Comments on: Basic logic — relationships between statements — converses and contrapositives http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/ Mathematics related discussions Wed, 15 May 2013 08:08:31 +0000 hourly 1 http://wordpress.com/ By: indirect proofs: contrapositives vs. proofs by contradiction « Sebastian Pokutta’s Blog http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12440 Tue, 18 Oct 2011 20:01:36 +0000 http://gowers.wordpress.com/?p=3346#comment-12440 [...] discussion on contrapositives vs. proofs by contradiction as part of Timothy Gowers’ Cambridge Math Tripos, mathoverflow, and Terry Tao’s blog. At first sight these two concepts, the contrapositive [...]

]]> By: gowers http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12280 Fri, 07 Oct 2011 20:32:10 +0000 http://gowers.wordpress.com/?p=3346#comment-12280 Damn. OK I’ll add in to (i) that n can be 1.

]]> By: Chris Purcell http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12279 Fri, 07 Oct 2011 19:18:14 +0000 http://gowers.wordpress.com/?p=3346#comment-12279 “For every positive integer n, the following are equivalent: (i) n is a prime; (ii) (n – 1)! + 1 is divisible by n.”

1 is a positive integer. 1 is not prime. 0! + 1 = 2, which is divisible by 1. Hence, 1 is a counterexample.

]]> By: Blog of the Month- October 2011 | MY DIGITAL NOTEBOOK http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12248 Thu, 06 Oct 2011 06:49:12 +0000 http://gowers.wordpress.com/?p=3346#comment-12248 [...] Gower’s Weblog: Blog of the Month for October 2011. Prof. Timothy Gowers (a.k.a. Tim Gowers) is a field medalist and an eminent mathematician. He is a member of the Department of Pure Mathematics and Mathematical Statistics at Cambridge University . He is ideal of many student and math majors including me. I was not a huge fan of his writing as he  has  written  only a few articles after I joined the web. But as he started his series on ‘CAMBRIDGE TEACHING‘, I have become a regular reader of his weblog. One math student must read his posts on Cambridge Teaching. Most Recent Post: Basic logic — relationships between statements — converses and contrapositives  [...]

]]> By: gowers http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12243 Wed, 05 Oct 2011 19:14:43 +0000 http://gowers.wordpress.com/?p=3346#comment-12243 I intended to write what I actually did write, but your suggestion is clearer, so I’ve adopted it.

]]> By: anonymous http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12242 Wed, 05 Oct 2011 18:24:29 +0000 http://gowers.wordpress.com/?p=3346#comment-12242 When I read this sentence,

”But then $latex| x – y |$ is a positive number that is not greater than $latex| x – y |$.”

it seems like you may have intended to write something else, like

‘But then $latex| x – y |$ is a positive number, which shows that…”

]]> By: gowers http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12237 Wed, 05 Oct 2011 14:28:11 +0000 http://gowers.wordpress.com/?p=3346#comment-12237 Thanks — I see your point about integrating now. To get maths you do exactly as you are doing but insert “latex” after the first dollar with no space between the dollar and “latex”. So if you want, for example, to write \int_0^1f(x)dx, then what you actually type in is £latex \int_0^1f(x)dx£ except that the pound sign should be a dollar sign.

]]> By: andrejbauer http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12236 Wed, 05 Oct 2011 14:15:45 +0000 http://gowers.wordpress.com/?p=3346#comment-12236 Uh, there are a lot of interesting points here. (By the way, what do I have to write to get math notation?)

We have a meta-theorem for first-order logic which says that P(x) with a free parameter x has a proof iff \forall x P(x) has a proof.
In other words, as far as proving a statement goes, it does not matter whether the outer universal quantifiers are there. This is the precise meaning of “x is implicitly quantified”, I think. Also, this meta-theorem is sometimes misunderstood as saying that P(x) is equivalent to \forall x P(x), which is nonsense.

Suppose I tell you that I integrated a function on [0,1] and got 42. Can you tell which function it was? No. Suppose I tell you that I universally quantified a statement over the domain [0,1] and I got true. Can you tell me which statement it was? No. We must not confuse the truth value of \forall x P(x) with the expression \forall x P(x). A fair analogy would be this: if I show you the expression \forall x P(x) then you can tell what P is. Likewise, if I show you the expression \int_0^1 f(x) dx then you can tell what f is.

Yes, what you say about having free parameters in the middle of the proof is essentially what I am trying to say. Speaking as a logician, you simply must have free parameters because the rule of inference for universal quantifiers requires you to use them. Namely, in order to prove \forall x \in A . P(x) you must do the following: pick a letter which has not been used so far, say z, assume z \in A and prove P(z). Here the fresh letter is a free parameter, and it cannot be eliminated without significant changes to how we write down things.

]]> By: Richard Baron http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12234 Wed, 05 Oct 2011 13:52:22 +0000 http://gowers.wordpress.com/?p=3346#comment-12234 If we are keen to tie down the logic that lies behind “Oh, and by the way, this deduction works for all n”, should it go like this?

1. Write out the proof. We start with P(n), then after several lines get to R(n), then after more lines to S(n), then eventually to Q(n). There is no binding of n in any of this.

2. Use the deduction theorem several times, with liberal application of brackets, to convert the lines of proof into one gigantic sequence of nested conditionals.

3. Now we have a single open formula, with free n. Put one last big pair of brackets round it.

4. Finally, put the universal quantifier, for all n, on the front.

Then no logician will be able to accuse a mathematician of any sloppiness. But we do need to be sure that the deduction theorem applies to the logic we implicitly invoke in writing proofs. (I guess it does.)

]]> By: gowers http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12233 Wed, 05 Oct 2011 13:33:58 +0000 http://gowers.wordpress.com/?p=3346#comment-12233 As a matter of fact, I do have a half-written post in which I discuss at some length the difference between free and bound variables. However, I’m not sure I agree with your second sentence. I think that when such a sentence is uttered, n is sometimes free and sometimes universally quantified — but only implicitly. For instance, if I say, “I discovered a cool fact yesterday: if p is a prime of the form 4m+1 then it can be written as a sum of two squares,” then I would maintain that what I actually mean is that every prime of the form 4m+1 can be written as a sum of two squares. However, if I start to prove this result and begin by saying, “Let p be a prime of the form 4m+1,” then p has that mysterious fixed-but-arbitrary status and perhaps it’s better to call it free.

I’m not sure I completely buy the function/integral analogy, but you may be able to persuade me. If I take a statement P that involves a parameter n (presumably the analogue of the function) and form the statement \forall n\ \ P(n) then I get a statement without parameters, just as if I integrate f between 0 and 1 then I get a number that doesn’t depend on the variable that the function takes. But in the first case I get a statement from which I can deduce all the individual statements P(n), whereas from the definite integral I can’t say anything about individual values of f. So it seems to me that integrating is throwing away information in a way that universally quantifying isn’t.

I’m also interested by your last three words. I can see that having free parameters in the middle of proofs is extremely natural, and of course I do it myself, but why can’t one avoid it by simply universally quantifying everything? For instance, if I want to prove that for every n\in\mathbb{N}\ \ P(n)\implies Q(n), I would normally say, “Let n\in\mathbb{N} and suppose that P(n),” and proceed to deduce Q(n). But …

Hmm … honesty compels me to leave that last paragraph there, but I think I now see the answer to my question. If, for example, my argument went P(n)\implies R(n)\implies S(n)\implies Q(n) then the first line of my proof basically has to be that P(n)\implies R(n). I certainly can’t start with \forall n\in\mathbb{N}\ \ P(n) as my initial assumption (since I need to deduce Q(n) from P(n) and not from the quite possibly false statement that P holds for all n). So if I “universally quantify over everything”, what I’m doing is taking P(n) as my premise and then saying, “Oh, and by the way, this deduction works for all n,” at the end of the proof. Is that what you mean when you say that free parameters are unavoidable?

]]> By: andrejbauer http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12232 Wed, 05 Oct 2011 13:06:02 +0000 http://gowers.wordpress.com/?p=3346#comment-12232 About the beginning of your post. Clearly, the $n$ in the two statements is a _free parameter_. It is _not_ quantified, and the two statements are clearly converses of each other. The trouble is that many mathematicians do not understand the difference between a free parameter and a universally quantified one. But the difference is there. It is the same as the difference between a function and its definite integral. Perhaps it would be a good idea to say something about free variables in these posts. They are necessary in hypothetical reasoning, i.e., in the middle of a proof we often do have free parameters, and unavoidably so.

]]> By: Simon Tatham http://gowers.wordpress.com/2011/10/05/basic-logic-relationships-between-statements-converses-and-contrapositives/#comment-12229 Wed, 05 Oct 2011 12:32:55 +0000 http://gowers.wordpress.com/?p=3346#comment-12229 ‘the two properties, “generate all of \mathbb{Z}” and “have a common factor bigger than 1″ are equivalent’: you meant the first one to be “do not generate all of \mathbb{Z}”, I think.

Many thanks — corrected now.

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