Mathematical Analysis Question?

Question

I'm stuck on a problem, because I don't really understand what mapping is so I don't know what I can and can't do:

The problem is this:
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Suppose that f:R -> R is a map such that for all x,y in R (the real numbers)

f(x+y) = f(x) + f(y) and f(xy) = f(x)f(y)

Prove that f is either:
1) the identity mapping for all x in R. i.e. f(x)=x,
or
2) the zero mapping for all x in R. i.e. f(x)=0
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he gave us one hint, that somewhere in our proof that the fact that every positive real number is a square will be very useful.

I have no idea what to do and any help would be wonderful!

Thanks in advance!

Answer 1

Ah, the old "prove there are no nontrivial automorphisms of R" problem. Don't let the word mapping confuse you, it's just another way of saying "function". Anyway, let f be a map with the above properties. There are two possibilities: either f(1)=0 or f(1)≠0. In the first case, f must be the zero function, and in the second, f must be the identity function, as we shall show.

Case 1: f(1)=0

Let x be any real number. We know that x*1=x, so f(x) = f(x*1) = f(x)*f(1) = f(x)*0 = 0 (since anything times 0 is 0). Since x was any real number, it must be the case that for all x, f(x)=0, and so in this case f is just the zero function.

Case 2: f(1)≠0

In this case, we know f is not the zero function, so we must prove that it is the identity function. Consider that 1*1=1. So f(1) = f(1*1) = f(1)*f(1). This is a quadratic equation in f(1), which has only two possible solutions: f(1)=1 and f(1)=0. Since we have stipulated that f(1)≠0, this means that f(1)=1.

Now, suppose n is a natural number. If n=1, then as previously shown, f(n)=n. Further, if f(n) = n, then f(n+1) = f(n) + f(1) = n + 1 (since f(1)=1 and by the inductive hypothesis f(n)=n). Thus by induction, f(n)=n for all natural numbers n.

Now, we wish to show that for any integer n, f(n)=n. We know that if n is positive then it is a natural number, and so f(n)=n by the previous paragraph. Let us consider the case n=0: we know that 0+0=0, so f(0) = f(0+0) = f(0) + f(0), and subtracting f(0) from both sides yields that f(0) = 0. Finally, suppose n is negative -- then -n is a natural number, and so f(-n)=-n. But n+(-n) = 0, so we have that f(n+(-n)) = f(0) = 0, and also that f(n+(-n)) = f(n) + f(-n) = f(n) + (-n). Therefore, f(n)+(-n) = 0, so f(n)=n. Thus for all integers n, f(n)=f(-n).

Now, we wish to show that for any rational number r, that f(r)=r. Suppose r=1/n, where n∈Z. We know that f(n*r) = f(n/n) = f(1) = 1. However, we also know that f(n*r) = f(n)*f(r) = n*f(r). Therefore we have that n*f(r) = 1, so f(r)=1/n=r. Finally, let r be any rational number -- then it can be written as p*1/q, where p and q are integers. So f(r) = f(p*1/q) = f(p)*f(1/q) = p*1/q = r (since p is an integer and 1/q is the reciprocal of an integer). Thus, for any rational number r, f(r)=r.

Finally, we wish to show that for any real number x, f(x)=x. Here is where we use your teacher's hint -- we will show that f must preserve the ordering on R. Consider the following:

If x≥0, then ∃y such that y²=x. Then f(x) = f(y²) = f(y)², so f(x)≥0. We also know that 0 = f(0) = f(x+(-x)) = f(x)+f(-x), so f(-x) = -f(x). Thus, if x≤0, then -x≥0, so -f(x) = f(-x) ≥ 0, thus f(x)≤0. Thus, f maps nonnegative number to nonnegative numbers and nonpositive numbers to nonpositive numbers. But this means that f preserves the ordering on R, because x≤y ⇒ x+(-y)≤0 ⇒ f(x+(-y))≤0, and f(x+(-y)) = f(x)+f(-y) = f(x)-f(y) (since f(-y)=-f(y)), thus x≤y ⇒f(x)-f(y)≤0 ⇒ f(x)≤f(y).

With that in mind, and the fact that f(r)=r for all rational numbers r, we prove that f(x)=x for all real numbers x. Suppose the contrary, that f(x)≠x for some real number x. Then either f(x)x. If f(x)

Answer 2

There is one, VERY IMPORTANT condition for the statement to be true - the map (function) "f" MUST be continuous. Let's make that assumption and proceed the following way: take
f(x+y) = f(x) + f(y) and let x = y, we obtain
f(2x) = f(x + x) = f(x) + f(x) = 2f(x), the same way
f(3x) = f(2x + x) = 3f(x), etc....., so we proved that
(*) f(nx) = n*f(x) for every natural n = 1,2,3,....
Now let's denote f(1) = c /this is some real constant/, then
f(n) = n*f(1) = cn, i.e. for all natural numbers f is a LINEAR HOMOGENEOUS function. The idea is to prove that the same is true for all real numbers. The approach we follow easily leads to a conclusion that we can expand the same for all rational numbers, indeed in (*) changing x to x/n produces
f(x/n) = (1/n)*f(x) and then f((m/n)x) = (m/n)*f(x);
f(0) = 2f(0), so f(0) = 0 and f(-x) = -f(x).
All these lead to f(rx) = r*f(x) /r - rational/ and f(r) = cr.
But for the irrational we need explicitly the assumption we made above for continuity /every irrational number is a limit of a sequence of rationals/. Having proven
f(x) = cx for all real numbers x, let's pay attention to
f(xy) = f(x)f(y). We have
f(xy) = cxy, f(x)f(y) = cx*cy = c^2*xy, so
cxy = c^2*xy and xy, being arbitrary, leads to the equation
c = c^2, or c(c-1) = 0, whose roots are c=0 and c=1.
Hence f(x) = cx is either f(x) = 0 /c=0/ or f(x) = x /c=1/