proof of an inequality

[triclino]

in proving the cancellation law in inequalities in real Nos we have the following proof:

 

 

Let : ~(a<b)

Let : a+c<b+c

 

Since ~(a<b) and using the law of trichotomy we have : a=b or b<a

 

for a=b => a+c=b+c.....................................................................(1)

 

for b<a => b+c<a+c......................................................................(2)

 

and substituting (1) into (2) we have : a+c<a+c ,contradiction.

 

Hence ~(a+c<b+c)

 

Thus by contrapositive we have: a+c<b+c => a<b

 

is that proof correct??

Edited January 28, 2015 by triclino

[fiveworlds]

Can you specify your definition of tilde in this case please.

[triclino]

it is not (negation)

 

So ~(a<b) means not (a<b)

 

Sorry in the 1st line of proof i sould have writen ~(a<b) instead of ~a<b

 

I corrected it

Edited January 28, 2015 by triclino

[MomentTheory]

 

 

and substituting (1) into (2) we have : a+c<a+c ,contradiction.

 

The contradiction arises only when you substituted "=<" with "<"

 

This should read a+c=<a+c, you forgot the "=" defined by (1) a+c=b+c

[John]

First, about your wording: instead of "Let ~(a<b)" and "Let (a+c)<(b+c)", it's more correct to say "Assume ~(a<b) and (a+c)<(b+c)."

Second, your proof involves the two cases a=b and b<a. Substitution of the first into the second isn't a valid step, since for all real numbers a and b, a cannot simultaneously be equal to b and greater than b. That is to say, these are entirely separate cases and should not be mixed as you've done here.

 

Third, what axioms/results are you allowed to use? If you're allowed to make use of the fact that (b<a) => (b+c)<(a+c), for instance, then assuming you can also make use of additive inverses and the associativity of addition, you can do a pretty short direct proof of this cancellation law without having to resort to an indirect proof of the contrapositive. I'd be a bit more clear and detailed here, but this seems like it might be a homework question, so I don't want to give too much away.

 

 

 

The contradiction arises only when you substituted "=<" with "<"

 

This should read a+c=<a+c, you forgot the "=" defined by (1) a+c=b+c

 

This is incorrect. As detailed before, the entire substitution is invalid, but even in the context of the OP's reasoning (the apparent point was to show the logical conflict between (a+c)=(b+c) and (b<a) => (b+c)<(a+c)), the equality shouldn't be carried over.

Edited January 28, 2015 by John

[triclino]

First, about your wording: instead of "Let ~(a<b)" and "Let (a+c)<(b+c)", it's more correct to say "Assume ~(a<b) and (a+c)<(b+c)."

 

Second, your proof involves the two cases a=b and b<a. Substitution of the first into the second isn't a valid step, since for all real numbers a and b, a cannot simultaneously be equal to b and greater than b. That is to say, these are entirely separate cases and should not be mixed as you've done here.

 

Third, what axioms/results are you allowed to use? If you're allowed to make use of the fact that (b<a) => (b+c)<(a+c), for instance, then assuming you can also make use of additive inverses and the associativity of addition, you can do a pretty short direct proof of this cancellation law without having to resort to an indirect proof of the contrapositive. I'd be a bit more clear and detailed here, but this seems like it might be a homework question, so I don't want to give too much away.

 

 

 

This is incorrect. As detailed before, the entire substitution is invalid, but even in the context of the OP's reasoning (the apparent point was to show the logical conflict between (a+c)=(b+c) and (b<a) => (b+c)<(a+c)), the equality shouldn't be carried over.

First i suggest you write to all those authors that they use the word "let" instead of suppose in their proofs to correct their style of writing

Second,on page104 in Suppes's book ,Introduction to Logic i find the following rule concerning substitution:

 

"if S is an open formula,from S and t_1 =t_2,or from S and t_2=t_1 we may deriveT,provided that T results from S by replacing 0ne or more occurrences of t_1 in S by t_2".

 

So in our case if we put:

 

S = b+c<a+c

 

t_1= a+c

 

t_2 = b+c

 

then the derivable formula T IS

 

b+c<b+c.

[John]

First i suggest you write to all those authors that they use the word "let" instead of suppose in their proofs to correct their style of writing

No need to get snippy. You were attempting a proof by contradiction, and in that sort of proof, we "assume" that certain things are true. In this case, you were assuming ~(a<b) and (a+c)<(b+c). I understood your meaning just fine, but using "let," it'd be more correct to say, in this case, "let it be the case that ~(a<b) and (a+c)<(b+c)." It's not a big deal, and "let" is fine overall, but the change does add clarity.

 

Second,on page104 in Suppes's book ,Introduction to Logic i find the following rule concerning substitution:

 

"if S is an open formula,from S and t_1 =t_2,or from S and t_2=t_1 we may deriveT,provided that T results from S by replacing 0ne or more occurrences of t_1 in S by t_2".

 

So in our case if we put:

 

S = b+c<a+c

 

t_1= a+c

 

t_2 = b+c

 

then the derivable formula T IS

 

b+c<b+c.

I understand that, but your proof involves the two separate (by the trichotomy law, as you mentioned) cases that a=b and b<a. You derived a result from the first case, which was perfectly fine, but then you substituted that result into the entirely separate second case, which is not correct.

 

Perhaps I can explain more clearly. In case 1, you assumed a=b and derived a result from that. In the second case, you assumed a>b, thus any results from case 1 do not apply.

 

Essentially, in the substitution line of your proof, you derived a contradiction from the idea that case 1 and case 2 are both true, whereas by the trichotomy law, exactly one is true.

 

Is this a homework exercise?

Edited January 28, 2015 by John

[studiot]

I agree with John, there is a difference between an assertion and an assumption.

The use of the word 'Let' is the beginning of an assertion.

Assertions are considered true for the entire proof.

 

Proof by contradiction is carried out by assuming that which you wish to prove and demonstrating that the assumption leads to the contradication of an axiom or assertion and so cannot be true.

[triclino]

Yes i understand ,but to convince me, you have to produce a law of logic or mathematics that will oppose my substitution

 

On the other hand Suppes's subtitution law is more convincing than your justification

 

Also that substitution law includes all cases ,even my case otherwise it wold be aspecific law applying only in certain cases.

 

But i do not detect that in the definition of the law.

[John]

The law is fine, but you're misapplying it here. I'm not sure how to explain better than I have already, but I will give it a shot.

In case 1, you have that a=b. You use this to immediately say (correctly) that (a+c) = (b+c). Now, in case 2, you have that b<a. Thus, it is not the case that a=b, therefore it is not the case that (a+c) = (b+c). This is why the substitution fails: because in case 2, the statement that (a+c) = (b+c) is no longer true.

Edited January 29, 2015 by John

[triclino]

i am sorry but I do not see how or where i misapply the definition

[iNow]

Situation 1: A = B

Situation 2: A > B

 

Given this difference, the answer found in Situation 1 cannot be used as a substitute in Situation 2. A cannot be equal to B and also greater than B at the same time.

[MomentTheory]

i am sorry but I do not see how or where i misapply the definition

 

That's funny because I gave you more lenience than these other guys, giving you the benefit of the doubt that you were trying to prove something beyond your assumptions and allowing your error to continue until you excluded the "=" (as I mentioned above); yet they have clearly given you a better definition of your fault than I have and you still do not see your error?

 

Even if it were true, what would you be proving except a logical inequality (a fault), surely you can't believe equal numbers can in fact be unequal? Maybe on the quantum level this can be true, but not with an a/b test.

[triclino]

Situation 1: A = B

Situation 2: A > B

 

Given this difference, the answer found in Situation 1 cannot be used as a substitute in Situation 2. A cannot be equal to B and also greater than B at the same time.

 

 

 

 

 

Is then the following proof without using contrapositive but only using contradiction, correct ??

 

Let, a+c<b+c...............................................................................................................................................................................(1)

 

Let, ~(a<b).......................................................................................................................................................................................(2)

 

Then by using (2) and and the law of trichotomy we have: a=b or b<a

 

For a=b => a+c<a+c ,by substituting into (1)

 

For b<a => b+c<a+c => a+c<a+c,by using the transitive property of inequalities and (1)

 

Hence a+c<a+c ,contradiction and

 

Thus ~~(a<b) => a<b

Edited February 2, 2015 by imatfaal
Fixing mis-quote

[triclino]

 

a logical inequality (a fault), surely you can't believe equal numbers can in fact be unequal? Maybe on the quantum level this can be true, but not with an a/b test.

 

Can you perhaps refer me to any mathematical or logical book that I can find the definition of the phrase "logical inequality" ??

 

No equal Nos are not unequal that is why we end up with a contradiction

[imatfaal]

Can you perhaps refer me to any mathematical or logical book that I can find the definition of the phrase "logical inequality" ??

 

No equal Nos are not unequal that is why we end up with a contradiction

"No equal Nos are not unequal"

 

2 and 2 - these are equal and are not unequal. Yours statement shown to be incorrect.

 

a=b, a>b, a<b are mutually exclusive (thus trichotomy) - there can be no pair of elements that share two of these characteristics.

 

You have assumed one of the characteristics then you have assumed a second of them - BUT you have then tried to combine the

results; this does not work. Your logic depends on the mis-asssumption that a=b AND a<b at the same time. This is an error in logic

[John]

 

Is then the following proof without using contrapositive but only using contradiction, correct ??

 

Let, a+c<b+c...............................................................................................................................................................................(1)

 

Let, ~(a<b).......................................................................................................................................................................................(2)

 

Then by using (2) and and the law of trichotomy we have: a=b or b<a

 

For a=b => a+c<a+c ,by substituting into (1)

 

For b<a => b+c<a+c => a+c<a+c,by using the transitive property of inequalities and (1)

 

Hence a+c<a+c ,contradiction and

 

Thus ~~(a<b) => a<b

This reasoning is actually correct, though if you were to present the proof (in a paper, or for an assignment, etc.), then you'd want to flesh it out a bit, and some of the wording is a bit odd.

 

And just for the record, your original post does include a correct proof. Specifically,

 

in proving the cancellation law in inequalities in real Nos we have the following proof:

 

 

Let : ~(a<b)

Let : a+c<b+c

 

Since ~(a<b) and using the law of trichotomy we have : a=b or b<a

 

for a=b => a+c=b+c.....................................................................(1)

 

for b<a => b+c<a+c......................................................................(2)

 

Thus by contrapositive we have: a+c<b+c => a<b

is sufficient to show what you want to show, though again, you'd probably want to flesh it out. It's only the substitution part that renders the proof invalid.

 

Of course, all of this using the contrapositive or arriving at a contradiction stuff is unnecessary, since there is a very simple direct proof, as mentioned earlier.

[triclino]

This reasoning is actually correct, though if you were to present the proof (in a paper, or for an assignment, etc.), then you'd want to flesh it out a bit, and some of the wording is a bit odd.

 

 

 

 

 

But my substitution in my 2nd proof isnt it the same with the substitution in my 1st proof??

 

BECAUSE when i substitute a=b in the inequality a+c<b+c (1) of my 2nd proof isnt the SAME like i combine a=b And a<b??

 

a+c<b+c is the result of adding c to both sides of the inequality a<b

 

Like b+c<a+c was the result of adding c to both sides of the inewuality b<a

[John]

Not quite. In your first proof, you substituted a result from case 1 into the entirely separate case 2, which was invalid. In your second attempt, you substituted your first assumption into each case, which is valid since your assumptions apply to the entire proof.

 

When I first read your second proof, I also thought you'd made the same mistake, but then I read it more closely and saw you hadn't.