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Noxate
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Homework Statement
Although the basic properties of inequalities were stated in terms of the collection
P of all positive numbers, and < was defined in terms of P, this
procedure can be reversed. Suppose that P10-P12 are replaced by
(P'10) For any numbers a and b one, and only one, of the
following holds:
(i) a = b,
(ii) a < b,
(iii) b < a.
(P'11) For any numbers a, b, and c, if a < b and b < c, then
a < c.
(P'12) For any numbers a, b, and c, if a < b, then
a+c<b+c.
(P'13) For any numbers a, b, and c, if a < b and 0 < c, then
ac < bc.
Show that P10-P12 can then be deduced as theorems.
Homework Equations
Addition properties:
P1: (Associative law of addition) (a+b)+c=a+(b+c)
P2: (Existence of additive identity) a+0=0+a=a
P3: (Existence of additive inverse) a+(-a)=(-a)+a=0
P4: (Commutative law of addition) a+b=b+a
Multiplication properties:
P5: (Associative law of multiplication) a x (b x c)=(a x b) x c
P6: (Existence of multiplicative identity) a x 1=1 x a=a
P7: (Existence of multipliative inverse) a x a-1=a-1 x a=1 (If a is different from zero)
P8: (Commutative law of multiplication) a x b=b x a
Distributive property
P9: (Distributive law) a x (b+c)=(a x b)+(a x c)
Order properties:
Let P be called the collection of positive numbers, then:
P10: (Trichotomy Law) For every a, exactly one of the following happens: a=0, a is in P or -a is in P
P11: (Closure under addition) If a and b are in P, a+b is in P
P12: (Closure under multiplication) If a and b are in P, a x b is in P
The Attempt at a Solution
I'm really not entirely sure what the question is asking me to do. I believe it is asking me to show that P'10-P'13 are true from the properties P1-P9 and thus the statements in P10-P12 are also true.
Here is my attempt at doing that.
Starting with:
(P'12) For any numbers a, b, and c, if a < b, then
a+c<b+c.
From P3, if a<b, then a+(-b)<b+(-b)
-> a-b<0
Thus a-b+c-c<0
->(a+c)+(-b-c)<0
->(a+c)-(b+c)+(b+c)<0+(b+c)
->a+c<b+c
Thus we have deduced P'12
Now P'11:
(P'11) For any numbers a, b, and c, if a < b and b < c, then
a < c.
From P'12, a+c-b-c<0
->a+c-b<c
c-b>0, then (c-b)+a>0+a
thus a<a+c-b<c
Therefore we have deduced P'11.
I'm fairly confident in these 2, the next 2 are where I don't feel as though my proof is very good.
P'10:
(P'10) For any numbers a and b one, and only one, of the
following holds:
(i) a = b,
(ii) a < b,
(iii) b < a.
Using P7, if a<b then 1<b/a, which would mean 1<1 if a=b. From (P3) a+(-a)=0, thus b+a+(-a)=0+b. So 1<1 and 1=1 cannot both be true and thus if a<b then it can't also be true that a=b.
Similarly for b<a, then 1<a/b, implies 1<1 if a=b. So i cannot be true if either ii or iii are true. Lastly if b<a is true and a<b is true then both b-a<0 and a-b<0 would have to be true.
Assuming a>0 and b>0, then from P9, a(b-a)<0 and a(a-b)<0 are both true, thus -a^2+ab<0->a^2>ab and from a(a-b)<0 a^2<ab. a^2<ab<a^2 cannot be true unless a=0.
Similarly we can find that b^2<ab<b^2 would only be true if b=0. Thus if both ii and iii were true it would cause contradictions in 0=0 and 0<0 and a=b, which we've already established cannot be true. So i, ii and iii are exclusive and P'10 is true.
P'13.
(P'13) For any numbers a, b, and c, if a < b and 0 < c, then
ac < bc.
a<b->a-b<0<c. Using P9, c(a-b)<0<c(c)
-> ca-cb<0<c^2.
Then applying P3, ac-bc+bc<0+bc
-> ac<bc
Therefore we have deduced P'13.
For P10, since we've shown that either a=b, a<b or b<a. Then if a=0 clearly b<a or (-b)<a as b=/=a and one of either b<a or (-b)<a must be true. So any number x must either be a=0, a<0 or (-a)>0.
I believe I still must somehow show that P11 and P12 statements are true somehow and then from that I would simply state that the set P can represent all numbers a>0 or (-a)>o, if a<0.
That's basically what I got, again I'm not entirely sure that I've understood the question properly.
