# The uniqueness of zero and other questions \[new title\]

**URL:** <https://boards.straightdope.com/t/the-uniqueness-of-zero-and-other-questions-new-title/209636>\
**Category:** Factual Questions\
**Created:** [October 26, 2003, 8:40am UTC](https://boards.straightdope.com/t/the-uniqueness-of-zero-and-other-questions-new-title/209636 "2003-10-26T08:40:34Z")\
**Posts on this page:** 2\
**Page:** 3

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**Author:** ![shaneomac](https://avatars.discourse-cdn.com/v4/letter/s/8c91f0/32.png) [@shaneomac](https://boards.straightdope.com/u/shaneomac)\
**Post date:** [October 26, 2003, 10:00pm UTC](https://boards.straightdope.com/t/the-uniqueness-of-zero-and-other-questions-new-title/209636/41 "2003-10-26T22:00:32Z")

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Here’s a “correct” statement on the uniqueness of zero:

There exists a number 0 in the set of real numbers such that for all real numbers x we have: 0_x=x_0=0. Assume we have another number in the set of real numbers, say that number is a, such that for all x in the set of real numbers a_x=x_a=0. Then we have 0_a=0 and we have 0_a=a therefore 0=0\*a=a and 0=a hence 0 is unique.

For why you can’t divide by zero:  
For all x,y,z in the set of real numbers we have x(y+z)=xy +xz. Now, assume for the sake of contradiction that 0/0 in the set of real numbers exists, let a denote this number.  
Consider 1_0=1_(0+0)=1_0+1_0. then we have 1\*(0/0)=1\*(0/0) + 1\*(0/0) or 1_a=1_a + 1\*a since a is a real number we have a/a=1 so 1=1 + 1 which can not be. Hence there does not exist a in the set of real numbers such that a=0/0.

On a side note showing that -2 has not real square root is easy but showing that the square root of 2 is a real number is not so easy. It is not sufficient to state that the square of the square root of 2 is real thus the square root of two must be real, because certainly -2 is real and it’s square root is not. If you wanted to rigorously show that every positive real number has a real square root you actually need a rather involved proof. I don’t really remember exactly how it would be shown but you should be able to show it using an infinite series of rational numbers Xm/Yn with m,n real numbers, and you want to construct the series in such a manner that the limit is the square root of two. You actually don’t need the concept of a limit to show it you only need something a little weaker, the concept that every set bounded from has a least upper bound.  
But as I said I don’t remember it exactly, analysis is not my strong point in math. If anybody else actually knows how it’s shown I would like to see it if you have the time.

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**Author:** ![Shade](https://avatars.discourse-cdn.com/v4/letter/s/2bfe46/32.png) [@Shade](https://boards.straightdope.com/u/Shade)\
**Post date:** [October 26, 2003, 11:05pm UTC](https://boards.straightdope.com/t/the-uniqueness-of-zero-and-other-questions-new-title/209636/42 "2003-10-26T23:05:05Z")

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Let r\>0 be a rational number.  
Let S={x in R | x[sup]2[/sup]\<r}  
S non-empty and has an u.b. max(1,r).  
Hence S has l.u.b. s.  
Then exist real numbers in S arbitrarily close to s (else exists lower u.b.), so can’t have s[sup]2[/sup]\>r.  
Suppose s[sup]2[/sup]\<r.  
Consider (s+e)[sup]2[/sup]=s[sup]2[/sup]+e[sup]2[/sup]+2se  
Choose e\<\<(r-s[sup]2[/sup])/s, and have (s+e)[sup]2[/sup]\<r.  
Hence (s+e) in S; contradition.  
Hence s[sup]2[/sup] in fact equals r.

That’s not quite rigorous, but does it make you happy? I can hopefully fill it in with epsilons and deltas if you’d like.

Of course, there are other ways to define R, some of which iirc give roots automatically.

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