Difference between revisions of "Talk:Elementary proof"
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| + | : Existence depends on what system you are operating in. Uniqueness is fairly easy if you are operating in a field. Proposition: in any field F, with c and element of F, the equation <math>x^2=c</math> has at most 2 solutions. Proof: Consider <math>x^2=c</math>. this implies that <math>x^2-c=0</math>. We may assume that c has at least one square root (call it <math>c^{1/2}</math>). So we have <math>(x+c^{1/2})(x-c^{1/2})</math>. Now, since fields have no zero divisors (easy excercise), we must have <math>x+c^{1/2}=0</math> or <math>x-c^{1/2}=0</math> which gives us only two choices. Q.E.D. [[User:JoshuaZ|JoshuaZ]] 20:44, 6 February 2007 (EST) | ||
Revision as of 01:44, February 7, 2007
Who uses this concept?
I don't think that it is correct that anyone calls a proof "elementary" just because it does not use the square root of (-1). Nor is it true that an elementary proof is necessarily preferred.
It is not an assumption that (-1) can have a square root. It can be proved. Nor is the square root unique. (-1) has 2 square roots, if it has any.
I suggest killing this article. It just isn't useful. RSchlafly 00:11, 5 February 2007 (EST)
If the test were what is "useful", then most math articles should be deleted!
I'd love to see a proof that the following exists and is unique (plus and minus roots):
--Aschlafly 20:12, 6 February 2007 (EST)
- Existence depends on what system you are operating in. Uniqueness is fairly easy if you are operating in a field. Proposition: in any field F, with c and element of F, the equation
has at most 2 solutions. Proof: Consider . this implies that 
. We may assume that c has at least one square root (call it 
). So we have 
. Now, since fields have no zero divisors (easy excercise), we must have 
or 
which gives us only two choices. Q.E.D. JoshuaZ 20:44, 6 February 2007 (EST)
