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Jedo the Jedi
Paragon in Training
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 Posted: Mon Oct 15, 2012 4:51 pm Post subject: 1 |
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I was pretty good at algebra and trigonometry in school, but I never took an advanced class involving proofs.
What's the proof that (-1)(-1) = 1? It's probably super simple...
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Lepton*
Guest
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| Posted: Mon Oct 15, 2012 6:08 pm Post subject: 2 |
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I was going to post about this a couple weeks ago, as it seems like an interesting problem.
One way to show it would be to write -1 = e^(i pi), using the Argand diagram.
Or...
-1 * -1 = -1 * -1 +1 -1 = -1 * ( -1 +1 ) +1 = -1 * 0 +1 = 1 |
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Nsof
Daedalian Member
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| Posted: Mon Oct 15, 2012 7:26 pm Post subject: 3 |
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01. 0=0*0. **
02. 1=1+0. 0 is the additive identity
03. 1=1+0*0. 01, 02
04. 1+(-1)=0. 1, -1 are negatives
05. (-1)+1=0. 04, Commutativity
06. 1=1+(1+(-1))*((-1)+1). 03, 04, 05
07. 1=1+1*(-1)+1*1+(-1)*(-1)+(-1)*1. 06, Distributivity (three times)
08. 1=[1+(-1)]+[1+(-1)]+(-1)*(-1). Commutativity, Associativity,
09. 1=0+0+(-1)*(-1). 04 twice
10. 1=(-1)*(-1). 02 (twice and it should be commuted)
** requires separate proof 
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Nsof
Daedalian Member
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| Posted: Mon Oct 15, 2012 7:36 pm Post subject: 4 |
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Lepton's proof is shorter. (I almost wrote "lepton's is shorter" but decided against )
That was a simultaneous post.
One that began before my daughter going to sleep and ended after.
Its a process far more complicated than proving math.
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Jedo the Jedi
Paragon in Training
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| Posted: Mon Oct 15, 2012 8:38 pm Post subject: 5 |
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Lepton's is shorter, but yours shows the steps I wasn't making in my head.
So basically, when you want to substitute things in proofs, you just have to show how the thing you are substituting out equals what you are subbing in...and you have to have separate proofs for those. Boy is math complicated.
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Nsof
Daedalian Member
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| Posted: Mon Oct 15, 2012 11:07 pm Post subject: 6 |
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How about: 1>0
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Thok
Oh, foe, the cursed teeth!
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| Posted: Mon Oct 15, 2012 11:25 pm Post subject: 7 |
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| Nsof wrote: |
| How about: 1>0 |
What are your axioms for <?
Assuming the standard:
Trichotomy: either x<y, or x=y, or x>y
and
Multiplication:
If a>0, and x<y, ax<ay. As a corollary, if a<0 and x<y, ax>ay,
Then we have:
Either -1<0 or -1>0. If -1<0, then (-1)*(-1)>(-1)*0, so 1>0. If -1>0, then (-1)*(-1)>(-1)*0, so 1>0. |
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Nsof
Daedalian Member
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| Posted: Wed Oct 17, 2012 4:59 am Post subject: 8 |
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| Quote: |
| Either -1<0 or -1>0. If -1<0, |
Or 1=0. Its not hard to prove that 1!=0 but it requires a proof.
| Quote: |
if a ≤ b then a + c ≤ b + c
if 0 ≤ a and 0 ≤ b then 0 ≤ a b
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Yes.
However, I am not sure the corollary is part of the field order definition (if a<0 and x<y, ax>ay ) so not sure about the following: If -1<0, then (-1)*(-1)>(-1)*0
Maybe I'm nitpicky and it follows easily from definitions.
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Jack_Ian
Big Endian
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| Posted: Wed Oct 17, 2012 10:33 am Post subject: 9 |
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| Nsof wrote: |
| How about: 1>0 |
Shouldn't this be true by definition?
The ">" relationship must be defined, otherwise it's like saying "Prove f(1, 0)=T".
How can you define the ">" relationship while maintaining some doubt about the truth of the 1>0 statement? |
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Thok
Oh, foe, the cursed teeth!
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| Posted: Wed Oct 17, 2012 10:34 am Post subject: 10 |
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| Nsof wrote: |
| Quote: |
| Either -1<0 or -1>0. If -1<0, |
Or 1=0. Its not hard to prove that 1!=0 but it requires a proof.
| Quote: |
if a ≤ b then a + c ≤ b + c
if 0 ≤ a and 0 ≤ b then 0 ≤ a b
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Yes.
However, I am not sure the corollary is part of the field order definition (if a<0 and x<y, ax>ay ) so not sure about the following: If -1<0, then (-1)*(-1)>(-1)*0
Maybe I'm nitpicky and it follows easily from definitions. |
I don't know for certain what your definitions of < are, which is why I asked.
If a<=0, then a+(-a)<=0+(-a), so 0<=-a. Then if x<y, (-a)x<(-a)y, and adding ax+ay to both sides gives ay<ax.
(More generally, if x<y and a>0, then 0<y-x, so 0<a(y-x), and ax<ay.)
I'll point out my proof actually proves the stronger statement that if there is a <, then 0<=x^2 for all x (all squares are positive). |
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Nsof
Daedalian Member
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| Posted: Thu Oct 18, 2012 12:08 am Post subject: 11 |
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| Jack_Ian wrote: |
| Nsof wrote: |
| How about: 1>0 |
Shouldn't this be true by definition?
The ">" relationship must be defined, otherwise it's like saying "Prove f(1, 0)=T".
How can you define the ">" relationship while maintaining some doubt about the truth of the 1>0 statement? |
From wikipedia article Field_axioms
| Existence of additive and multiplicative identity elements wrote: |
| There exists an element of F, called the additive identity element and denoted by 0, such that for all a in F, a + 0 = a. Likewise, there is an element, called the multiplicative identity element and denoted by 1, such that for all a in F, a · 1 = a. To exclude the trivial ring, the additive identity and the multiplicative identity are required to be distinct. |
The definition of 0 and 1 put the requirement that they differ but does not say anything about their order.
In fact there does not need to be an order to them unless you define what is an order (the properties written by Thok)
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Jack_Ian
Big Endian
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| Posted: Thu Oct 18, 2012 9:55 am Post subject: 12 |
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| Nsof wrote: |
| The definition of 0 and 1 put the requirement that they differ but does not say anything about their order. |
OK, but the question was, how do you define the ">" relationship. |
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Nsof
Daedalian Member
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| Posted: Fri Oct 19, 2012 12:53 am Post subject: 13 |
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Its actually two sets of definitions. I'll borrow from two places.
Total ordered set
| Wolfram wrote: |
A total order (or "totally ordered set," or "linearly ordered set") is a set plus a relation on the set (called a total order) that satisfies the conditions for a partial order plus an additional condition known as the comparability condition. A relation <= is a total order on a set S ("<= totally orders S") if the following properties hold.
1. Reflexivity: a<=a for all a in S.
2. Antisymmetry: a<=b and b<=a implies a=b.
3. Transitivity: a<=b and b<=c implies a<=c.
4. Comparability (trichotomy law): For any a,b in S, either a<=b or b<=a. |
To apply this on a field rather than a general set, two more properties are required
Ordered field
| Wikipedia wrote: |
A field (F,+,*) together with a total order ≤ on F is an ordered field if the order satisfies the following properties:
if a ≤ b then a + c ≤ b + c
if 0 ≤ a and 0 ≤ b then 0 ≤ a b
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I quoted the two field-relevant properties in an earlier post but didnt bother with the total order definition itself because its properties are "uninteresting" in this field-specific proof.
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ralphmerridew
Daedalian Member
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| Posted: Mon Apr 01, 2013 9:13 pm Post subject: 14 |
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| Code: |
1) 0 + 0 = 0 additive identity
2) 0*(0+0) = 0*0 multiply both sides of 1 by 0
3) 0*0 + 0*0 = 0*0 distribute 2
4) (0*0 + 0*0) + -(0*0) = 0*0 + -(0*0) add -(0*0) to both sides of 3
5) 0*0 + (0*0 + -(0*0)) = 0*0 + -(0*0) associative + on 4
6) 0*0 + 0 = 0 additive inverse on 5
7) 0*0 = 0 additive identity on 6
8) 1 + (-1) = 0 additive inverse
9) (1 + (-1))*(1 + (-1)) = 0 substitute 8 into 7
10) (1 + (-1))*1 + (1 + (-1))*(-1) = 0 distribute 9
11) (1*(1+(-1))) + ((-1)*(1+ (-1))) = 0 commute * on 10
12) (1*1 + 1*(-1)) + ((-1)*1 + (-1)*(-1)) = 0 distribute 11
13) (1 + (-1)) + ((-1) + (-1)*(-1)) = 0 multiplicative identity on 12
14) 0 + ((-1) + (-1)*(-1)) = 0 additive inverse on 13
15) (-1) + (-1)*(-1) = 0 additive identity on 14
16) 1 + ((-1) + (-1)*(-1)) = 1 + 0 add 1 to both sides on 15
17) (1 + (-1)) + (-1)*(-1) = 1 + 0 commute + on 16
18) 0 + (-1)*(-1) = 1 + 0 additive inverse on 17
19) (-1)*(-1) = 1 additive identity on 18 |
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