Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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bob bundy wrote:

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??

bob bundy wrote:

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

bob bundy wrote:

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 ??

bob bundy wrote:

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??

**chrislav**- Replies: 10

Prove ;

1) a>=b and b>=c => a>=c

2) a>=b and b>=a => a=b

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