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Question

If $f(x+y)=f(x)f(y)$ and $f(0)\ne0$, what is $f(0)?$

My thought process

I've decided to set $f(0)=f(x)f(y)$

and also as a side note i put $f(0)=f(0)f(0)$ incase if X and Y were both 0

But that's all I have gotten so far and I needed help with this

  • 3
    How about 1. Certainly, the square of 1 is also 1 – 2017-01-19
  • 0
    How is the operation $*$ defined, is it the usual multiplication of real field? if it is then $(f(0))^2=f(0)\implies f(0)=1$. – 2017-01-19
  • 0
    How do you get $f(0)=f(x)\times f(y)$? I could see $f(0)=f(x)\times f(-x)$.... – 2017-01-19
  • 2
    You derived $f(0) = f(0) \ast f(0)$. Now divide both sides by $f(0)$. What do you get? – 2017-01-19

2 Answers 2

1

It is assumed that $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb{R}$.

So you can choose, as a particular case, $x=0$ and $y=0$.

Then you get : $f(0) = f(0)^2$, that is $f(0)\left(1-f(0)\right)=0$.

Since it is assumed that $f(0)\neq0$, then you can divide both sides of the last equality by $f(0)$, and you get $1-f(0)=0$. Finally $f(0)=1$.

Note also that, knowing that, you can now choose $y=-x$ and get $f(0)=f(x)f(-x)$, so that :

$$\forall x\in\mathbb{R},\,f(-x)=\frac{1}{f(x)}$$

I suggest that, starting from that point you try to see what else can be derived concerning $f$ ...

2

Let $t=f(0)$. Then $t^2=t$. This is equivalent to

$$t(t-1)=0,$$

hence $t=0$ or $t=1$.