Ok. I'm refreshed and I've had a few new thoughts on how to pursue this topic. I have been noticing that people seem to use patterns as proof that 2^0 = 1. Defining exponents as repetitive multiplication, and multiplication as repetitive addition, it would seem that 2^0 is undefined:

2^3 = 2 * 2 * 2 = (2+2)+(2+2) = 8

2^2 = 2 * 2 = (2+2) = 4

2^1 = 2 = 2 = 2

2^0 = = = ???

2^-1= 1/2 = 1/2 = 1/2

Although the empty product would explain it as:

2^3 = 1 * 2 * 2 * 2 = 1*((2+2)+(2+2)) = 8

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

2^1 = 1 * 2 = 1* (2) = 2

2^0 = 1 = 1 = 1

2^-1= 1 * 1/2 = 1* (1/2) = 1/2

But when considering 0^0, the empty product would say:

0^3 = 1 * 0 * 0 * 0 = 1*0 = 0

0^2 = 1 * 0 * 0 = 1*0 = 0

0^1 = 1 * 0 = 1*0 = 0

0^0 = 1 = 1 = 1

0^-1= 1 * 1/0 = 1*0 = 0 <-- ?

Also defining 0^0 as equal to 1.

Now, the 0^-1 leads to the div-by-0 question, especially 0/0?

Using the pattern, x/x = 1 for all x, why wouldn't 0/0 be 1? x/x, for every number, even those infinitely close to 0 (Lim x->0), yields 1, yet not 0, which leaves an undefined point on the graph.

The same argument is used for z^0 = 1(where z not equal 0). I use z here since I'm talking about a graph and want the exponent to be the x axis for clarification. The graph of z^x looks like it would go right through 1 (except for z=0) there for it must go through 1, right? Well, we know that, just because something seems like it should be, doesn't make it so, but let's go with it for the time being.

If that is the case, then let's consider all z for z^x from inf as it approaches 0. When z = inf, z^x = inf for all positive x and infinitely close to 0 for all negative x. Does it equal 1 when x = 0?

As z falls from inf and approaches 1, the exponential curve broadens until, at z=1, it becomes a perfectly horizontal line at y=1. As z continues down from 1 towards 0, the curve changes so that the negative exponents increase towards infinity, and the positives fall towards 0. When we are infinitely close to z=0, we have the mirror of z=inf, so what about when z = 0?

0^x represents the true impulse function. It is 0 for all x's except x=0 where it equals 1.

0^0 is 1, unless all other x^0 are also considered undefined. We have a choice to either use the empty product or not.

Now, going back to div-by-zero; 0 can divide 0 into 0 parts evenly. My friend Mal made the next logical leap: 0 can not divide 1 evenly into 0 parts, there for the answer is 0 remainder 1. Here's a few examples:

3/0 = 0 reaminder 3

2/0 = 0 reaminder 2

1/0 = 0 reaminder 1

0/0 = 0 reaminder 0 or 1 remainder 0

-1/0 = 0 reaminder -1

This is supported by the long division process:

_00_

0 ) 10

0

----

10

0

----

10

Let's follow the procedure in words:

The first question can be phrased a couple of ways: "How many times can 0 go into 1?", but also "How many parts can 0 divide 1 into evenly?"

One might think the answer is infinity, or 0, but 0 is the only correct answer, and here's why. Zero represents nothing. Even if you chopped 1 into an infinite number of pieces, there would still be pieces, not nothing, so the answer is simply, you can not divide 1 into 0 parts which means that the answer to put above the 1, is 0.

Doing the remainder for the first part, we subtract off 0*0, which leaves 1. Bring down the next most significant digit, the 0, and now we have to divide 10, in its entirety, by zero. Well, the same logic applies as before. 10 can not be divided by 0, so a 0 goes above the 0 also. That completes the quotient part of the process, and now we have to determine how we want to handle the remainder. The remainder is 10, so we can either put that into fraction form (10/0) or we can continue the division process into the decimal places, which would yield infinite 0's.

Let's look at it a different way. Mal described div-by-zero as removing pennies from a pile on a table. I would like to use the examples of pies. When you serve pie, you divide it up and put it onto plates to be served. If you have one pie, and 3 people want it, each plate will get 1/3 of the pie put on their plate. If two people wanted the pie divided between them, then 1/2 the pie would be removed from the pie pan and placed on 1 plate and the other 1/2 would be put on a second plate.

Now here is were an interesting concept get's introduced, the remainder. If only 1 person wanted 1/2 the pie, then 1/2 the pie would be taken out of the pie pan and placed on the plate, leaving 1/2 remaining in the pan. What ever is left in the pan after the pie is divided represents the remainder.

So next comes the idea that no one wants any pie. No pie will be removed from the pie pan and put onto 0 plates. The answer to 1 pie divided for zero servings is 0 servings with 1 whole pie remaining, thus 1/0 = 0 remainder 1. This applies to 2 pies, three pie and even x number of pies divided into 0 servings.

The debatable point to this pattern of x/0 = 0 remainder x, is that of the infamous 0/0. 0*0 = 0 therefore 0/0 = 0, but x/x = 1 therefore 0/0=1. Is one right, is neither right, or are both right? The rules of math must evenly apply to all aspects of math, ultimately, without exception. This then says that the rules of multiplication and division must be followed, and the statement x/x = 1 is a pattern, not a rule. This ultimately tells us that 0/0 = 0 remainder 0, and that x/x is an inverse impulse function. It is 1, everywhere except x=0.

But what does that remainder do for math? Well, let's look at another principle of math:

2/4 = 1/(4/2)

This is provable. I'm using easily divisible numbers so the proof to follow doesn't get huge:

2/4 = 1/4 = 0.5

and:

1/(4/2) = 1/(2/1) = 1/2 = 0.5

so what about 0/1?

0/1 = 1/(1/0) = 1/(0 remainder 1) = 0

This tells us that the remainder is meaningless. Perhaps this falls into the realm of imaginary numbers, or something very much like it.

So, if x/0 is now defined, then 0^x is defined for all x's, even negatives:

0^1 = 1 * 0 = 1*0 = 0

0^0 = 1 = 1 = 1

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

0^-2= 1 * 1/(0*0) = 1*(0+1/0) = 0

0^-3= 1 * 1/(0*0*0) = 1*(0+1/0) = 0

Demystifying π (pt2)

1 year ago

Using the pattern, x/y=1 for all x...

ReplyDeletewhat if we think about that as a slope... if we do that instead as {x,y} where x=y; then the slope of the line is 1 even at the origin {0,0}. It would seem unlikely that the "correct" mathematical solution would be a normal straight 45* line with an infinitesimally small hole at the origin.

0^x represents the true impulse function. It is 0 for all x's except x=0 where it equals 1.

ReplyDeleteIf we also think of that as a line; what we are really saying is {x,y} where x=0; ie: {0,y}

thus for y=0, it is the origin {0,0}

thus f(x,y) where x=0 would be the straight 90* line that crosses the origin... again, unlikely that the origin would be an infinitesimally small hole in the line.

This tells us that the remainder is meaningless....

ReplyDelete0/1 = 1/(1/0) = 1/(0 remainder 1)=0

This is exactly where continued fractions come in... you still have 1 left over so it is carried over into the next iteration