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I know there are equations for odd numbers . But is there an equation for positive integers.

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closed as not a real question by Antonio Vargas, DonAntonio, JSchlather, Ittay Weiss, Stefan Hansen Mar 18 '13 at 6:26

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Is this a real question? If so, can you be more specific? What equations for odd numbers are you thinking of? It will be hard to make sense of this. –  1015 Mar 18 '13 at 3:12
    
Well, you must be looking for $|x|$ ? Where $x$ is an integer.? –  Inceptio Mar 18 '13 at 3:13
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Do you mean odd positve $x$: $x = 2n - 1$ for some $n \in \mathbb Z, n\geq 1$? If so, then even positive $y: y = 2n$ for some $n\in \mathbb Z n\geq 1$. –  amWhy Mar 18 '13 at 3:15
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I doubt this is homework... –  L. F. Mar 18 '13 at 3:16
    
You need to define what structure you are working in. In a ring or field like $\mathbb Z_7$ you can't define the positive integers at all. You need an ordered (ring, field) or whatever. –  Ross Millikan Mar 18 '13 at 3:35

3 Answers 3

up vote 3 down vote accepted

I'll answer based on my comment.

We can use the following notation to denote a positive odd integer:

  • "$x$ is an odd positive integer $\iff x=2n−1\,$ for some $\,n\in \mathbb Z,n\geq 1$.

We can use the following notation to denote a positive even integer so,

  • "$y$ is a positve even integer" $\iff y = 2n\,$ for some $\,n \in \mathbb Z,\,\,n\geq 1$

These notations reflect the definitions of odd positive integers, and even positive integers, respectively:

If we change the notation to exclude the qualifications $n\geq 1$ (so that $n$ is some integer), then we have defined odd and even integers, respectively, though when defining odd integers, in general, it is more common to define them as the set of all integers $x$ that can be expressed in the form $\;x = 2n +1$ for some $n \in \mathbb Z$.

Essentially, the notation reflects that an even integer, by definition, is an integer $x$ that is divisible by $2$, and odd integer is not.

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An integer $m$ is odd $\iff$ there exists an integer $n$ such that $m=2n+1$. I don't think there is something similar to this for positive integers. Well, this is slightly more advanced, but an integer $n$ is nonnegative (positive or zero) $\iff$ there exist integers $a, b, c$, and $d$ so that $n=a^{2}+b^{2}+c^{2}+d^{2}$, and $n$ is positive $\iff$ not all four of them are zero. This result is Lagrange's four square theorem, but I imagine it wasn't what you're looking for.

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So, every $n$ can be represented in sum of $4$ squares? –  Inceptio Mar 18 '13 at 3:18
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@Inceptio: every positive $n$ if the squares are allowed to be the same and include $0$ –  Ross Millikan Mar 18 '13 at 3:19
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@RossMillikan: Oh well! That was an Olympiad question in our country. :) –  Inceptio Mar 18 '13 at 3:24
    
@Inceptio, I don't have much Olympiad experience, but that's impressive considering that result is very nontrivial. –  Dylan Yott Mar 18 '13 at 3:26
    
Maybe you were expected to know the four square theorem and quote it for this problem, but not prove it. –  Ross Millikan Mar 18 '13 at 3:28

You can just say $x > 0$. As far as I know there isn't an algebraic statement that enforces positiveness throughout a calculation in quite the same manner that $2x$ enforces evenness.

Anyways, be careful not to confuse math notation for math.

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However,x can equal 0.01 which is not a positive integer. I want the equation for the set (1,2,3,4,5,6,7,8). Thanks anyway. –  yes man Mar 18 '13 at 4:28
    
@yesman in that case you just say $x \in \mathbb N$ or "x is a positive integer" or "x is an integer between 1 and 8 inclusive" etc. again there isn't any algebraic statement that enforces integer-ness throughout a calculation –  amr Mar 18 '13 at 5:22

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