# Spivak' Calculus Chapter 7 Problem 19(b) [continuity]

The question is from Spivak's Calculus 3rd Ed:

Suppose $0<a<1$, but that $a$ is not equal to $1/n$ for any natural number $n$. Find a function $f$ which is continuous on $[0,1]$ and which satisfies $f(0) = f(1)$, but which does not satisfy $f(x) = f(x+a)$ for any $x$.

The solution from Spivak's solutions book is as follows:

In general if ${1 \over n+1} <a< {1 \over n}$ then define $f$ arbitrarily on $[0,a]$, subject only to $f(0)=0 , f(a)>0$ and $f(1-na)=-nf(a).$ Then define $f$ on $[ka,(k+1)a]$ by $f(ka+x)=f(x)+ ka.$ In particular we have, we have $f(1) = f(na + (1-na))= na+f(1-na)=0$ but $f(x+a)-f(a)=f(a)>0$ for all $x$.

I don't see how we get $na+f(1-na)=0$ and the final $f(x+a)-f(a)=f(a)>0$ for all $x$.

EDIT: Post below found an error in the answer book. Instead we require $f(ka+x)=k\color{red}{f(a)}+f(x)$, then we have $f(1)=0$ and the final statement also follows.

• This is impossible to type up, as the solution is a graph, rather than a formula. It's a beautiful, classic problem. You should make your post a reasonable post by stating specifically what in particular makes you dissatisfied. If you want a formula for a function, you should work on that. Jul 22 '17 at 17:14
• Don't answer your question with an answer in which you post the solution given in your answer book. What you can do, and I strongly suggest you do, is to include the solution in your answer book: by editing within your question field to include it there, along with an explanation addressing why you're not satisfied with it. Jul 22 '17 at 17:17
• @amWhy Ok I will fix it now, but I wanted to see alternative soln's please undo any downvote. Jul 22 '17 at 17:22
• @helios321 The downvotes are probably because the question doesn't follow the rules mentioned in the help centre Jul 22 '17 at 17:28
• There is often a rush to downvote. Who knows why... Jul 22 '17 at 17:33

First of all, note that $$0<1-na<1-\frac n{n+1}=\frac 1{n+1}$$, so $$1-na\in (0,a)$$. Next, there is a typo in what you typed: It should be that $$f(1)=f(na+(1-\color{red}{n}a))$$. But there is also a typo in Spivak's solution: He should have set $$f(ka+x)=k\color{red}{f(a)}+f(x)$$. [A consequence of this is that $$f(ka) = kf(a)$$ for all $$k=1,\dots,n$$.] So then $$f(1) = nf(a) + f(1-na) = nf(a) + (-nf(a)) = 0$$.
Last, note that when $$t\in [ka,(k+1)a]$$ for some $$k=0,1,\dots,n-1$$, we have $$t=ka+x$$ for some $$x\in [0,a]$$, so $$f(t)=kf(a)+f(x)$$. On the other hand, $$t+a = (k+1)a+x$$, so $$f(t+a)= (k+1)f(a)+f(x)$$, from which we conclude that $$f(t+a)-f(t)=f(a)>0$$ for all $$t\in [0,1-a]$$.
• Thanks for your answer. But it seems you too have made a typo! $t+a = (k+1)+x$ should be $t+a = \color{red}{a}(k+1)+x$. Jul 23 '17 at 4:39
• Possible (but not really relevant) typo: $1 - \frac{n}{n+1}$ = $\frac{n+1}{n+1}-\frac{n}{n+1}$ EQUALS $\frac{1}{n+1}$. Jun 9 at 0:49