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I have looked through some of the previous questions posted on this topic, and I think mine is different. Is there a flaw in defining division by zero? For example, define $\frac{a}{0} = \infty_a$ it would seem like things work now, for example, $\frac{a/0}{b/0}=\frac{\infty_a}{\infty_b}=a/b$. What could go wrong with this idea, or more specifically, is it defined in some branch of mathematics? |
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Since $0+0=0$, we have $a=0\cdot \infty_a=(0+0)\cdot\infty_a=0\cdot\infty_a+0\cdot\infty_a=a+a$. So $a=a+a$, and therefore $a=0$. It certainly makes life simpler to have everything equal to $0$. |
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I'm not going to consider $a=0$ (i.e. the quotient $\frac{0}{0}$) as we can construct functions in calculus limit initially tends to $\frac{0}{0}$ after some work turns out to be any real number we'd like. It may be the case that allowing $\infty_0$ could be useful, for example we have $0 = 0 \cdot \infty_0$, but I don't consider it here. I noticed in André Nicolas' post, he showed that allowing division by zero ends up being trivial when you assume that the new elements you add obey the distributive law. So I conclude that if these new numbers are well-defined, then we can't use the distributive property with them. What follows are a few thoughts in that regard. Assume that $a \neq 0$ and define $\frac{a}{0} = \infty_a$ for some element $\infty_a$. Certainly $\infty_a$ is not a real number, so let us extend $\mathbb{R}$ to a new set $\mathbb{R}^{\dagger}$ which includes every $\infty_x$ for all $x \in \mathbb{R} \setminus \{0\}$. What properties will this expanded set $\mathbb{R}^{\dagger}$ have? We decided above that it should not have the distributive property. Well, we know that $\frac{a}{0} = \infty_a$ so perhaps $a = 0 \cdot \infty_a$. This strikes us as odd, because we know for any real number, multiplication by zero always yields zero. So we are at a crossroads. We can do one of two things: (1): We can say "the new set $\mathbb{R}^{\dagger}$ must follows the rules of multiplying by zero" in which case we would derive $a=0$, which would be a contradiction (remember, we assumed $a \neq 0$ in the beginning). If we enforced this restriction, we would find our new set of numbers paradoxical and then throw them out. (2): Allow this strange property of zero in this new set and accept all the consequences for its use. Here is one consequence of (2): Proposition: If $\mathbb{R}^{\dagger}$ is associative and commutative, then it contains only three elements. Proof: Let $a, b \in \mathbb{R} \setminus \{0\}$. Now since we assumed (2), we know $a=0 \cdot \infty_a$ and $b = 0 \cdot \infty_b$, so we consider the product $ab = (0 \cdot \infty_a) (0 \cdot \infty_b)$. We can write this product in two ways: $$(i): (0 \cdot \infty_a) (0 \cdot \infty_b) = (a \cdot 0) \cdot \infty_b = 0 \cdot \infty_b = b, $$ but on the other hand $$(ii): (0 \cdot \infty_a) (0 \cdot \infty_b) = \infty_a (b \cdot 0) = \infty_a \cdot 0 = a.$$ We conclude that $a=b$. So we have $\mathbb{R}^{\dagger} = \{0, a, \infty_a \}$. Perhaps we should not assume $\mathbb{R}^{\dagger}$ is not commutative or not associative or not both, then... |
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Presumably for non-zero $a$ you have $\frac{1}{\infty_a} = \frac{0}{a} = 0$, so this suggests for non-zero $a$ and $b$ that $\frac{1}{\infty_a} = \frac{1}{\infty_b}$ which in turn suggests ${\infty_a} = {\infty_b}$ and so perhaps $a=b$. More briefly $\frac{1}{\infty_a} = 0$ suggests $\infty_a = \frac{1}{0} = \infty_1$. I doubt you intend this, so at some stage the distinctions you are trying to create or the manipulations you hope to preserve are lost. |
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What you want to look at is called nonstandard analysis. You don't divide by zero but by infintesimals. See http://en.wikipedia.org/wiki/Non-standard_analysis |
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http://en.wikipedia.org/wiki/Surreal_number http://en.wikipedia.org/wiki/Hyperreal_number and maybe Non-Standard Analysis. The point is how to make a system that can have division by 0 defined without encountering any inconsistencies. But another question to ask is are the all the 0's the same? for example is 0 bananas same as 0 apples? what about limit $x$ going to 0 and $x^2$ going to zero ? are both the limits the same thing in every context? |
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