- #1
solakis1
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Given the following axioms:
For all a,b,c we have:
1) a+b = b+a
2) a+(b+c)=(a+b)+c
3) ab = ba
4) a(bc) = (ab)c
5) a(b+c) =ab+ac
NOTE,here the multiplication sign (.) between the variables have been ommited
6) There ia a number called 0 such that for all a,
a+0 =a
7)For each a, there is a number -a such that for any a,
a+(-a) = 0
8)There is a number called 1(diofferent from 0) such that for any a,
a1 = a
9)For each a which is different than 0there exists a number called 1/a such that;
a.(1/a)= 1.
10) exactly one of a>b,b>a or a=b holds
11) if a>b ,b>c then a>c
12) if c>0 ,a>b then ac>bc
13) if a>b then a+c>b+c for any c
The definitions:
14) a/b = a(1/b)
15) $a\geq 0\Longrightarrow |a|=a$ and $ a<0\Longrightarrow |a|=-a$.
Then by using only the axioms and the definitions above prove:A) If a>0 and b>0 then a>b iff aa>bbB) If $x\neq 0$ then (|x||x|)/x=x for all x
For all a,b,c we have:
1) a+b = b+a
2) a+(b+c)=(a+b)+c
3) ab = ba
4) a(bc) = (ab)c
5) a(b+c) =ab+ac
NOTE,here the multiplication sign (.) between the variables have been ommited
6) There ia a number called 0 such that for all a,
a+0 =a
7)For each a, there is a number -a such that for any a,
a+(-a) = 0
8)There is a number called 1(diofferent from 0) such that for any a,
a1 = a
9)For each a which is different than 0there exists a number called 1/a such that;
a.(1/a)= 1.
10) exactly one of a>b,b>a or a=b holds
11) if a>b ,b>c then a>c
12) if c>0 ,a>b then ac>bc
13) if a>b then a+c>b+c for any c
The definitions:
14) a/b = a(1/b)
15) $a\geq 0\Longrightarrow |a|=a$ and $ a<0\Longrightarrow |a|=-a$.
Then by using only the axioms and the definitions above prove:A) If a>0 and b>0 then a>b iff aa>bbB) If $x\neq 0$ then (|x||x|)/x=x for all x
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