Exactly how many functions f: mathbb{Q} right | vedclass

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(B) The given conditions are $f(x+y) = f(x) + f(y)$ (Cauchy's functional equation) and $f(xy) = f(x)f(y)$.For $x, y \in \mathbb{Q}$,the only solutions

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Two

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(B) The given conditions are $f(x+y) = f(x) + f(y)$ (Cauchy's functional equation) and $f(xy) = f(x)f(y)$.For $x, y \in \mathbb{Q}$,the only solutions to these equations are the identity function $f(x) = x$ and the zero function $f(x) = 0$.$1$. If $f(1) = 0$,then $f(x) = f(x \cdot 1) = f(x)f(1) = 0$ for all $x$.$2$. If $f(1) 
eq 0$,then $f(1) = f(1 \cdot 1) = f(1)^2$,which implies $f(1) = 1$. By induction,$f(n) = n$ for all $n \in \mathbb{N}$,and consequently $f(x) = x$ for all $x \in \mathbb{Q}$.Thus,there are exactly $2$ such functions.Hence,option $B$ is correct.

Exactly how many functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ exist such that $f(x+y) = f(x) + f(y)$ and $f(xy) = f(x)f(y)$ for all $x, y \in \mathbb{Q}$?

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