a function f:R-R is said to preserve subtraction if, for all x,y in R, f(x-y)=f(x)-f(y)?

Question

(The R's are real numbers)
a) Prove that if f:R-R preserves subtraction, then f(0)=0
b) Give any example of a function that preserves subtraction
c)Prove that if f:R-R preserves subtraction, and if 0 is the only number that f maps to 0, then f must be injective

Answer 1

a) f(0) = f(0-0) = f(0) - f(0) = 0.

b) Fix k in the reals, and define f(x) = kx.

c) Suppose you have f(x)=f(y) for some reals x and y. Then f(x)-f(y) = 0, and since f preserves subtraction, f(x-y) = 0. Since 0 is the only element which gets sent to zero, this implies that x-y = 0, and thus x = y.

I have a challenge for you: find a nontrivial function f which preserves subtraction but is not injective. There are such functions, but they are not necessarily easy to find. Cheers!

Answer 2

A function f:R-R is said to preserve subtraction if, for all x,y in R, f(x-y)=f(x)-f(y) (The R's are real numbers)

a) Since f(x - y) = f(x) - f(y)
Then f(x - x) = f(0) = f(x) - f(x) = 0
ie f(0) = 0

b) f(x) = ax
f(x - y) = a(x - y) = ax - ay = f(x) - f(y)

c) If f(x) is injective then it is 1-1
So if f(x) = p and f(x) = q

Then f(x) - f(x) = p - q
BUT f(x) - f(x) = f(x - x) (as it is injective)
= f(0)
= 0
Therefore p - q = 0 so p = q
whence there is only one value for f(x)

Whoops sorry ..did not copy but you were all here when i came back with my solution

Answer 3

a) f(0) = f(x-x) = f(x) - f(x) = 0
b) any linear function. say f(x) = 3x
....then f(x-y) = 3(x-y) = 3x - 3y = f(x) - f(y)
c) you might want to define injective for those of us who've been out of abstract algebra for most of half a century. I see from Wikipedia that f is injective if f(a) = f(b) implies a=b. So suppose f preserves subtraction, and f(a) = f(b). Then f(a) - f(b) = 0 and f(a) - f(b) = f(a-b). so f(a-b) = 0, and if f(x) = 0 → x = 0, then a-b = 0 and so a = b, ie, f is injective.

Answer 4

a) Given the function, then let x-y=0, so x=y, so f(x)=f(y).
Then , f(x-y)= f(x)-f(y) = f(x)-f(x)=0.

b) f(x)=3x

c)I think, then f(x-y)=f(x)-f(y)=0.
So, f(x)=f(y), so x=y, so x-y=x-x=0, So, since 0 just maps to 0, its injective?