Question 8.7
3 months ago
By Exercise 3.16, if f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) for all xxx and yyy, then there exists ccc such that f(x)=cxf(x)=cxf(x)=cx for all rational numbers xxx. Since Q\mathbb{Q}Q is a dense set, this means that by Exercise 6b, f(x)=cxf(x)=cxf(x)=cx for all xxx.
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Q 8.7