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# x^0 and x^1 Power

TokyoJunkie

1
Tue 1 Sep, 2009 12:03 pm
@ebrown p,
I have questioned the division by zero rule, and found it to ultimately be a wild goose chase, just as I find it a wild goose chase to question the simple truth of the existence of one of a quantity when you have that quantity. I really am wondering if we're not talking in circles. I don't know for sure.

Mathematics is self-referential and tautological. Can you prove to me that 2 + 2 = 4 without using the axiom of infinity? No, you can't, because without that mathematical axiom, there is no reason to believe 2 or 4 even exists, much less come to the otherwise baseless conclusion that 2 + 2 = 4. Mathematics is very pure and rigid, which earns it its crown as the purest science.

Since it is an axiom that 1 times any number is that number, it therefore holds that every quantity (regardless of whether it is in itself or a product of an operation or series of operations) has an implicit multiplication by 1.
FreeDuck

1
Tue 1 Sep, 2009 12:15 pm
@TokyoJunkie,
I don't know how I feel about that. Would you also say that any number has an implicit addition of zero, since zero is the additive identity? The implicit presence of a 1 doesn't sound quite right. Why not take it even further and say there is an implicit 1^∞? I'm thinking that if my math teacher told me that there was an invisible 1 in every problem I would become quite confused and think it a little to magical for mathematics. I think the first couple of explanations were pretty good. I would probably use division since most students can easily understand why a number divided by itself is 1. Fresco, I believe, posted an example of this.

0
Tue 1 Sep, 2009 12:22 pm
@TokyoJunkie,
I can think of examples where 2+2=11 and 2+2=10. (base 3 and base 4)
TokyoJunkie

1
Tue 1 Sep, 2009 12:22 pm
@FreeDuck,
Well, both multiplication by 1 and addition of 0 are essentially not doing anything to a quantity. I know this can in fact be useful in some situations ... I forget those, though, it's been years. *Scratches head*

You know, there can actually be an arbitrary number of implied 1s (in multiplication) and implied 0s (in addition). It doesn't matter how many times you do the op; in the end, nothing changes.
TokyoJunkie

1
Tue 1 Sep, 2009 12:25 pm
Quote:
I can think of examples where 2+2=11 and 2+2=10. (base 3 and base 4)

Changing the symbolism doesn't change the quantity. Symbols have no bearing on the numbers or operations they represent. I think anyone who's ever learned algebra knows this.
0 Replies

0
Tue 1 Sep, 2009 12:25 pm
@TokyoJunkie,
TokyoJunkie wrote:
You know, there can actually be an arbitrary number of implied 1s (in multiplication) and implied 0s (in addition). It doesn't matter how many times you do the op; in the end, nothing changes.

Exactly. There can be zero implied ones, and it doesn't change a thing.

You're searching a coal cellar at midnight for a black cat that isn't there.
TokyoJunkie

1
Tue 1 Sep, 2009 12:31 pm
Quote:
Exactly. There can be zero implied ones, and it doesn't change a thing.

You're searching a coal cellar at midnight for a black cat that isn't there.

That doesn't invalidate my point, though. You have to be stark raving mad to assert that when you have a quantity you don't necessarily have one of that quantity, if you excuse the appeal to common sense.

0
Tue 1 Sep, 2009 12:51 pm
@TokyoJunkie,
My (and I think ebrown's and Freeduck's) point is that your "implied 1" doesn't bring any value to the discussion. It introduces complexity, without resolving anything.

It's a digression into the identity property (and, in fact, an infinite regression as Freeduck and I have pointed out), but it doesn't help in describing powers or exponents.
0 Replies

TokyoJunkie

1
Tue 1 Sep, 2009 12:56 pm
Okay, folks, if it's a more rigorous proof you want, then a more rigorous proof I shall deliver, although I've said more than I actually need to explain the concept ...

Well, we know that multiplying exponents means adding them, x^m * x^n = x^m+n. And when we divide exponents, we subtract them, x^m/x^n = x^m-n. By this reasoning, we can conclude that x^2/x^2 = x^2-2 = x^0. Because it's the same number divided by itself, the value is equal to 1.

Now, if x^2 means that there are 2 x's being multiplied, if we divide that by x^1, we subtract 1 from the 2 and get x^1, or just one x by itself. If we divide that by x^2, in which we can only possibly get a value of 1, we cross out all the x's and are left with ... something. What is the value?

... 1.

As in my original assertion, but to avoid the taboo "implicit 1", x^n = x(1) * x(2) ... x(n); x^2 = x(1) * x(2); x^1 = x; x^0 = ?

In order for these axioms of exponentiation to hold true, there has to be an implicit 1, which in itself does not violate any other rules or axioms, and is therefore a valid assertion.
FreeDuck

1
Tue 1 Sep, 2009 01:12 pm
@TokyoJunkie,
TokyoJunkie wrote:

Now, if x^2 means that there are 2 x's being multiplied, if we divide that by x^1, we subtract 1 from the 2 and get x^1, or just one x by itself. If we divide that by x^2, in which we can only possibly get a value of 1, we cross out all the x's and are left with ... something. What is the value?

... 1.

I think you might have meant "now if we divide that by x^1"... because otherwise you get x^-1 which is another discussion. But what you're really saying is that x^0 is the same as x^1/x^1 which is the same as x/x which we know to be 1 because a number divided by itself is 1. I think that was also fresco's example, or something close to it.
TokyoJunkie

1
Tue 1 Sep, 2009 01:26 pm
@FreeDuck,
Yeah, I structured that last post weirdly. It should've read "divide x^2 by x^2 to get x^0." Humble apologies.

That proof is all dandy and fine, but it doesn't explain how you cross out all factors in a multiplication and end up with 1, other than "well, you get 1 from the number divided by itself." What are you going to say, that it's "self-evident" that a number over itself is 1? Isn't this just a reversal of the implicit 1 that comes with every number? If 5/5 = 1, then it follows that 1 * 5 = 5.

What is so hard to understand, folks?
TokyoJunkie

1
Tue 1 Sep, 2009 01:33 pm
And to put the kibosh on the "infinite regress" argument, an infinite amount of 1s multiplied by themselves is still 1, so in the end it is still only an implied 1. Same goes for the 0; an infinite number of added 0s will still just be an implicit 0.

1
Tue 1 Sep, 2009 01:39 pm
@TokyoJunkie,
But that puts the kibosh on your implicit one as well. If one multiplied by one is just one, then two multiplied by one is just two. "Implicit one" disappears in a puff of logic.
TokyoJunkie

1
Tue 1 Sep, 2009 01:43 pm
Wrong. That is equivalent to saying that you don't have one of a quantity, or in other words, without the 1, the whole number disappears. You can't get around the identity property no matter how many semantics games you try.
FreeDuck

1
Tue 1 Sep, 2009 02:05 pm
@TokyoJunkie,
TokyoJunkie wrote:

Yeah, I structured that last post weirdly. It should've read "divide x^2 by x^2 to get x^0." Humble apologies.

That proof is all dandy and fine, but it doesn't explain how you cross out all factors in a multiplication and end up with 1, other than "well, you get 1 from the number divided by itself." What are you going to say, that it's "self-evident" that a number over itself is 1? Isn't this just a reversal of the implicit 1 that comes with every number? If 5/5 = 1, then it follows that 1 * 5 = 5.

What is so hard to understand, folks?

Well, if your students are not comfortable with division yet then you could have a problem, yes. Hopefully they would be taught that before going into exponents.
FreeDuck

1
Tue 1 Sep, 2009 02:26 pm
@FreeDuck,
BTW, while learning division and multiplication they should have learned about the multiplicative identity, and explaining why a number divided by itself is exactly where it is the least ambiguous. By the definition it means that x*1 = x. By doing the inverse you get 1 = x/x. This is what it means to be the identity.
0 Replies

0
Tue 1 Sep, 2009 02:27 pm
@TokyoJunkie,
Sorry, dude. You can't have it both ways.

Multiplicative identity property of the number 1: Any number multiplied by 1 remains the same.

You do understand that "any number" includes "one", right?

So either you get into an infinite regression, or your "implicit one" is just so much smoke and mirrors.
TokyoJunkie

1
Tue 1 Sep, 2009 02:43 pm
Your argument stems from a basic misunderstanding of the identity property, so I shall treat it properly.

Although that statement "remains the same" is true, it doesn't disprove an implicit 1. The implicit 1 doesn't change the number, but it is there regardless because, as the identity goes, x/x = 1, ergo 1 * x = x

Even if you say 2 is just 2, it has to be 2 of something. It's a scalar quantity ... it's 2 of 1. That also makes it 1 of 2. 1 is the basic numeric unit ... unity. No other number makes sense without first making sense of 1.

You can't equivocate 1 and other numbers like that; while there is only one 1 of 2 ... 1 of 2 of 1 of 2 is 4. By comparison, there can be an arbitrary string 1 of 1 of 1 of 1 of 1, and it still is 1.

If my implicit 1 is only smoke and mirrors, how do you describe something? What sense does the indefinite article "a" make then?

0
Tue 1 Sep, 2009 02:54 pm
@TokyoJunkie,
TokyoJunkie wrote:

Your argument stems from a basic misunderstanding of the identity property,

'Fraid not. You're the one trying to twist the rule into creative little origami shapes.
0 Replies

ebrown p

0
Tue 1 Sep, 2009 03:30 pm
"Implicit one" disappears in a puff of logic.

Well said ((chuckles))

((I am, however, still stubbornly holding on to my implicit 2))

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