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Reading about hyperreals I learned that a serious problem with such systems is their undefinability.

So I tried to construct a definable system of hyperreals by introducing an infinitismal element $\varepsilon$ such that

$$0^\varepsilon=\frac{1}e$$ by definition.

It seems to me that such system would inevitably include divisors of zero but I am not sure.

I wonder whether such system viable and whether it in fact includes divisors of zero.

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Have you seen this? –  Ian Mateus Oct 30 '12 at 23:08
    
@Ian Mateus yes this is exactly what I saw before making this question. –  Anixx Oct 30 '12 at 23:10

2 Answers 2

I would like to challenge a premise of the original question, namely the assumption that the hyperreal number system is "undefinable". Most researchers in the field expected this to be the case, but Kanovei and Shelah proved in 2003 the existence of a definable model (see http://arxiv.org/abs/math/0311165).

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The real numbers $\mathbb R$ is(usually thought of as) a field which means it's got operations 0,1,+,-,*,/ with some conditions (associativity, distributivity, etc.)

But you can also view $\mathbb R$ as a ring by throwing away division: that means it's just got operations +,-,*.

When you do this, it's possible to add an element $e$ satisfying $e^2 = 0$ (so $e$ is a zero divisor) to get a new ring $\mathbb R[e]/(e^2).$

Inside this ring you can define $D(f) = f(x+e) - f(x)$ and find e.g. $D(x^2) = x^2 + 2xe + e^2 - x^2 = 2xe$ which is the derivative of $f$ times $e$.

The problem is this isn't a field so you can't do division.. it's much harder to build a field with infinitesimals in it but it can be done using techniques from logic called nonstandard arithmetic.

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It's actually quite easy to build a hyperreal field. Simply take an ultraproduct of $\mathbb R$, say by the natural numbers, and show that any strictly increasing function from the natural numbers to $\mathbb R$ is not the image of a real number in the canonical embedding; and therefore has to be an infinite number. –  Asaf Karagila Oct 30 '12 at 21:56
    
@Asaf, To be sure, one doesn't need ultraproducts if all one wants is an elementary extension. One can for example start with Skolem's model of the integers which is completely constructive, and pass to the field of fractions. –  user72694 Feb 11 at 18:46

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