Proof for sum of two reals

Question

Let $$f(x), \ f(y),\ f(x+y),$$ for $\ x,y \in \mathbb{R}.$ Consider that: $$I) \ \ f(x)=0, \forall x \in \mathbb{R}$$ my professor said that if $I)$ was true, then: $$f(x)=0 \implies f(y)= f(x+y)=0 \ \forall x,y \in \mathbb{R},$$ I didn't see that implication so clearly so I questioned him. He proved that: $$f(y)=0, \forall y \in \mathbb{R}$$ Rather easily, but said that proving $$f(x+y)=0, \forall x,y \in \mathbb{R}$$ was not so trivial, and I should start proving that all reals are a sum of other two.

My question is:

a) Is that the easiest approach or there's a more intuitive one?

b) How would one prove that the sum of two real numbers must be a real number?

PS: This was an entry-level college course, so despite working with functions, reals were not well defined for us yet. Also, I studied polynomials and complex numbers in high school, despite not formally knowing what a function was. So don't just conclude this is trivial for anyone.

EDIT: Sorry for editing/reviving such an old thread, but this post had to be fixed, I didn't realize how much of a wreck my grammar was at that time.

Answer 1

This is more of a "how to think about it" type hint.

"$f(x) = 0 \ \ \ \forall x \in \Bbb R$" is completely independent of the fact that "$x$" is used to denote the input to $f$.

Think of what "$f(x) = 0 \ \ \ \forall x \in \Bbb R$" means in words:

"This function $f$ gives me back $0$ no matter what real number I give to $f$."

To expand on that...

"This function $f$ gives me back $0$ no matter what real number I give to $f$. And it doesn't matter if I call that real number $x$, or $y$, or even $x+y$. The point is that if I feed a real number to $f$, then $f$ always spits out $0$."

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Note: This is essentially the idea of a "dummy variable" which is commonly discussed when introducing integrals in calculus. For example, $\int_0^1 2x \, dx = 1$. And this is completely independent of the fact that $x$ is the variable of integration. We could just easily say $\int_0^1 2y \, dy = 1$, because it doesn't matter what we call the variables. What matters is that we use the variables correctly.

Answer 2

If you know that $f(x)=0$ for all $x\in\mathbb{R}$, then it is completely trivial that $f(x+y)=0$ for all $x,y\in\mathbb{R}$. Indeed, let $x$ and $y$ be real numbers. Then $x+y$ is just some real number, and you already know that $f$ of any real number is $0$ (that is the meaning of "$f(x)=0$ for all $x\in\mathbb{R}$"). Thus $f(x+y)=0$.

Answer 3

It was hard to understand the question:

But I think the questions are:

1) if you are given that $f(x) = 0$ for all $x \in \mathbb R$, prove $f(y) = f(x+y) = 0$ for all $x,y \in \mathbb R$.

Proof: If $x,y \in \mathbb R$ then $k= x+y \in \mathbb R$ and $f(k) = f(x+y) = 0$. And $y \in \mathbb R$ so $f(y) =0$. And $0=0$ so $f(y) = f(x+y) = 0$.

2) if you are given that $f(x+y) = 0$ for all $x, y \in \mathbb R$ prove that $f(y) = 0$ for all $y \in \mathbb R$.

Your teacher claims this is not so easy. It's not much harder. $0 \in \mathbb R$ and $y + 0 = y$ so $f(y) = f(y+ 0) = 0$.