Can a reductio ad adsurdum argument be rearranged into a direct argument that does not require a supposition? In short, are suppositions necessary for logical arguments?
On 2/1/2017 6:19 PM, David Wilson wrote:
Can a reductio ad adsurdum argument be rearranged into a direct argument that does not require a supposition?
As well as any logical argument. Any logical argument proceeds from premises and shows they entail a conclusion. Whether the premises are suppositions or facts is not relevant to the logic. An argument from absurdity is of the form X=>Y, Y is absurd and therefore false, not-Y => not-X ... which seems direct to me. Do you just mean you want it to prove a positive? in which case you can substitute Z = not-X so it becomes not-Z=>Y, Y is absurd and therefore false, not-Y=>Z. Brent Meeker
In short, are suppositions necessary for logical arguments?
The reduction ad absurdum argument to prove X generally goes not-X => ... => false, a contradiction, therefore X is true. e.g. not-X => A, not-X => not-A, hence x => A and not-A => F, therefore X is true. I've always found such arguments, while effective, in a way unsatisfying, because by the time you find your contradiction, you have wander far from your premise. I was wondering if there might always be a way to turn such an argument around, so that a proof of not-X => false, can be flipped to prove the contrapositive, true => X. I'm obviously not talking about building the argument around not-X => false and then invoking the contrapositive, but rather inverting the entire argument into a direct argument where X appears only as the last statement proven, not as part of an earlier supposition.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Brent Meeker Sent: Wednesday, February 01, 2017 10:00 PM To: math-fun Subject: Re: [math-fun] Logic question
On 2/1/2017 6:19 PM, David Wilson wrote:
Can a reductio ad adsurdum argument be rearranged into a direct argument that does not require a supposition?
As well as any logical argument. Any logical argument proceeds from premises and shows they entail a conclusion. Whether the premises are suppositions or facts is not relevant to the logic. An argument from absurdity is of the form X=>Y, Y is absurd and therefore false, not-Y => not-X ... which seems direct to me. Do you just mean you want it to prove a positive? in which case you can substitute Z = not-X so it becomes not-Z=>Y, Y is absurd and therefore false, not-Y=>Z.
Brent Meeker
In short, are suppositions necessary for logical arguments?
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If the 'reducto' proof is: not-X => A => B => ... => Y => Z, but Z is clearly false, hence not-not-X, ie X you should be able to invert that to be not-Z (which is clearly true) => not-Y => ... => not-B => not-A => X but good luck trying to find that from starting at the not-Z end. On 2017-02-02 17:15, David Wilson wrote:
The reduction ad absurdum argument to prove X generally goes not-X => ... => false, a contradiction, therefore X is true. e.g. not-X => A, not-X => not-A, hence x => A and not-A => F, therefore X is true. I've always found such arguments, while effective, in a way unsatisfying, because by the time you find your contradiction, you have wander far from your premise. I was wondering if there might always be a way to turn such an argument around, so that a proof of not-X => false, can be flipped to prove the contrapositive, true => X. I'm obviously not talking about building the argument around not-X => false and then invoking the contrapositive, but rather inverting the entire argument into a direct argument where X appears only as the last statement proven, not as part of an earlier supposition.
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Right. And it's even harder than finding the chain of implication; it's finding the a Z to start from. That's because you only know not-Z because Z is absurd. But a false proposition implies anything. So /*logically*/ you can start from any absurdity but to get anywhere you have to get to a Z that's implied by not-Y, which is implied by not-W, etc. Brent On 2/2/2017 5:56 PM, William R Somsky wrote:
If the 'reducto' proof is:
not-X => A => B => ... => Y => Z, but Z is clearly false, hence not-not-X, ie X
you should be able to invert that to be
not-Z (which is clearly true) => not-Y => ... => not-B => not-A => X
but good luck trying to find that from starting at the not-Z end.
On 2017-02-02 17:15, David Wilson wrote:
The reduction ad absurdum argument to prove X generally goes not-X => ... => false, a contradiction, therefore X is true. e.g. not-X => A, not-X => not-A, hence x => A and not-A => F, therefore X is true. I've always found such arguments, while effective, in a way unsatisfying, because by the time you find your contradiction, you have wander far from your premise. I was wondering if there might always be a way to turn such an argument around, so that a proof of not-X => false, can be flipped to prove the contrapositive, true => X. I'm obviously not talking about building the argument around not-X => false and then invoking the contrapositive, but rather inverting the entire argument into a direct argument where X appears only as the last statement proven, not as part of an earlier supposition.
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The reason for the “discomfort” with the proof by contradiction rule (RAA, reduction ad absurdum) is that it is not “constructive” (technically, not intuitionistic). The best way to understand this is by formalizing logic in natural deduction style. In intuitionistic logic (Brouwer) in natural deduction style (Gentzen, Prawitz), there is a rule that allows the deduction of ANY proposition from a deduction (from premises) that yields falsity (F). There is also a rule (implication introduction) that allows you to deduce P => Q from a deduction of Q under the premise P. The negation of a proposition neg P is an abbreviation for P => F (F is the symbol for falsity). Then the implication introduction rule yields the rule that says that if you can deduce F from P then P => F, that is neg P, is provable. In particular, if you can deduce F from neg P then you can deduce neg neg P. But, in intuitionistic logic P is NOT provable from neg neg P in general (the law of the excluded middle)! This is equivalent to the fact that P or neg P is NOT provable in general in intuitionistic logic. In fact, the proof by contradiction rule (RAA) is equivalent to adding P or neg P as axioms to intuitionistic logic, or to add axioms neg neg P => P in intuitionistic logic. I explain all this in chapter 2 of my book on discrete math (Section 2.4) see second edition in progress http://www.cis.upenn.edu/~jean/discmath-root-b.pdf Best, -- Jean Gallier
On Feb 2, 2017, at 9:08 PM, Brent Meeker <meekerdb@verizon.net> wrote:
Right. And it's even harder than finding the chain of implication; it's finding the a Z to start from. That's because you only know not-Z because Z is absurd. But a false proposition implies anything. So /*logically*/ you can start from any absurdity but to get anywhere you have to get to a Z that's implied by not-Y, which is implied by not-W, etc.
Brent
On 2/2/2017 5:56 PM, William R Somsky wrote:
If the 'reducto' proof is:
not-X => A => B => ... => Y => Z, but Z is clearly false, hence not-not-X, ie X
you should be able to invert that to be
not-Z (which is clearly true) => not-Y => ... => not-B => not-A => X
but good luck trying to find that from starting at the not-Z end.
On 2017-02-02 17:15, David Wilson wrote:
The reduction ad absurdum argument to prove X generally goes not-X => ... => false, a contradiction, therefore X is true. e.g. not-X => A, not-X => not-A, hence x => A and not-A => F, therefore X is true. I've always found such arguments, while effective, in a way unsatisfying, because by the time you find your contradiction, you have wander far from your premise. I was wondering if there might always be a way to turn such an argument around, so that a proof of not-X => false, can be flipped to prove the contrapositive, true => X. I'm obviously not talking about building the argument around not-X => false and then invoking the contrapositive, but rather inverting the entire argument into a direct argument where X appears only as the last statement proven, not as part of an earlier supposition.
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The mediant of any 2 positive rational numbers p/q and r/s is med(p/q, r/s) = (p+q)/(r+s). This can be thought of as a mapping med: Q+ x Q+ —> R from the cartesian product of the positive rationals with itself to the reals. Question: --------- Is med the restriction to Q+ x Q+ of some continuous function Med: R x R —> R ???
I've tried without success to look this up, but: The graph of the gamma function for negative reals has a critical point — call it x_n — between every pair of adjacent integers {-n, -n+1}. See e.g. https://en.wikipedia.org/wiki/Gamma_function#/media/File:Gamma_plot.svg <https://en.wikipedia.org/wiki/Gamma_function#/media/File:Gamma_plot.svg>. It appears that the points (x_n, |gamma(x_n)|) lie on the graph of some nice convex function y = f(x). Is there in some sense a "nicest" convex f that works here? * * * Alternatively, it likewise appears that the graph of y = |gamma(x)| for negative x has one tangency in each interval (-n, -n+1) (n > 0) with the graph of some nice convex function y = g(x) that lies very close to the graph of y = f(x) hinted at above. Sort of an "envelope". Perhaps it's more natural to nail the function g ??? If so, is there a unique "nicest" g that works here? —Dan
A kind soul has pointed out that I spaced out when typing the mediant. For positive fractions p/q and r/s assumed to be in lowest terms, their mediant (CORRECTED) is med(p/q, r/s) = (p+r)/(q+s) . (It's well known that med(p/q, r/s) always lies strictly between p/q and r/s.) —Dan
On Feb 3, 2017, at 1:08 AM, Dan Asimov <asimov@msri.org> wrote:
The mediant of any 2 positive rational numbers p/q and r/s is
[WRONG:] med(p/q, r/s) = (p+q)/(r+s).
This can be thought of as a mapping
med: Q+ x Q+ —> R
from the cartesian product of the positive rationals with itself to the reals.
Question: --------- Is med the restriction to Q+ x Q+ of some continuous function
Med: R x R —> R
???
I don't think med(.,.) is continuous as a map from QxQ to Q (which rules out the possibility of a continuous extension to RxR). Specifically, I claim that on every neighborhood of (1,2) in QxQ, med(p/q,r/s) takes on values arbitrarily close to 1 (when p,q are much larger than r,s) and values arbitrarily close to 2 (when r,s are much larger than p,q). But maybe the moral here is that we should use a different topology on QxQ. Jim On Friday, February 3, 2017, Dan Asimov <asimov@msri.org> wrote:
A kind soul has pointed out that I spaced out when typing the mediant.
For positive fractions p/q and r/s assumed to be in lowest terms, their mediant (CORRECTED) is
med(p/q, r/s) = (p+r)/(q+s)
. (It's well known that med(p/q, r/s) always lies strictly between p/q and r/s.)
—Dan
On Feb 3, 2017, at 1:08 AM, Dan Asimov <asimov@msri.org <javascript:;>> wrote:
The mediant of any 2 positive rational numbers p/q and r/s is
[WRONG:] med(p/q, r/s) = (p+q)/(r+s).
This can be thought of as a mapping
med: Q+ x Q+ —> R
from the cartesian product of the positive rationals with itself to the reals.
Question: --------- Is med the restriction to Q+ x Q+ of some continuous function
Med: R x R —> R
???
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participants (6)
-
Brent Meeker -
Dan Asimov -
David Wilson -
James Propp -
Jean Gallier -
William R Somsky