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I'm relearning mathematics by reading through What is Mathematics?. It begins by explaining the natural numbers, introducing the following property:

$a.0=0$

To make $a$ the subject, I would divide both sides by zero, leaving:

$a=\frac{0}{0}$

I'm aware, but unsure why, that division by zero isn't allowed. I also look at that equation and read it to mean that any natural number $a$ is given by $\frac{0}{0}$, which can't be right. Moreover, cutting a non-existant cake into no pieces makes my head spin.

Is it correct to say that the mistake is the division by zero, or is there more to this than meets the eye?

Jon
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    By dividing by zero, you in fact multiply both sides by $0^{-1}$. Unforuntately, $0^{-1}$ does not exist, as $0$ is not a unit in $\mathbb{R}$, i.e., there is no $y$ such that $0\cdot y=1$ (in fact, your book tells you $0\cdot y=0$ for all $y$). – mathse May 25 '14 at 07:25
  • there is a video on youtube by numberphile http://www.youtube.com/watch?v=BRRolKTlF6Q&feature=kp – S L May 25 '14 at 07:46
  • "I would divide both sides by zero, leaving:

    $a=\dfrac00$"

    This seems a bit informal. You should say if $a\cdot 0=0$ then $\dfrac00$ is a rational number which is the solution of $a0=0$. But since the fraction$\dfrac 00$(which is ought to be equivalent to division of 0 by 0) comes out to be indeterminate so it doesn't follow the law of determinateness. Similarly $\dfrac a0 $is a ration number, ought to be a solution of $\dfrac a0 0=a$,but the def of multi of rationals say $\dfrac a0 0=\dfrac{a0}{0}=\dfrac00 = $ indeterminate.We want a number system having division determinateness.

     – user103816 May 25 '14 at 08:13
  • @user31782 What is the law of determinateness? Never heard of this, also has no google hits (in a math context). Can you provide a link? – user144248 May 25 '14 at 09:03
  • @user144248 http://djm.cc/library/Number_System_of_Algebra_Fine_edited.pdf Article-9, 17 and 22 are related. Although it doesn't seem to be a law but since natural numbers follow this law(whatever), we ought to make a system of rationals also to follow this law, so that the structure of rules on fractions can be embedded into Naturals. Moreover the book is quite old so modern literature might not use this term. – user103816 May 25 '14 at 09:37
  • @user31782 Your previous comment is non-sense, to be honest. $\frac{0}{0}$ is not a rational number (you claim it is; only, you say, it may be indeterminate) because $\frac{1}{0}$ just doesn't exist. By the way, the link you sent says, in section 17, that the 'law of indeterminateness' does not hold when one of the numbers is $0$. – user144248 May 25 '14 at 09:56
  • @user144248 I agree that $\frac 00$ is not a rational number. What I am tryin to say is: Suppose one is defining rationals and their rules. One will start with, if $a=xb$ then $\frac ab$ is a rational number which should come out to be a solution of $a=xb$(we do not presume that $\frac ab$ is the solution of $a=xb$). Then one construct such type of rules of multi and division etc so that $\frac ab$ infact become the sol of $a=xb$. After defining the rules one[cont...] – user103816 May 25 '14 at 10:57
  • [...cont] would expect the symbol $\frac 00$ to be the solution of $0=x0$. Before recognizing that $\frac 00$ is indeterminate, one should expect that $\frac 00$ is a rational number which could give a determinate solution of $0=x0$. In this sense my previous comment make some sense. The OP, uptil the 3rd and 4th line of his question had not yet recognized that $\frac 00$ is indeterminate so he should say:"if $a⋅0=0$ then $\frac 00$ is a rational number which is the solution of $a0=0$" [cont...] – user103816 May 25 '14 at 10:58
  • [...cont]Moreover the OP says "...I would divide both sides by zero, leaving..." - Wrong! He would say:"Acc. to the definition of division" ( Or if he had been dealing with rations then: "Acc. to the definition of rationals") – user103816 May 25 '14 at 10:59
  • @user31782 [...] It's no problem to multiply both sides of an equation by $b^{-1}$ as long as $b$ is a unit. In $\mathbb{R}$ all elements are units except for $b=0$. (ed ajf) – user144248 May 25 '14 at 13:43
  • @user144248 I thought OP was talking of $\mathbb{N}$ – user103816 May 25 '14 at 13:52
  • @user31782 Same problem. He can still multiply both sides by $b^{-1}$ as long as $b^{-1}$ exists (it doesn't if $b=0$). If he wishes to multiply only by elements that are in $\mathbb{N}$, this only works when $b=1$. – user144248 May 25 '14 at 13:57
  • @user144248 since uptil $\S$1 "Calculation with integers" subpart "Law of arithmetic" chapter "Natural Numbers", we are working in $\mathbb{N}$ yet, division, fractions, their rules and structure is not defined so it is nonsense to talk about $\mathbb{R}$. Whatever I said in my previous comments were under the assumption that the OP is talking of article-1. – user103816 May 25 '14 at 14:16

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