http://www.bbc.co.uk/berkshire/content/art...o_feature.shtml I personally think it's nifty and can work. :s

But isnt anything divided by zero 0 ? i mean lets get down to the basics of math 5x4 ----- 5 ----- 5 ----- 5 ----- 5 add em up its 20 5 x 0 there are 0 5s so there are no dashes so its 0

No, we've always been taught that 0 is an infinite number, and you can't really define an infinite number now, can you?

That actually pretty sexy, I like it. question: Will it also be compatible in the real world rather than just the digital world?

I think I heart about this. So what happens when nullity is divided by 0? Again nullity? And what about nullity or something else divided by nullity?

umm well lets find limit as x=>0 for (x-1) / [(e^x) - 1] in this scenario, you get a situation where you must find a limit of a number which approaches 0/0 using L'Hopitols rule, you can take the derivative of both the numerator and the denominator to get => 1/ e^x =1 basically 0/0 can equal anything =================== now..... for something like |1/x| as x approaches 0 the answer is infinity(graph it) for -|1/x| as x approaches 0 the answer is negative infinity(graph it) for 1/x as x approaches 0 the value does not exist(graph it) ==================== if a problem is simply defined as an equation /0 then this si basically what has been said for the past 1200 years, it can equal anything rational irrational or imaginary. ============ ============ ============ I just went over the videos and actually read the article. basically his argument is flawed in this simple way. he states 0^0=0^1*0^-1 if that's true then 0^0=0^2*0^-2 and that's where it falls apart (0/1)^2*(0/1)^ would = nullity^2 based off of substitution from his method where (0/1)*(0/1)^-1=nullity so by his theory nullity would have to be a number which exists such that nullity^n=nullity where n is any number n. the only number which holds true for this would be 1 and as such nullity would then have to equal 1. Or else be undefined... which is right back where we were but now with a new symbol. furthermore 1/0 != infinity -1/0 != -infinity the limit of 1/|x| as x approaches 0 is infinity the limit of -1/|x| as x approaches 0 is -infinity but 1/0 is still undefined.

We learned these problems in Calc 2 close to the beginning of the school year...some of those comments are great

I messed with my teacher on a math test with this, for a problem I didn't understand... She thought it was pretty funny.

>_> I'm only in algebra 2 so thinking logically 0 is a complete absence of value, so dividing by zero is impossible because 0 does not exist, being used as a placeholder is the same as putting a smiley face there and saying it is nothing, looking at a problem one divided by "absence of value" would not be a valid equation because the second variable isn't present, thats like dividing 3 by pencil, because its not a numerical value it wouldn't have a proper solution, I didn't bother to try to understand some of the previous posts but technically speaking why does x^0=1, being any number multiplied by itself 0 times would yield nothing because the number technically doesnt exist due to the exponent being a absence of value, after everything its used for a 0 still comes down to being a place holder, as an example 1,000 could be written in scientific notation but it would continue in an endless cycle because of the 10^x.. in short 0 sucks >0