I need a proof for why any number times 0 equals 0.

I would also like to know where this falls apart:

take 5*0=0

If I have a pile of zero apples, and add another pile of zero and
another and so on, so I have 5 piles of zero apples, then the total is
zero apples. Fine. I can make "x*5" into x+x+x+x+x.

But what if I start with a pile of 5 apples (since 5*0=0*5)- what is
the addition equivalent to multiplying by zero?

Not sure about the rest 
- but where it falls down is pretty simple

   5 * 0 = 0

=> 5 = 0/0 = Infinity (or a divide by zero error)

Best think of multiplying by zero as a 'zapper'

" 5 times 3" means add 3 five times.
So, 5*3 = 3+3+3+3+3 = 15

5*0
= 0 +0 +0 +0 +0
= 0
Proven.

What about 0*5 ?
It says "zero times 5"
Since zero is nothing, then do nothing to 5. Do not add 5 to nothing.
Just don't do anything on 5. Nothing. Nada. Zero.
If that doesn't convince you, then do you believe 5*0 = 0*5?
If yes, there!

frde, um, I want something more than the word "zapper".

ticbol, if you do nothing to 5, you get 5. Which is wrong.

Thank you for the comments, but I want something a bit more heavyweight.

No, if you do nothing to 5, you get nothing.
If you add nothing to 5, you get 5.

Remember, "times" means add in many times.
So, "0 times..." means nothing, don't move, don't do addition.

5 x n = 5 x (n + 1) - 5

5 x 0 = 5 x (0 + 1) - 5
= 5 x 1 - 5
= 0
QED

Now, why does 0+1 = 1?
Because 0 is the additive identity element.
Why?
By definition.

Thank you xarqi !

> But what if I start with a pile of 5 apples (since 5*0=0*5)- what 
> is the addition equivalent to multiplying by zero?

It would be zero piles of 5 apples.

Sorry for my broken English!

I shall use the system of axioms:
http://math.la.asu.edu/~kawski/classes/mat371/handouts/realaxioms.pdf

"[-" = "belong to".
"R" is Set of Real Numbers.

For all x, 0 [- R:

x*0 + x*0 = 0*x + 0*x = (0 + 0)*x = 0*x = x*0  -  F8, F9, F2, F8.
x*0 + x*0 = x*0  -  (*).
x*0 = y  -  (**).
(y + y = y) <=> ( (*), (**) ).
(y + y) + (-y) = y + (-y)  -  since "+" is Function.
y + (y + (-y)) = 0  -  F1, F3.
y + 0 = 0  -  F3.
y = 0  -  F2.
x*0 = 0.

Thank you for the interesting problem!

> If I have a pile of zero apples, and add another pile of zero and
another and so on, so I have 5 piles of zero apples, then the total is
zero apples.

From the beginning you have the pile of zero apples! Next you add a
pile of zero apples to your starting pile 5 times. And then you obtain
the pile of zero apples.

> But what if I start with a pile of 5 apples (since 5*0=0*5)- what is
the addition equivalent to multiplying by zero?

From the beginning you have the pile of zero apples. Next you add a
pile of 5 apples to your starting pile zero times. And then you obtain
the pile of zero apples.

You cannot prove that. The multiplication in R (set of real numbers)
defines 0 as its absorbing element, that is x * 0 = 0 whatever x is.
The same way the multiplication defines 1 as the identity element.


François.

Thank you everyone!