inequality proof

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kanali

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I tried to prove that:

\(\displaystyle a\geq b\) and \(\displaystyle b\geq a\Longrightarrow a=b\). for all a,b belonging to reals

Proof:

\(\displaystyle a\geq b\) and \(\displaystyle b\geq a\) =(a>b or a=b)and (b>a or b=a) which according to logic it is equal to:

(a>b and b>a) or (a>b and b=a) or ( a=b and b>a) or ( a=b and b=a).

And now how do we curry on??
 
What does "according to logic" mean?

Try this...

If a >= b, then a = b + c for some c >= 0
 
According to the distributive property of propositional logic we have that :

(a>b or a=b) and ( b>a or b=a) = (a>b and b>a) or ( a>b and b=a) or (a=b and b>a) or ( a=b and b=a).

Now examining each case we have:

1) (a>b and b>a) => a>a

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

3) (a=b and b>a) => a>a

4) (a=b and b=a) => a=b

So we see that only the last case give as the right result.

Can we ignore the other cases ?? NO .Because that would be against logic.

So what to do??
 
tkhunny said:
What does "according to logic" mean?

Try this...

If a >= b, then a = b + c for some c >= 0
Click to expand...

O.K

FOR c>0 => b+c>b => a>b

For c=0 => a=b

HENCE \(\displaystyle a\geq b\).

But we want only a=b
 
Is this according to logic? Use tkhunny's hint, then add the following equations.

a = b + c
b = a + c

where c ? 0

The resulting equation shows that c must be zero.
 
kanali said:
According to the distributive property of propositional logic we have that :

(a>b or a=b) and ( b>a or b=a) = (a>b and b>a) or ( a>b and b=a) or (a=b and b>a) or ( a=b and b=a).

Now examining each case we have:

1) (a>b and b>a) => a>a

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

3) (a=b and b>a) => a>a

4) (a=b and b=a) => a=b



So we see that only the last case give as the right result.

Can we ignore the other cases ?? NO ( YES we can ignore those possibilities ? a>a is a contradiction - not allowed.).Because that would be against logic.

So what to do??
Click to expand...
 
Subhotosh Khan said:
( YES we can ignore those possibilities ? a>a is a contradiction - not allowed.).
Click to expand...

According to what axiom ,theorem,definition or a law of logic can you say that??

Is a mathematical proof an entity that you can subtract add or do whatever you like without any reasonig??
 
kanali said:
According to what axiom ,theorem,definition or a law of logic can you say that??
Click to expand...
The axiom of trichotomy states that if each of a & b is a real number then exactly one of the following is true \(\displaystyle a=b,\;a<b,\text{ or }b<a\)

If \(\displaystyle a<b\) then that contradicts the given, \(\displaystyle b\le a\).
If \(\displaystyle b<a\) then that contradicts the given, \(\displaystyle a\le b\).
That leaves only one possibility \(\displaystyle a=b\)
 
A proof is a concept or demonstration following rules developed from an axiomatic base. It has no life of its own. It is not an "entity".

A common axiom is Reflexivity. The concept of equality usually has this property. IF a exists, then a = a.

If you define a concept such as ">", it is very unlikely to include "=". If you can do it, I'd like to see it.

Therefore, a > a is not consistent with your set of axioms or the rules developed from those axioms. If you get a > a, you have a problem and you have either disproved what you were trying to disprove or you wadered off, somewhere.

Try this simpler explanation:

Bob: Dude, bring all your friends who have blue cars. We'll have a party.
Steve: Schweet!!
... a while later ...
Bob: Yo, Steve! Your car is red.
Steve: So?!
Bob: This party is for blue cars only.
Steve: I would like to stay.
Bob: Sorry, you must go.
Steve: {gumble}{grumble}
Bob: Oh, come on, Steve, we'll have a blue and red car party tomorrow.
... a short time later...
Bob: Ralph, That's really an SUV. It is blue, but is it really a car?
Ralph: Technically, it's a "crossover". This makes it both car and SUV.
Bob: Fair enough. Hey, everyone! Ralph is here!
Everyone: Hey, Ralph...

Here's what we just witnessed:
Ralph: Bob didn't see why Ralph could attend. With a little discussion, we had a new theorem. "Crossovers are cars and SUVs." Ralph can stay.
Steve: Steve was simply outside the definition and could not be included. Can we ignore Steve? We MUST ignore steve or we must write him a new theorem.

Making sense and following logic have no room for things outside the framework. Let's assume the Natural Numbers exist. Do you have a cow? The Natural Numbers don't care about your cow. If it's not in the system, throw it out. If you REALLY REALLY want to include it, you'll have to make a new system - possibly by expanding the present system. One more example. Let's assume an extension of the natural numbers, so that we can talk about the Natural Numbers and cows. Do you have a cow? Oh, that is interesting. The cow is now in the system. We'll need a little more work defining addition, if we need that, but we have a system.
 
Maths is not imagination and writing stories of what ever comes into our mind.

Definition of mathematical proof: A mathematical proof is a sequence of statements linked logically together and leading to a final statement which what we actually want to show or prove.

Example : prove for all real ,x : 0x =0

Proof : (1.x = x) => [(1+0)x = x+0] => [1.x +0x = x+0] => [ x +0x = x+0] => 0x = 0.

The statements are:

1) 1.x = x

2) (1+0)x = x+0

3) 1x +0x =x+0

4) x+0x = x +0

5) 0x =0

Those statements are linked together by the logical implication sign "=>"

And leading to the final result : 0x = 0 ,which is what we want to prove.
 
pka said:
kanali said:
According to what axiom ,theorem,definition or a law of logic can you say that??
Click to expand...
The axiom of trichotomy states that if each of a & b is a real number then exactly one of the following is true \(\displaystyle a=b,\;a<b,\text{ or }b<a\)

If \(\displaystyle a<b\) then that contradicts the given, \(\displaystyle b\le a\).
If \(\displaystyle b<a\) then that contradicts the given, \(\displaystyle a\le b\).
That leaves only one possibility \(\displaystyle a=b\)
Click to expand...


Are you sure that the law of trichotomy allow us to throw away the logically implied fact in my proof: ( a>a or a=b) ???

You mixing up the ordinary by contradiction proof ,where in assuming \(\displaystyle a\neq b\) and by using the trichotomy law we end up in a contradiction ,with my proof. Which is entirely different
 
kanali said:
pka said:
kanali said:
According to what axiom ,theorem,definition or a law of logic can you say that??
Click to expand...
The axiom of trichotomy states that if each of a & b is a real number then exactly one of the following is true \(\displaystyle a=b,\;a<b,\text{ or }b<a\)

If \(\displaystyle a<b\) then that contradicts the given, \(\displaystyle b\le a\).
If \(\displaystyle b<a\) then that contradicts the given, \(\displaystyle a\le b\).
That leaves only one possibility \(\displaystyle a=b\)
Click to expand...


Are you sure that the law of trichotomy allow us to throw away the logically implied fact in my proof: ( a>a or a=b) ???

You mixing up the ordinary by contradiction proof ,where in assuming \(\displaystyle a\neq b\) and by using the trichotomy law we end up in a contradiction ,with my proof. Which is entirely different
Click to expand...

I think this is a Troll - trying to waste our time with useless arguements.
 
I like my "proof" via tkhunny's suggestion.

(I wonder whether or not pka would give me partial credit.)

:)
 
kanali said:
I tried to prove that:

\(\displaystyle a\geq b\) and \(\displaystyle b\geq a\Longrightarrow a=b\).
for all a,b belonging to reals
Click to expand...

\(\displaystyle a \geq b\)

\(\displaystyle a - b \geq 0\) . . . . . \(\displaystyle (*)\)

------------------------------------------------------------------

\(\displaystyle b \geq a\)

\(\displaystyle b - a \geq 0\)

\(\displaystyle -1(b - a) \leq -1(0)\)

\(\displaystyle -b + a \leq 0\)

\(\displaystyle a - b \leq 0\) . . . . . \(\displaystyle (**)\)

-----------------------------------------------------------------

Putting \(\displaystyle (*)\) and \(\displaystyle (**)\) together:

\(\displaystyle 0 \leq a - b \leq 0\)

\(\displaystyle (a - b)\) is sandwiched between \(\displaystyle 0\), so it equals \(\displaystyle 0\).

Then \(\displaystyle a - b = 0\)

Result: \(\displaystyle a = b\)
 
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