Why x*0 = x

[robvalue]

By the way, there's nothing stopping people making up their own "number systems" with whatever rules they want. Abstract mathematics doesn't even have to correlate with reality in any way at all, it generally just has to be internally consistent. But if you want to make up your own system that isn't internally consistent, that's up to you really. Although a lot of the concepts obviously are inspired by reality, and hence applicable to them, the maths itself remains entirely abstract.

However, there are more ways you can visualize why these things should be true (apologies if I'm repeating other examples):

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A bucket holds 10 apples. I have no bucketfulls. How many apples do I have? I have none. There is no bucket, hence no apples. 0 * 10 = 0

I have 10 buckets, each with 0 apples in them. How many apples do I have? I have none. No apples anywhere. 10 * 0 = 0

What is the area of a shape which is 10cm long by 0cm wide? It has no area, it takes up no 2D space.

Or a pattern...

4 * 5 = 20
3 * 5 = 15
2 * 5 = 10
1 * 5 = 5
0 * 5 = ?

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I really like the example of putting 10 books into 0 piles. I never thought of that as a teacher!

You can think of X / 0 as the limit of X / N as N tends to zero (say X and N are positive for simplicity). For any finite number K, we can find an N small enough so that X / N is larger than K. The limit is infinite.

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A ninth, 1/9 = 0.11111111.....

1 = 9 * (1/9) = 9 * 0.1111111.... = 0.999999.....

Or

X = 0.99999......

10X = 9.99999.....

10X - X = 9.99999..... - 0.999999....

9X = 9

X =1

[Alex K]

You mess up your own pattern at the end. Let's define the pattern using algebra:

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x * y = x + ((x - 1) * y) + (y - x)

Plugging in x=5, and y=6:

5 * 6 = 5 + (4 * 6) + 1
5 * 6 = 5 + (6 + 6 + 6 + 6) + 1

Plugging in x=2, and y=4:

2 * 4 = 2 + (1 * 4) + 2
2 * 4 = 2 + 4 + 2

Seems to match your pattern, right?

Let's try x=0, y=1

0 * 1 = 0 + (-1 * 1) + 1
0 * 1 = 0 + -1 + 1
0 * 1 = 0

Tada! Math still works, the universe isn't broken. If you forget the middle part of your pattern, which is essentially "add m to itself n-1 times" then you don't get a correct answer.

By the way, your pattern simplifies to:

x * y = x + ((x - 1) * y) + (y - x)
x * y = x + (xy - y) + y - x
x * y = xy

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So you can prove its the correct pattern using algebra anyway.

Oh my dog! I was going to do this exact construction, and now you done did it. Chapeau

[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.

[Tiberius]

Another way of looking at it is to look at multiplications of 1:

5*1 = 5
4*1 = 4
3*1 = 3
2*1 = 2
1*1 = 1
0*1 = ?

There's a clear pattern emerging. Anything multiplied by 1 is itself. So the only logical value for 0*1 is 0.

[Edwardo Piet]

5*1 = 5
4*1 = 4
3*1 = 3
2*1 = 2
1*1 = 1
0*1 = ?

I think that's the most concisely lucid explanation yet. If he doesn't get that one I suggest giving up on explaining it to him

[robvalue]

I just worked 0 hours, and I'm paid £10 an hour. How much should I get paid?

If I get paid 0 * £10= £10, then I've also earned another 10 by working another 0 hours right after that. In fact, I'll pull a million shifts of zero hours right after one another, and get paid 10 mil for doing absolutely nothing.

Problem?

Also, try this:

1 * 10 = 10
0.5 * 10 = 5
0.1 * 10 = 1
0.001 * 10 = 0.01
0 * 10 = 10!?

[ErGingerbreadMandude]

Holy fuck I was wrong

[ErGingerbreadMandude]

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.

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

Corrected it

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

[Tiberius]

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

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

[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.