Nullity?

MathsIsFun · 2006-12-07 10:25:49

A way to divide by zero?

http://www.bbc.co.uk/berkshire/content/ … ture.shtml

I haven't watched the RealAudio presentation, so I cannot comment (except that "nullity" already means something: http://en.wikipedia.org/wiki/Nullity)

numen · 2006-12-07 11:17:06

That symbol, uppercase phi, is the reciprocal of the golden ratio. I think he missed something there

mathsyperson · 2006-12-07 11:59:57

Ah, he can probably get away with that. Φ is also used in statistics as something to do with the normal distribution, so giving it another use again isn't hugely controversial.

It doesn't seem like he's done anything too profound though. All he's done, really, is define 0/0 as being this 'nullity'. You don't need to do anything to make up a definition, and a definition by itself isn't really that impressive.

The way that he showed that 0/0 = 0^0 was quite interesting though.

Incidentally, couldn't all those plane crashes and pacemaker failures be solved by just changing the formula they use a bit?

In Excel-speak, =1/x would become =IF(x=0,0,1/x). Considering the effect of that kind of failure and the apparent simplicity of the solution, it's surprising that the plane crashes ever happened, really. Maybe I'm missing something.

MathsIsFun · 2006-12-07 12:57:08

Programming-wise, you should anticipate and handle all possible conditions. So your Excel example is good. In fact Excel itself "handles" a divide by zero, by returning "#DIV/0!" possibly a better name than "Nullity".

I still haven't got around to installing RealPlayer to hear what the actual idea is, though.

mikau · 2006-12-07 13:19:56

I must say his theory is total idiocy.

Its about as good as 5*0 = 0 so 5 = 0/0

For those who can't watch the video, his proof is as follows:

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

Pfft! he merely proves that one undefined number is equal to another undefined number. He also assumes 0^-1 = 1/0. Proove that!
He said "any number to the negative 1 is its reciprocal". Actually, thats not true. Any number but zero. He might as well say that any number to the power of zero is 1 thus 0^0 = 1.

He also states that 1/0 is EQUAL to infinity. That is also not true. The fraction approaches infinity as x approaches zero, but we all know, the limit of f(x) as x approaches a is not necessarily equal to f(a).

He just uses clever (actually not so clever) algebra and proves his "theroem" by treating zero like a nonzero number. Its about as valid as that 1 = 2 proof.

Last edited by mikau (2006-12-07 13:26:56)

mikau · 2006-12-07 13:49:34

Proof he is incorrect using his own methods.
He assumes that the rules exponants apply in that addition and subtraction can be used for multiplication and division.

He also said that any number to the power of 1 is itself.

This is the first part of his theorem and I'll branch off from there:

0^0 = 0^(1 1) since 0 = 1 - 1

0^0 = 0^1 * 0^-1 (according to him)

0^0 = 0^1 * 0^-1 any number to the power of one is itself (also his words)

so 0^0 = 0 * 0^-1.

Now here's where my twist comes in. 0*0 = 0 in otherwords 0^2 = 0, making this substitution:

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

thus 0^0 = 0^(2-1) = 0^1 = 0.

Thus 0^0 is in fact equal to 0 and therefore this "NEW" number does not exist.

I did this using only properties that he used. EXCEPT I said 0*0 = 0 and made the substitution.

MathsIsFun · 2006-12-07 13:56:59

And of course 0^0 is "indeterminate". You could also claim that any number to the power 0 is 1 (as I note at the bottom of this page Laws of Exponents)

I am thinking now that this poor guy might get a roasting.

mikau · 2006-12-07 14:01:46

Aye! Though it appears he's seeking to or thinks he proved it is not indeterminate.

What do you think of my proof? I'm trying to use only the same methods he used to show that his process is flawed by using the same reasoning to derive alternate answers.

Zhylliolom · 2006-12-07 15:35:02

Hey guys look

Then

Therefore

It really works!!!!!!11!1!1cos² x + sin² x!1!!1!!1!

*most horrific disemboweling of a theory ever witnessed by mankind removed for the sake of the young ones on the forum*

Just read one of his papers and shake your head, I think it'll be enough for you.

http://www.bookofparagon.com/Mathematic … ineVII.pdf

mikau · 2006-12-07 15:44:58

I agree.

Whats worse is they're already teaching this fallacy to kids.

Zhylliolom · 2006-12-07 16:41:02

Yes, what upset me most was not his treatment of my faithful wife mathematics, but how he rushed to teach it to impressionable youngsters, who may have just had their lives ruined.

Toast · 2006-12-07 19:25:25

Interesting how he gets from

to , but aren't they basically just the same indeterminate expressions?
And yes, he really shouldn't be teaching this to youngsters without the wider maths community accepting it first. Their lives could indeed be ruined forever...

Last edited by Toast (2006-12-07 19:25:43)

Patrick · 2006-12-08 07:18:47

Oh my g.. I'm never going to believe in anything without prooving it myself first from now on

Anyway, Dr Anderson.. DR?! How in the world can a person with that degree fail so hard at understanding 0 and infinity?

edit: Okay, this doctor degree seems to be different from the danish doctor degree. In Denmark, it's the highest academic degree possible to obtain, but apparently this is the same as what's called "licentiat" in Danish(aka P.hD). Still, teaching this to children without the aproval of the math community is a very bad career choice in my opinion.

Last edited by Patrick (2006-12-08 07:45:00)

simron · 2006-12-08 14:19:10

Nullity is null and void.

Ricky · 2006-12-08 15:06:55

As soon as you talk about division by 0, you are assuming that 0 has a multiplicative inverse. That's all division really is, multiplying by the multiplicative inverse. And as soon as you assume that such an inverse exist, you lose the fact that the reals are a field. Being a field gives the reals all those nice properties.

But it's worse than that. The rational numbers are defined as an expansion of the integers where:

a / b = c /d if and only if ad = bc.

Sub in 0/0 for such a number and you get 0*d = 0*c. Now if we accept the integers are a ring, we know that x(y+z) = xy + xz. But this also means that 0 = xw - xw = x(w - w) = x(0). Thus, 0*d = 0*c = 0 for all c and d. So 0/0 is equal to any and all rational numbers.

What does this mean? Well, it's quite simple. 0/0 + a = any and all rational numbers. Ack! Addition is no longer a binary operation since it isn't 1-1. Nor is multiplication.

Congratulations. A want to solve a non-existent problem destroyed all math as we know it.

Sheesh!

Math Is Fun Forum

#1 2006-12-07 10:25:49

Nullity?

#2 2006-12-07 11:17:06

Re: Nullity?

#3 2006-12-07 11:59:57

Re: Nullity?

#4 2006-12-07 12:57:08

Re: Nullity?

#5 2006-12-07 13:19:56

Re: Nullity?

#6 2006-12-07 13:49:34

Re: Nullity?

#7 2006-12-07 13:56:59

Re: Nullity?

#8 2006-12-07 14:01:46

Re: Nullity?

#9 2006-12-07 15:35:02

Re: Nullity?

#10 2006-12-07 15:44:58

Re: Nullity?

#11 2006-12-07 16:41:02

Re: Nullity?

#12 2006-12-07 19:25:25

Re: Nullity?

#13 2006-12-08 07:18:47

Re: Nullity?

#14 2006-12-08 14:19:10

Re: Nullity?

#15 2006-12-08 15:06:55

Re: Nullity?

Board footer