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$100$ is a real number or we could call it a hyperreal number as every element of $\mathbb R$ is also an element of $\mathbb R^*$. If we add an infinitesimal say $\epsilon$ to $100$, the new number will be $100 + \epsilon$. We cannot give a numeric value to the number $100 + \epsilon$, because $\epsilon$ is not a real number or we can't ever suggest even an approximated value for $\epsilon$. The number $100 + \epsilon$ only tells how much close $100 + \epsilon$ is to $100$, but is not equal to $100$. Here, by 'numeric value' I mean that if $\epsilon$ would have some value like $\pi$ has a value, that is a non-terminating decimal. We cannot say that $100 + \epsilon$ is an infinitesimal, because in order for it to be an infinitesimal it must fulfill the condition $-a<\epsilon<a$ for all positive real numbers $a$. Can we call $100 + \epsilon$ a hyperreal number? I just want to confirm the name for such numbers, and share my understanding of infinitesimals, and want some suggestions if I understand them correctly.

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i think you should review the definitions, as this concept of "numeric value" is very problematic when considering hyperreals. i think goldblatt's Lectures On The Hyperreals is a very lucid exposition –  citedcorpse Jun 13 '13 at 18:06
    
Thanks for the suggestion! –  Samama Fahim Jun 13 '13 at 18:17
    
Your "numeric value" is the standard part (en.wikipedia.org/wiki/Standard_part_function)? –  Martín-Blas Pérez Pinilla Jan 24 at 10:09
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up vote 4 down vote accepted

Things get a little fuzzy when you say "numeric value". But I do think I understand what you are saying: we cannot "pin down" $\epsilon$ to a particular numeric real value, given the standard definition of the reals.

If $\epsilon$ is an infinitesimal, then yes, we call $100 + \epsilon$ a (non-real but) hyper-real number.

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Yes! by numeric value I mean the same what you understand.. like we can give an approximated value to $\pi$ to three decimal places $3.146$, and then add $100$ to it. We can say that $100 + \pi \approx 103.146$. But, we cannot do the same for $100 + \epsilon$. But I want to know why is it not appropriate to say 'numeric value'? –  Samama Fahim Jun 13 '13 at 18:15
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Simply because "numeric" is ambiguous, and is the adjective which corresponds to the noun "number", so one could say that $\epsilon$ does not have a real numeric value ($\iff$ is not a real number), but it does have a hyperreal/infinitesimal numeric value ($\iff$ it is an infinitesimal number). –  amWhy Jun 13 '13 at 18:21
    
Thanks for the clarification! I think I've misused the term 'numeric'. However, whatever be the hyperreal value of $\epsilon$ we cannot give it an approximation in real number system, because $\epsilon$ is not an element of $\mathbb R$, and thanks again. –  Samama Fahim Jun 13 '13 at 18:25
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Yes, and that's very well said, Samama. –  amWhy Jun 13 '13 at 18:27
    
Thank you amWhy.. :) –  Samama Fahim Jun 13 '13 at 18:31
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Every hyperreal number has a decimal expansion. You just have to think hard about what a decimal expansion means here.

What makes hyperreal numbers useful is not just that they contain infinitesimals, it is the fact that a transfer principle holds that allows us to translate correct statements about real numbers to statements about hyperreal numbers. Some care has to be taken about which statements one can transfer. Loosely speaking, we can transfer everything that does not talk about sets of real numbers. But if you want to actually work with hyperreals, you have to learn this material from a proper text.

For simplicity, we restrict ourselves to numbers between $0$ and $1$. Now what does it mean that every real number has a decimal expansion? It means that if we know all digits, we know the real number. Stated differently, two different real numbers can be told apart from their decimal expansion. Since decimal expansions are not unique, we have to take care in defining what it means that real numbers have decimal expansions. Here is a definition that works: A decimal expansion of $r$ specifies for each natural number $n$ in which of the intervals $[k 10^{-n},(k+1) 10^{-n}]$ for for $k=0,\ldots 9$ the number $r\mod 10^{-(n-1)}$ lies. That real numbers have decimal expansions means that we can recover $r$ from its decimal expansion. Now can we do this with hyperreal numbers? We have to find a statement that says we can find only one real number with a given decimal expansion that is written in a form that we can apply the transfer principle to:

If $x\neq y$ are real numbers, then there exists a natural number $n$ such that $|x\mod 10^{-(n-1)}-y\mod 10^{-(n-1)}|<10^{-n}$.

It should be noted that one can define $\mod 10^{-n}$ in terms of a statement that one can apply transfer to. Applying transfer, we get

If $x\neq y$ are hyperreal numbers, then there exists a hypernatural number $n$ such that $|x\mod 10^{-(n-1)}-y\mod 10^{-(n-1)}|<10^{-n}$.

The point is that hyperreals contain versions of natural numbers $n$ with $n$ infinitely large (relative to ordinary real numbers). If we specify the decimal expansion even in such places, we have specified the hyperreal number completely.

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