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the number of not identically zero functions $f:\mathbb{R} \to \mathbb{R}$ satisfying the equation $f(xy)=f(x)f(y)$ and $f(x+z)=f(x)+f(z)$ for some $z$ not equal to zero

  1. one
  2. finite
  3. countable
  4. uncountable

it seems like the question asks about the number of homomorphisams from R to R.but I think there are uncountable such homomorphisams. how ever the answer give is one..whats my mistake in thinking.how should I proceed..please help

José Carlos Santos
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blesson babu
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2 Answers2

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The mistake in your thinking is, primarily, that guessing the answer and not doing any actual mathematical work will help you in finding the solution.

If you think there are uncountable many functions, then try to show that. Construct at least a couple of them, and then possibly "construct" uncountably many of them. It probably won't be as easy as you think.

To actually solve the problem, however, I suggest you try to transform the two limitations you have into something else. See, for example, what happens if you plug in $x=y=1$ or $x=z=0$, and work from there.

5xum
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  • i used above hint as follows.please correct me if i done some mistake. f(x)=f(x−1+1)=f(x−1)+1=f(x−2)+2=....=f(x−x)+x=x hence f(x)=x.hence there exists only one such function.is it okey – blesson babu Feb 25 '18 at 05:43
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Since $f$ is not identically zero, take $y$ such that $f(y)\ne0$. Then $f(y)=f(1y)=f(1)f(y)$ implies $f(1)=1$.

Take $x=0$ in $f(x+z)=f(x)+f(z)$ and get $f(z)=f(0)+f(z)$, which implies $f(0)=0$.

Therefore, $f$ is a ring homomorphism $\mathbb{R} \to \mathbb{R}$.

Now see this question.

lhf
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  • @blessonbabu, in $f(x+z)=f(x)+f(z)$, you cannot choose $z$, it's given and fixed. – lhf Feb 22 '18 at 17:36