- #1

- 911

- 18

## Homework Statement

[/B]

Suppose that the function ##f## satisfies the two properties. ##f(x+y)=f(x)+f(y)## and ##f(x\cdot y)=f(x)\cdot f(y)##, but that ##f## is not always ##0##. Prove that ##\forall x~ f(x) = x##, as follows:

(a) Prove that ##f(1)=1##

(b) Prove that ##f(x)=x## if ##x## is rational

(c) Prove that ##f(x) > 0## if ##x>0##

(d) Prove that ##f(x) > f(y) ## if ##x > y##

(e) Prove that ##f(x) =x ## for all ##x##. (Hint: Y=Use the fact that between any two numbers there is a rational number.)

This is problem 17 from chapter 3 of

**Spivak's Calculus**.

## Homework Equations

Definition of a function, equality of two numbers

## The Attempt at a Solution

[/B]

(a) Now ##f(x) = f(x\cdot 1) = f(x)\cdot f(1)##. Now if ##f(1)=0##, then ##f(x) = 0## for all ##x##. But we are told that this is not the case. So ##f(1)\ne 0##. From the second property, ##f(1) = f(1)\cdot f(1)##. Since ##f(1)\ne 0##, we have ##f(1)=1##

(b) First we will prove that ##f(x) = x## if ##x \in \mathbb{N}## using induction. We have already proven the Base case in (a). Now let ##P(x)## be the statement ##f(x)=x##. So base case, ##P(1)##, is true. Let ##k \geqslant 1 ## be arbitrary element in ##\mathbb{N}##. Suppose ##P(k)## is true. Now ##f(k+1) = f(k)+1=k+1## because ##P(k)## is true. So ##P(k+1)## is true. Since ##k \in \mathbb{N}## is arbitrary, the result is true for all ##x \in \mathbb{N}##. Hence ##\forall x ~\in \mathbb{N} f(x) =x##. Now let ##c## be an arbitrary rational number. So ##c = \frac{m}{n}##, for some ##m \in\mathbb{Z}## and for some ##n \in \mathbb{N}##. So ##n \ne 0##. Hence ##\frac{1}{n} \ne 0##. Now ##f(1) = f(n \cdot \frac{1}{n}) = f(n)\cdot f(\frac{1}{n})##. This leads to ##f(\frac{1}{n}) = \frac{1}{f(n)} = \frac{1}{n}##, since ##n \in \mathbb{N}##. Now since ##m \in\mathbb{Z}##, we can have ##m## to be nonpositive. There might be cases where ##m \leqslant 0##. Now ##f(0) = f(0+0) = f(0) + f(0)##, so ##f(0) = 0##. So if ##m=0##, ##f(m)=m##. Next if ##m<0##, then ##-m >0 ## and ##-m \in \mathbb{N}##. Using our earlier result for natural numbers, ##f(-m)=-m##. Now ## 0 =f(0) = f(m+(-m)) = f(m) + f(-m)##, hence ##f(-m) = -f(m)##. This means that ##-m = f(-m) = -f(m)##. So if ##m<0##, we have ##f(m)=m##. So we proved that in cases where ##m\leqslant 0##, we have ##f(m) = m##. Now since ##c=\frac{m}{n}##, we have ##f(c) = f(m\cdot \frac{1}{n}) = f(m)\cdot f(\frac{1}{n})##. Using the results we have proven for ##m## and ##n##, we get ##f(c) = \frac{m}{n}=c##. So proved that if ##x## is a rational number, ##f(x)=x##

(c) Suppose ##x>0##. Now ##\sqrt{x}## is the unique positive square root of ##x##, such that ##x = (\sqrt{x})^2##. Hence ##f(x) = f((\sqrt{x})\cdot (\sqrt{x})) =\left[ f(\sqrt{x})\right]^2##. But ##f(\sqrt{x}) \in \mathbb{R}##, so ##\left[ f(\sqrt{x})\right]^2 \geqslant 0##. This means that ##f(x)\geqslant 0##. Now we will prove that if ##f(x) = 0##, we reach a contradiction. Suppose ##f(x)=0##. Since ##x>0##,we have ##\frac{1}{x} >0##. Now ##1=f(1) = f(x\cdot \frac{1}{x}) = f(x)\cdot f(\frac{1}{x})##. Now since ##f(x)=0##, this leads to ##1=0##, which is a contradiction. So ##f(x)\ne 0## and since ##f(x)\geqslant 0##, we have ##f(x) > 0## if ##x>0##.

(d) Suppose ##x>y##. Then ##(x-y) > 0##. Using the previous result, we have ##f(x-y) > 0##. Now ##f(x-y) = f(x) + f(-y)##. Now ##0=f(0) = f(y + (-y)) = f(y) + f(-y)##. Hence ##f(-y) = -f(y)## and ##f(x-y) = f(x)-f(y)##. As ##f(x-y) > 0##, this leads us to ##f(x) > f(y)##.

(e) Let ##x## be arbitrary. Let ##\varepsilon >0## be arbitrary. There exists a rational number ##q## in ##(x, x+\varepsilon)##. Now ## x < q < x + \varepsilon##. Using the result in (d) and (b), we have ##f(x) < q < f(x) + f(\varepsilon)##. Hence we have ##f(x) - x < q-x ##. But ##q < x + \varepsilon##, this means that ##q-x < \varepsilon##. So we have ##f(x) - x < \varepsilon##. Similarly, there exists a rational number ##p## in ##(x-\varepsilon, x)##. So we have ##x-\varepsilon < p < x##. Again using the results of (d) and (b), we have ##f(x) - f(\varepsilon) < p < f(x)##. So we have, ##p - x < f(x) -x##. But we have ##-\varepsilon < p-x##. So this leads to ##-\varepsilon < f(x) - x##. Combining the two inequalities, we get ##-\varepsilon < f(x) - x < \varepsilon##. This is same as ## |f(x) - x| < \varepsilon##. Since ##\varepsilon >0## is arbitrary, this means that ##f(x)=x##. And since ##x## is arbitrary, the result is true for all ##x##.

Does this sound a good proof ? It took me 2 days to think through all of this. But solving Spivak's problems is rewarding exercise. But he assumes some knowledge of analysis it seems. For example, I had to use the definition of equality of two numbers in part (e), which is given in some another Analysis book I have. Spivak doesn't mention this in first 3 chapters.