from associativity, rest are definitions and inductive hypothesis

user160738
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  • I understand that $k-1<m$ but how did u replace the f^m term, ($f^{k−1}∘f^{m−k+1}$) – TheGamer Jul 31 '17 at 19:51
  • I think it's a proof by induction on $m$ that for each $k < m$ the equality holds. It's always very helpful to state this explicitly so the induction structure is clear. – Adayah Jul 31 '17 at 19:54
  • @TheGamer your inductive hypothesis applies to "every" $k<m$ (you assume this for general $m$, and showed it to be true for $m=2$). Once you know that $k-1<m$, you can apply inductive hypothesis to $k-1$ – user160738 Jul 31 '17 at 19:58