Treadstone Posted October 8, 2004 Posted October 8, 2004 123Rock's biggest problem is that you cant look at a rational number, ie 1/0 as having two independent quantities of 1 and 0 and then trying to compare that to 2/0 or 2 and 0. Rational numbers are used to exactly express a quotiant (sp?) of a real number. 1/0 has no value because (useing the def'n of divisor) there is no q such that 1 = q*0, and you cant compare something that lacks a value. Any number divided by zero does no exist because there isnt any number, a, such that 0|a so your comparing nothing when you write '1/0=2/0'. Hence 1/0=2/0 is meaningless edit - also you need to justify several of the algebraic steps in your proof
123rock Posted October 9, 2004 Author Posted October 9, 2004 I never said that you have to evaluate 1/0, or x/0. While still a fraction, x/0 applies to the same laws of arithmetic as all other fractions. 0/0=x and x * 0=0 are not the same equations because multiplying each side by zero in 0/0 (0)=x *0 implies that 0/0 is 1. If 0/0=0/0, which would be the only way that it can be undefined, otherwise 0/0 can equal whatever we want it to equal, then 1-1/0=0/0
ydoaPs Posted October 9, 2004 Posted October 9, 2004 I never said that you have to evaluate 1/0' date=' or x/0. While still a fraction, x/0 applies to the same laws of arithmetic as all other fractions. 0/0=x and x * 0=0 are not the same equations because multiplying each side by zero in 0/0 (0)=x *0 implies that 0/0 is 1. If 0/0=0/0, which would be the only way that it can be undefined, otherwise 0/0 can equal whatever we want it to equal, then 1-1/0=0/0[/quote'] you could put 1, 7, or 43 billion in for x, and it would still be undefined.
Treadstone Posted October 11, 2004 Posted October 11, 2004 fraction, x/0 applies to the same laws of arithmetic as all other fractions. you cannot use arithmetic laws on something that does not exist. x/0 is not a fraction because fractions have a quantity and can therefore be compared. Do you see what i'm saying? It has no value because it does not exist, there is no quantity so even though you can write in a fraction form it is still not a fraction. Saying x/0 is a fraction means that you are expressing some number, a, whereby 0*x=a. Remember, fractions are rational numbers which means the denominator and the numerator have a ratio equating them. Look over divisior rules as they apply number theory and the def'n of rational numbers....or just look at my previous post.
Guest jaydeschizo Posted October 11, 2004 Posted October 11, 2004 err... afaik using normal algabraic(sp?) mathematics you cant divide by 0 in any case, and any way you made a mess up in your example: a = b a^2 = b^2 a^2 - b^2 = ab - b^2 (a+b)(a-b) = b(a-b) here you divide both sides by a-b, but if a=b a-b is always 0, thus you say that 0/0 = 1 because you: (a+b)(a-b)/(a-b)=b(a-b) => (a+b)*1 = b*1 a+b=b b+b=b 2b=b 2=1; so to cut the cake: to prove that 0/0 doesnt exist you asume that 0/0 = 1 thats useless, in calulating with 0 u can only use limit cases... lim 1/0 = infinite... lim 0/0 = 1 is afaik unproven but most logical... but ill have to ask my teachers to be sure... maybe this has already been said, but my scientific english is a disaster anyway
Treadstone Posted October 12, 2004 Posted October 12, 2004 i think its easier to prove x/0 doesnt exist using number theory, simply put 1. x/0 => there exists q, an element of R, such that x = q*0, x!=0 2. since there is no such q x/0 does not exist 3. for x=0, 0/0 is trivial for 123's example and indeed for anything else
123rock Posted October 13, 2004 Author Posted October 13, 2004 i think its easier to prove x/0 doesnt exist using number theory' date=' simply put 1. x/0 => there exists q, an element of R, such that x = q*0, x!=0 2. since there is no such q x/0 does not exist 3. for x=0, 0/0 is trivial for 123's example and indeed for anything else[/quote'] if x/0=q, x*q=0 is not a valid reevaluation, because you are multiplying each side by 0, and thus assuming that 0/0 is 1.
123rock Posted October 13, 2004 Author Posted October 13, 2004 err... afaik using normal algabraic(sp?) mathematics you cant divide by 0 in any case' date=' and any way you made a mess up in your example:a = b a^2 = b^2 a^2 - b^2 = ab - b^2 (a+b)(a-b) = b(a-b) here you divide both sides by a-b, but if a=b a-b is always 0, thus you say that 0/0 = 1 because you: (a+b)(a-b)/(a-b)=b(a-b) => (a+b)*1 = b*1 a+b=b b+b=b 2b=b 2=1; so to cut the cake: to prove that 0/0 doesnt exist you asume that 0/0 = 1 thats useless, in calulating with 0 u can only use limit cases... lim 1/0 = infinite... lim 0/0 = 1 is afaik unproven but most logical... but ill have to ask my teachers to be sure... maybe this has already been said, but my scientific english is a disaster anyway[/quote'] That's only true if you evaluate x/0.
123rock Posted October 13, 2004 Author Posted October 13, 2004 you cannot use arithmetic laws on something that does not exist. x/0 is not a fraction because fractions have a quantity and can therefore be compared. Do you see what i'm saying? It has no value because it does not exist' date=' there is no quantity so even though you can write in a fraction form it is still not a fraction. Saying x/0 is a fraction means that you are expressing some number, a, whereby 0*x=a. Remember, fractions are rational numbers which means the denominator and the numerator have a ratio equating them. Look over divisior rules as they apply number theory and the def'n of rational numbers....or just look at my previous post.[/quote'] Everything has to be proven, except of course axioms. It was proven by arithmetic laws that x/0 does not exist, but these same arithmetic laws cannot be used when we are trying to get a different fraction? is 1-1/0 not equal to 0/0? Yes, because 1-1=0. 0/0(2/2)=0/0; 0/0(x)=0/0 Again, if arithmetic laws could not be used on 0/0, then 0/0 does not equal 0/0, which is not something that the axiom x=x accepts, or 1=1
Treadstone Posted October 13, 2004 Posted October 13, 2004 if x/0=q, x*q=0 is not a valid reevaluation, because you are multiplying each side by 0, and thus assuming that 0/0 is 1. thats not what i wrote, look again...i am useing number theory, the def'n of divisor in specific....i proved that x/0 does not exist and because it does not exist it cannot be compared to other expressions nor is it a fraction.
Dave Posted October 13, 2004 Posted October 13, 2004 is 1-1/0 not equal to 0/0? Yes, because 1-1=0. 0/0(2/2)=0/0; 0/0(x)=0/0 The initial statement is non-logical since 0/0 is undefined. How are you supposed to manipulate something which has not been defined?
Treadstone Posted October 13, 2004 Posted October 13, 2004 The initial statement is non-logical since 0/0 is undefined. How are you supposed to manipulate something which has not been defined? thats what i am also asking
ydoaPs Posted October 13, 2004 Posted October 13, 2004 err... afaik using normal algabraic(sp?) mathematics you cant divide by 0 in any case' date=' and any way you made a mess up in your example:a = b a^2 = b^2 a^2 - b^2 = ab - b^2 (a+b)(a-b) = b(a-b) here you divide both sides by a-b, but if a=b a-b is always 0, thus you say that 0/0 = 1 because you: (a+b)(a-b)/(a-b)=b(a-b) => (a+b)*1 = b*1 a+b=b b+b=b 2b=b 2=1; so to cut the cake: to prove that 0/0 doesnt exist you asume that 0/0 = 1 thats useless, in calulating with 0 u can only use limit cases... lim 1/0 = infinite... lim 0/0 = 1 is afaik unproven but most logical... but ill have to ask my teachers to be sure... maybe this has already been said, but my scientific english is a disaster anyway[/quote'] your "proof" is over before it starts if a=b, then [math]a^2-b^2=0[/math]
Treadstone Posted October 13, 2004 Posted October 13, 2004 your "proof" is over before it starts if a=b' date=' then [math']a^2-b^2=0[/math] this proof was 123's...his first post
Treadstone Posted October 14, 2004 Posted October 14, 2004 that doesn't make it any less invalid right, i also drew attention to it in my first post
Dave Posted October 14, 2004 Posted October 14, 2004 It's a trick proof anyway; nobody eas really going to take that seriously.
123rock Posted October 24, 2004 Author Posted October 24, 2004 your "proof" is over before it starts if a=b' date=' then [math']a^2-b^2=0[/math] That's the disproof that 0/0=1. If 0/0 doesn't equal 1, then 0/0(x) does not equal x, which is true for all numbers except for x=0. I never said that was a proof. Why not take me seriously Dave? The simple arithmetic is right there.
123rock Posted October 24, 2004 Author Posted October 24, 2004 The initial statement is non-logical since 0/0 is undefined. How are you supposed to manipulate something which has not been defined? Something has to be proven to be undefined before you can say it is. Besides, it's not a relatively hard question to imagine, of how many times 0 fits into 0, or how many monkeys can put zero bananas into zero baskets. I guess in that case it's undefined, but tautology is not mathematics, and before you can state that something is undefined solely based on a mathematical dictionary definition, you have to show it. Edit: Also, if you state that 0/0=0 and no other number, it leads to absolutely no contradictions whatsoever
matt grime Posted October 24, 2004 Posted October 24, 2004 That all depends on what you are trying to do by claiming a definition of 0/0. NB thinking about 0 fitting into 0 some number of times is not mathematics, and has no place in the discussion at all. Simply there is no way to define 0/0 in side any field, or ring, that is consistent with the axioms of a field, where consistent has the obvious meaning.
ed84c Posted October 24, 2004 Posted October 24, 2004 Ive spent many hours concidering this one before now and I have deicded it was 1
ed84c Posted October 24, 2004 Posted October 24, 2004 well it seems like a compromise between 0 and inft. and i figured 0 fits into 0 1 time.
AL Posted October 24, 2004 Posted October 24, 2004 Something has to be proven to be undefined before you can say it is. Besides, it's not a relatively hard question to imagine, of how many times 0 fits into 0, or how many monkeys can put zero bananas into zero baskets. I guess in that case it's undefined, but tautology is not mathematics, and before you can state that something is undefined solely based on a mathematical dictionary definition, you have to show it. Definitions are arbitrary. You do not "prove" definitions, nor do you "prove" something that has been arbitrarily designated as undefined. The only requirement in math is that a definition be "well-defined," meaning it doesn't lead to some logical inconsistency or absurdity. 0/0 cannot be well-defined, so it is left as undefined. Edit: Also, if you state that 0/0=0 and no other number, it leads to absolutely no contradictions whatsoever It leads to all sorts of contradictions. Here, have fun with this one: Let 0/0 = 0. Then: -infinity = ln(0) = ln(0/0) = ln(0) - ln(0) = -infinity - -infinity = infinity - infinity
albertlee Posted October 24, 2004 Posted October 24, 2004 Another simple example that something divide by 0 is not logical 3*0 = 0 if you devide both side by 0, what do you get???? 3 = 0/0 Aha does not make sense!!!
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now