Proove that e^x is always positive

[aaaa202]

May sound trivial but I don't find it trivial from the things I am given as a definition of the exponential function:

The exponential function satisfies f(a+b)=f(a)f(b), is the inverse to the ln and is restricted by the condition:
exp(x)≥1+x

I can't see how I can from this only proove that it is always positive. Can anyone help?

 

[Dick]

May sound trivial but I don't find it trivial from the things I am given as a definition of the exponential function:

The exponential function satisfies f(a+b)=f(a)f(b), is the inverse to the ln and is restricted by the condition:
exp(x)≥1+x

I can't see how I can from this only proove that it is always positive. Can anyone help?

If you are given exp(x)≥1+x then you already know exp(x)>0 if x≥0, right? So try and think of a way to show it if x<0.

 

[aaaa202]

ugh yeah that is where the hard work comes in for my part.

I have tried: exp(-x) ≥ 1 - x

So exp(x) ≤ 1/(1-x)

but that, as you can see, didn't get me anywhere.

But maybe this will do: for x>0

exp(x)exp(-x) = 1 which means exp(x) must be positive for x<0?

 

[Ray Vickson]

ugh yeah that is where the hard work comes in for my part.

I have tried: exp(-x) ≥ 1 - x

So exp(x) ≤ 1/(1-x)

but that, as you can see, didn't get me anywhere.

But maybe this will do: for x>0

exp(x)exp(-x) = 1 which means exp(x) must be positive for x<0?

Well, what do YOU think?

 

[aaaa202]

Yes!

 

[Fredrik]

I just want to say that there's a much easier way than what's been discussed so far.

 

[aaaa202]

and that is?

 

[Dick]

Yes!

There's probably a lot of ways to do it. But since you have those givens, I can't think of anything much easier. BTW you weren't really given that exp(0)=1. You just know exp(0)>0, which is enough. You can, of course, prove exp(0)=1 from what you are given.

 

[Fredrik]

We're only supposed to give hints, not complete solutions, but since you have found one solution I guess it's OK. What I had in mind is to use that 1=1/2+1/2.
$$e^x=e^{\frac 1 2 x+\frac 1 2 x}=\left(e^{\frac 1 2 x}\right)^2\geq 0.$$ Of course, you still have to prove that $e^x\neq 0$ for all x. But this is easy. Suppose that it's not true. Let y be a real number such that $e^y=0$. Let x be an arbitrary real number.
$$0=e^{x-y}0=e^{x-y}e^y=e^{x-y+y}=e^x.$$ Since x is arbitrary, this contradicts that $e^0\geq 1$.

When I said that this is "much" easier, I didn't realize that we still had to prove that $e^x\neq 0$ after the first step.

 

[Dick]

We're only supposed to give hints, not complete solutions, but since you have found one solution I guess it's OK. What I had in mind is to use that 1=1/2+1/2.
$$e^x=e^{\frac 1 2 x+\frac 1 2 x}=\left(e^{\frac 1 2 x}\right)^2\geq 0.$$ Of course, you still have to prove that $e^x\neq 0$ for all x. But this is easy. Suppose that it's not true. Let y be a real number such that $e^y=0$. Let x be an arbitrary real number.
$$0=e^{x-y}0=e^{x-y}e^y=e^{x-y+y}=e^x.$$ Since x is arbitrary, this contradicts that $e^0\geq 1$.

When I said that this is "much" easier, I didn't realize that we still had to prove that $e^x\neq 0$ after the first step.

Sure, f(0)=0 also satisfies f(a+b)=f(a)f(b). You need some other input. f(0) not equal to 0 would have been adequate. They gave you more than you really needed.