Why x*0 = x

[Tiberius]

Algebra version is correct if and only if x*0=0 so of course it's going to give a wrong result if x*0=x because the algebraic version is designed to give the correct result only if x*0=0.
That's like going into an open minded argument convinced you're the one that's right.
There can't be any * 's on the R.H.S on the algebra

I disagree, the algebra version doesn't depend on x*0=0, it's based on the pattern in your post.

Just look at what you are doing. You are trying to multiply two numbers, 5 and 6. You take the first number (5), then you add the second number to it repeatedly until you have added the second number 1 less times than the first number (in this case, you add 6 exactly 4 times), and then finally you add the difference between the numbers.

So if we do the same thing with 0 and 1, you take the first number (0) and then you add the second number to it -1 times, which means you subtract 1 (unless you are arguing that -1 * 1 is not -1), and then finally you add 1, which is the difference between 0 and 1. The result is 0 + -1 + 1, which is 0.

Either way, I disagree that I can't use * on the R.H.S, because multiplication is a standard function in algebra, but the way I've done it above just uses addition.

You are correct that I'm not going into this open minded, but that's because I understand the math and you apparently don't. It's a known fact that 0*x = 0. You are trying to show that this is wrong, but it's clearly not.

I mean, look at the following:

You've stated in your post that 1*1 = 1+ 0 and 0*1 =+ 1. To simplify, 1*1 = 1, and 0*1=1

So if both these statements equal 1, we can substitute them to get:

1*1 = 0*1

Now divide each side by 1:

1 = 0

You have a massive contradiction, 1 is not equal to 0, and there are many reasons why it isn't. So either 1*1 =/= 1, or 0*1 =/= 1. There are many proofs for the former, so the latter is clearly incorrect.