Another way of looking at it:

Anything zero times=0

But any non-zero number that isn't multiplied by anything at all (as opposed to multiplying it by zero) is the original non-zero number of course, and not the number zero... because you've done nothing with it.

Mentioning "zero" is not the same as using the number zero in the sum.

The whole thing is simply a Use-Mention Error.

https://en.wikipedia.org/wiki/Use%E2%80%...istinction

Just like rational numbers can be represented by many fractions, real numbers can also have more than one decimal representation. 0.9999... is one of the representations of the real number 1.

(December 14, 2015 at 9:03 am)LastPoet Wrote: Just like rational numbers can be represented by many fractions, real numbers can also have more than one decimal representation. 0.9999... is one of the representations of the real number 1.

Exactly. The construction of the real numbers from the rational numbers which I alluded to above is really quite fascinating, and there being several representations is almost an automatic prediction in this approach. So the idea is that in the rational numbers, there are infinite sequences a_n which seem to converge because following elements are closer and closer together, but some of them don't seem to converge to any element of the rational numbers. Such sequences are called Cauchy. The idea, which I find stunning, is to say the following: We now look at the set of all those Cauchy-sequences, and do one more thing: any pair of sequences the difference of which converges to zero we call equivalent. One then gets classes of equivalent Cauchy-sequences. This set of equivalence classes is the real numbers. Let that sink in

There's also this pattern I found..

5*5 = 5+5+5+5+5+ 0
5*6 = 5+6+6+6+6+ 1
5*7 = 5+7+7+7+7+ 2

4*4 = 4+4+4+4+ 0
4*5 = 4+5+5+5+ 1
4*6 = 4+6+6+6+ 2

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

2*2 = 2+2+ 0
2*3 = 2+3+ 1
2*4 = 2+4+ 2

1*1 = 1+ 0
1*2 = 1+ 1
1*3 = 1+ 2

0*0 =+ 0
0*1 =+ 1
0*2 =+ 2

I don't know what the frag is going on here, but I'll use myself as an authority (with a maths degree) to confirm that, for a finite number X:

0 * X = 0

X / 0 is undefined, the operation cannot be carried out

0.9999... (recurring) = 1

These are mathematical facts.

(December 13, 2015 at 9:59 am)Vic Wrote: You can't put 10 books in zero piles.

I can if I remove gravity.

(December 19, 2015 at 8:13 am)pool Wrote: There's also this pattern I found..

5*5 = 5+5+5+5+5+ 0
5*6 = 5+6+6+6+6+ 1
5*7 = 5+7+7+7+7+ 2

4*4 = 4+4+4+4+ 0
4*5 = 4+5+5+5+ 1
4*6 = 4+6+6+6+ 2

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

2*2 = 2+2+ 0
2*3 = 2+3+ 1
2*4 = 2+4+ 2

1*1 = 1+ 0
1*2 = 1+ 1
1*3 = 1+ 2

0*0 =+ 0
0*1 =+ 1
0*2 =+ 2

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

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

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

Also a proof by contradiction:

Let's assume you are correct with you definition of 0 * 1 = 1, and we've already proved that the algebra version of your pattern is correct.

Now let's try to do 1 * 1.

Your pattern states this should be: 1 + (0 * 1) + 0

Hey! That 0 * 1 looks familiar, we have the answer for it as well, so let's substitute: 1 + 1 + 0

Now looks what you've done, 1 * 1 = 2

But we all know 1 * 1 = 1, therefore contradiction, so 0 * 1 cannot equal 1.

(December 19, 2015 at 10:07 am)Tiberius Wrote: Also a proof by contradiction:

Let's assume you are correct with you definition of 0 * 1 = 1, and we've already proved that the algebra version of your pattern is correct.

Now let's try to do 1 * 1.

Your pattern states this should be: 1 + (0 * 1) + 0

Hey! That 0 * 1 looks familiar, we have the answer for it as well, so let's substitute: 1 + 1 + 0

Now looks what you've done, 1 * 1 = 2

But we all know 1 * 1 = 1, therefore contradiction, so 0 * 1 cannot equal 1.

Technically its proof by reductio ad absurdum. By taking the absurd, it resolves into a contradiction. Some 'intuitional' logicians do not accept proof by RA+.

(December 19, 2015 at 10:07 am)Tiberius Wrote: Also a proof by contradiction:

Let's assume you are correct with you definition of 0 * 1 = 1, and we've already proved that the algebra version of your pattern is correct(granted x*0=0).

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Now let's try to do 1 * 1.

Your pattern states this should be: 1 + (0 * 1) + 0

Hey! That 0 * 1 looks familiar, we have the answer for it as well, so let's substitute: 1 + 1 + 0

Now looks what you've done, 1 * 1 = 2

But we all know 1 * 1 = 1, therefore contradiction, so 0 * 1 cannot equal 1.

---

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