Negation of Mathematical Expressions

[delusia]

This is homework-related, so hints should suffice.

 

The text discusses how to negate conjunctions, disjunctions, conditionals, and biconditionals. The examples given are in symbolic logic and English expressions. However, some of the exercises require negating compound statements whose component statements are mathematical expressions. A sample problem is thus [math]3 < 5 \mbox{ or } 7 \geq 8[/math], whose answer at the end of the text is given as [math]3 \geq 5 \mbox{ and } 7 < 8[/math], which happens to be the only answer given.

 

This seems to make some sense if one notes a De Morgan's Law that previously is discussed, which is [math]\neg (P \vee Q ) \Longleftrightarrow \neg P \wedge \neg Q[/math]. The confusion comes from having to negate some of the component mathematical expressions for which there is no guide. What's the intuitive explanation? It also would need to be applied to compound statements such as [math]w - 3 > 0 \mbox{ implies } w^2 + 9 > 6w[/math] and [math]a - b = c \mbox{ iff } a = b + c[/math].

 

Any help with this would be greatly appreciated.

[matt grime]

so we have

 

(w-3 > 0) => (implies) (w^2+9>6w)

 

well X=>Y is the same as {not(X)} OR {Y} so its negation is?

[delusia]

The logic is understood. What's puzzling is negating mathematical expressions, such as [math]w - 3 > 0[/math] in this case. By chance would it be [math]w - 3 \leq 0[/math], and for [math]=[/math] would it be [math]\neq[/math], with regard only to equality and disregard to the terms?

[DQW]

If you are okay woth {NOT(3 < 5)} = {3 >= 5}, why should there be a problem in say, replacing 5 with w ?

[delusia]

So true. Thanks.

[ElijahJones]

X=>Y is the same as {not(X)} OR {Y} so its negation is?

 

You lost me on that one bro. I thought it was equivalent to notY => notX (the contrapositive). You must mean that symbol to be greater than or equal to. So x>=y has logical compliment x<y. I agree with that but I don't see how you can turn an order operation into a Boolean, the ordering implies more structure than a boolean algebra can account for does it not?

[matt grime]

=> is standard for imples. proposition A implies B is exactly the same as not(A) or B, exactly as i wrote.

[ElijahJones]

A=>B also is interpreted as A is a subset of B. In which case notA OR B = U. If A and B are disjoint it similarly makes no sense since B is a subset of notA yielding notA OR B = notA.

 

If the neither are subsets of the other but they have a non-empty intersection then A=>B could mean some A are B and some B are A but that is not the usual meaning of implies.

 

A=>B is If A Then B so that the truth statement of the contrapositive If notB Then notA is exactly the same as If A Then B and is a common form of proof.

 

The proof of infinitely many primes uses the same format. Assume there are finitely many. So write the last one P, now form the product of all primes plus one. This is relatively prime to all the primes by assumption, so it must be prime, contradicting our assumption of finitely many primes.

 

The form of the argument is

 

If "there are finitely many primes" Then "we can take the product of all primes and add one"

 

the contrapositive which proves our theorem is

 

If "we cannot take the product of all primes" Then "there are infinitely many primes"

 

So I see that thinking of this as sets is not really the way to go. In some cases it makes sense to think of A and B as sets but in general the statements can be much more complicated. Let me look at that again from a different perspective.

 

A=>B

 

notA OR B, no dude come one. No on this one I'm not just gonna sit back and say ok.

 

What is your definition of notA, B are these sets? The operator OR is a boolean, it makes no sense.

 

Suppose we are simply talking truth of the propositions A and B. Then notA OR B might be that either notA is true or B is true. Nope, sorry. Well wait a minute!

 

The truth table for implications is as follows:

 

A B A=>B

 

T T T

F T T

F F T

T F F

 

Its kind of goofy because you can argue from a false premise to a false conclusion and then the whole statement is taken as true. The only one of these I have seen used in real proofs is T, T = T. The false premise "Truths" are odd, so apparently a liar who gets it right in the end is afforded some credit and a liar from start to finish is to be believed.

 

That a true premise cannot imply a false conclusion explains T,F = F. But F, T=T and F,F=T I never have understood, I think it is a convention. Realistically would'nt F,F be undecidable? F,T also should be undecidable, if the premise is false but the conclusion true.

 

There are four basic implication arguments: implication, inverse, converse and contrapositive. Of course if the implication and its converse are true we have that most important "If and only if" A <=> B.

 

So we have

 

A=>B implication

notB=>notA contrapositive

B=>A converse

notA=>notB inverse

 

I only just now see that this does not exhaust the possiblities. What about

 

notA=>B

A=>notB

 

I guess these are redundant. Yeah. I'm sure Grimes will have something interesting to say about this post. I suppose if you look at the true values for implication they would be summed up by notA OR B because that would be the set {(T,T),(F,T),(F,F)}. Right because if B is true then the implication is taken as true (not without some reservation on my part for the false premises) but these are the cases {(T,T),(F,T)} now if B is false we only want to include a false premise {(F,F)} and of course this is not exclusive OR but it is redundant in its inclusion of {(F,F)}. Ok Grimes, I buy it but what is the point of summing up the truth table in that way?

 

EJ

[matt grime]

My name isn't Grimes.

 

What on earth are you rambling about? this is a discussion of negation of propositions, nothing to do with set, please, don't confuse things pointlessly (and the set equivalent of not is complement).

 

Moreobve, > and < are not usualy used for sets. containment of sets is round brackets. It would be corect for subgroup, not subset.

 

Evidently you need some somple examples to see why the truth values of impliction are as they are. Simply put you are confusing "causality" with imlpication.

 

For instance, thinking about natural numbers, take the propostion

 

If 4 divides x then x is even.

 

This is clearly a true proposition, though individually the case x=2 shows that we have "false implies true" ie the premise is false but the conclusion is true. x=7 shows that false impies false is true. Obvisoly we cannot find a case where the premise is true and the antecedent is false. That is the whole point of propositions.

 

Now, why don't you write out the truth taable for not(A) or B and see what you get; hopefully now the mists will lieft and yuo will see why A=>B is the same as not(A) or B.

 

 

I like the way you also think that there can be one and only one equivalent expression fro A implies B (not B implies not A). fortunately maths is quite a lot more varied than that, and you shuldn't think that only what you know is what is true.

[ElijahJones]

Evidently you need some somple examples to see why the truth values of impliction are as they are. Simply put you are confusing "causality" with imlpication.

 

Your arrogance is starting to bother me again.

 

For instance, thinking about natural numbers, take the propostion

 

If 4 divides x then x is even.

 

This is clearly a true proposition

 

Sorry studly but the proposition cannot be true by fiat, it must be determined true by examining the clauses.

 

Insert x = 2 and you get

 

If 4 divides 2 then 2 is even.

 

Since four does not divide 2 it is impossible to arrive at the conclusion that 2 is even. The statement is bullshit.

 

Insert x = 7 and you get

 

If 4 divides 7 then 7 is even

 

Again the statement is bullshit, and I gurantee that if you develop proofs bye taking this as a true statement you will be producing bullshit. This probably explains why much of modern mathematics is useless to the real world.

 

Since you are so adept at proving things perhaps you can derive us a solution to the oil crisis. If I use your reasoning I could start with,

 

"If elephants are pink then invading Iraq will solve our energy problems."

 

By the way the wall paper on your website has some symbols that appear to meaningless. Its sort of like you said f(x)=f(x).

[ElijahJones]

As much as it pains me to admit it, you are right.

 

If 4 divides 7 then 7 is even

 

If 4 divides 2 then 2 is even

 

Are true because they are examples of

 

If 4 divides x then x is even

 

But notice you cannot form TRUE and FALSE = FALSE

 

Also what happens if we do this

 

If 4 divides x then x is odd

 

Now of course the statement is always false no matter what the status of the prepositional parts.

 

The cases that arise are chracterized by

 

If 4 divides 2 then 2 is odd (False and False = False)

 

If 4 divides 2n+1 then 2n+1 is odd (False and True = False)

 

If 4 divides 4n then 4n is odd (True and False = False)

 

TRUE and TRUE cannot be formed.

 

I must humble myself and ask you to forgive me. But I still think it is arrogant to deride people who request explanations for the things you say. Noone is to be so trusted as to be taken without question. It is at just that point that pride leads us to a fall.

[matt grime]

You appear to be saying that it isn't obvious that "if 4 divides x then x is even" is true. i can't beleive you expect me to provide a proof of that... i didn't say you were stupid to not understand the reasoning behind "implies" being the way it is, i said that evidently you needed understandable examples to illustrate it where you can clearly see that a proposition is true despite it being "of the type" F => T ro F=>F.

 

again, your pink elephant comment shows that you are using the common language maening of implies with its causality connotation instead of the mathematical one. the comical proposition you give is a true proposition (well, probably, depends what's acceptable as a pink elephant) albeit nonsensical.

 

 

 

Thw wall papaer you refer to is the distinguished triangle in a traiangulated category that defines the homotompy colimit of a sequences of objects and morphisms

 

X_1 --> x_2 ---> X_3---> X_4 .....

 

it is the triangulated version of the cokernel of an injection in an ableian category defining the colimit. There is no reason for you to know what this is supposed to be. i doubt that more than a handful of people would understand it. not because it is hard but because it is obscure.