Suppose f: R → R is a function such that for all x, y ≤ R, f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y). We proved in Tutorial 4B that then we also have f(0) = 0, f(1) = 1, f(-1) = -1, and the implication a

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Suppose f: R → R is a function such that for all x, y ≤ R, f(x + y) = f(x) + f(y) and
f(xy) = f(x)f(y). We proved in Tutorial 4B that then we also have f(0) = 0, f(1) = 1,
f(-1) = -1, and the implication a<b⇒ f(a) < f(b). You may freely use these facts in
solving this question.
(a) Show that for each n = Z, we have f(n) = n.
Transcribed Image Text:Suppose f: R → R is a function such that for all x, y ≤ R, f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y). We proved in Tutorial 4B that then we also have f(0) = 0, f(1) = 1, f(-1) = -1, and the implication a<b⇒ f(a) < f(b). You may freely use these facts in solving this question. (a) Show that for each n = Z, we have f(n) = n.
(b) Show that for each z E Q, we have f(x) = x.
Transcribed Image Text:(b) Show that for each z E Q, we have f(x) = x.
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