Why x*0 = x

[ErGingerbreadMandude]

That algebra does depend on x*0=0
Because:
x * y = x + ((x - 1) * y) + (y - x)
where x=1 and y=-1

You simply cannot show that x*0=x using an algebra that has * on the R.H.S

Tell me, and I don't mean this to come across as mean, but how much experience with algebra do you have? The algebra I created was based on the pattern you stated, and it can contain multiplication of any numbers of the R.H.S provided there is no multiplication of 0, because that's what we are trying to show.

If x is 0 and y is 1, the algebra comes to: 0 * 1 = 0 + (-1 * 1) + (1 - 0)

The thing we are trying to work out is on the L.H.S. It does not appear in any form on the R.H.S, so the algebra is valid. I believe you are getting confused over what we are trying to work out. We aren't trying to prove how multiplication works, you accept that 1*1 = 1, that 5*5 = 25. What we are trying to work out what 0 * 1 is, using the pattern that you generated. My point was that you used the pattern wrong, you applied it incorrectly when it came to doing multiplication with 0, because you completely missed the (-1 * 1) part.

As I've also demonstrated, the algebra is correct because the R.H.S simplifies to exactly the L.H.S. Once you simplify your pattern, you get x * y. It's great, the pattern you detected actually exists, but you just applied it incorrectly in the final instance.

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
1 = 1

Now divide each side by 1:

1 = 1

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Again, I don't believe you understand how algebra works, because in algebra, you don't have to substitute anything back in if you don't want to, you can leave it in expanded form if it helps you to re-arrange the equation.

You are saying that 1*1 = 0*1

If we put x = 0 and y = 1, you are saying that y*y = x*y

Divide both sides by y, and you get: y = x

Now substitute x and y for the values you gave them: 1 = 0

See? The point is, an valid algebraic equation is always valid after you have modified it, as long as you do the same thing to each side of the equation, and you don't break any mathematical rules (i.e. dividing by zero).

So if 1*1 really does equal 0*1, as you claim, then you should be able to rearrange the equation and it should still be valid. If you divide both sides by 1, you cancel a 1 on each side (since both sides have multiplication by 1), and you are left with 1 = 0, which is clearly invalid, thus proving that the original equation was also invalid (because dividing by 1 is a perfectly acceptable mathematical operation).

Here's an example that does work:

2 = 1/4 * 8

Divide each side by 8: 2/8 = 1/4

Simplify: 1/4 = 1/4

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I have 0 experience with any form of math. Teeeeeeehee
You win this round.