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I am in middle on solving problem 17 from Chapter 3 of Spivak's Calculus. We have a function [itex]f(x)[/itex], which is a non-zero function and it obeys the following properties.

[tex]\forall \, x \, y \, [f(x+y) = f(x) + f(y)][/tex]

[tex]\forall \,x \, y \, \left[f(x \cdot y) = f(x)\cdot f(y)\right] [/tex]

We have to prove that [itex]f(x) = x[/itex] for all x. Using the quantifier, the goal becomes

[tex]\forall\, x [ x \in \mathbb{R} \Longrightarrow (f(x) = x)] [/tex]

Now here is my proof, which I am sure is not correct, but I am trying to understand my flaw.

Let [itex]x[/itex] be arbitrary. Assume [itex]x \in \mathbb{R}[/itex]. Now using the first given property of f , we can see that [itex]f(0+0) = f(0) = f(0) + f(0)[/itex]. It follows that [itex]f(0) = 0[/itex]. Now our goal is to prove that [itex]f(x) = x[/itex] for some arbitrary x. I am going to try indirect proof here. Assume [itex]f(x) \ne x[/itex]. Then due to the property of trichotomy, we have either [itex]f(x) > x[/itex] or [itex]f(x) < x[/itex]. For the case 1, we will assume [itex]f(x) > x[/itex]. Since [itex]x[/itex] is arbitrary, this is also valid for [itex]x=0[/itex]. So we have

[itex]f(0) > 0[/itex]. Since [itex]f(0) = 0[/itex], it follows that [itex]0 > 0[/itex]. We reach a contradiction here, so our assumption that [itex]f(x) > x[/itex] is wrong. Similarly, we can prove that [itex]f(x) \nless x[/itex]. So it must follow that [itex]f(x) = x[/itex]. The problem with this proof is that I have proven [itex]f(x) = x[/itex] only with the knowledge that [itex]f(0) = 0[/itex]. But all kinds of functions have the property that [itex]f(0) = 0[/itex], and it should not necessarily follow that [itex]f(x) = x[/itex].

I think I am using the universal quantifier in a wrong way here. Any guidance will help...