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hi chrislav
I suppose this comes from basic logic theory, or from (probability) trees.
Bob
Thanks bob,sorry for being so inquisitive .
But, i wonder, if we do not know the basic laws of logic and how are they ivolved in a mathemetical proof how can we be sure for the correctness of that proof??
hi chrislav
That wasn't quite what I meant. There are 4 cases:
case 1. a > b AND b > c
case 2. a > b AND b = c
case 3. a = b AND b > c
case 4. a = b AND b = cIf you are able to prove the required result for each of the above, then you're done.
Bob
O.K BOB
But i would like to know,if possible, what ,mathematical rule dictates that there must be those 4 cases
hi chrislav
Thanks for the quick reply. Yes, you can use LaTex. Look here: http://www.mathisfunforum.com/viewtopic.php?id=4397
That thread goes on for a long time but you'll read enough in the first few posts to get started.
Definition : a>=b <=> a>b or a=b
I think that definition is the place to start. If you consider separate cases, eg a>b AND b=c, you can show the required result for each case.
Hope that helps
Bob
O.K .Let us start:
using the definitionWhich is equal to:
But if this is correct which is the rule in mathematics we use to get the above ??
hi chrislav
Welcome to the forum.
Some people, looking at these, might say 'isn't it obvious?' So I'm guessing this is a proof from first principles analysis. Tp start you need to look at the definition of >=
If you post this back I'll see if I can help.
Bob
Thanks Bob ,you are right.
So the axioms and the definition needed for the above proofs are:
Axioms:
1) the trichotomy law for ">"
2) a>b and b>c => a>c
3) a>b => a+c>b+c
4) a>b and c>0 => ac>bc
Definition : a>=b <=> a>b or a=b
By the way isnt there any LaTex we can use??
Prove ;
1) a>=b and b>=c => a>=c
2) a>=b and b>=a => a=b
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