[math-fun] Theorem-dependance: possibly trivial question
Let us say that, if Theorem 1 (T1) can be used to prove Theorem 2 (T2), then T1 --> T2. Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children? Are there cases of longer loops T1 --> ... --> Tn --> T1 which cannot, or cannot naturally, be shortened ? Schoolchildren-compatable preferred again ! Thanks - Guy
Let us say that, if Theorem 1 (T1) can be used to prove Theorem 2 (T2), then T1 --> T2. Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children? Are there cases of longer loops T1 --> ... --> Tn --> T1 which cannot, or cannot naturally, be shortened ? Schoolchildren-compatable preferred again !
What is a school child? Victor miller told me he read Hogben's Mathematics for the Million at an age that put me to shame, five or eight or eleven or something like that. (Perhaps he will chime in and correct.) Ignoring the school child requirement, I think you find constructions of that sort in Halmos Naive Set Theory in regard to a variety of things equivalent to the axiom of choice. It also seems familiar from algebra --- ``the following definitions of a group are all equivalent'' ---- and perhaps topology. I doubt the shortest-chain requirement can be formalized. If T1 --> T2 --> T3, you can usually rewrite the proof to avoid explicit statement of T2. This comes up at ``problem-set level.'' I recall being asked to proved something without the mean-value theorem: no one in class could do it but we sure found some ways of disguising it. I suspect it also occurs at ``Fields-Medal level.'' There is a paper by someone Reed in the '50 proving (if I recall it right) that differentiable equals analytic without the Cauchy integral theorem. This may be a case where it took a century to squeeze out T2. There is another Minsky thing that works like this. Minsky said that the way to handle trigonometric-identities problems on exams is to write the left hand side at the top and the right hand side at the bottom. You then work forward from the top and backward from the bottom and put an equal sign in the middle. Since it is not clear what the rules of inference are in trigonometric identities, you are most likely to be graded correct. (And you get untll the exam is returned to figure out an explanation of why the middle equal is obvious.) The point is that if you put a trigonometric identities proof in your form, who can say how many intermediate Ts there are. Whit
Date: Mon, 23 Mar 2015 09:49:50 -0700 From: Whitfield Diffie <whitfield.diffie@gmail.com>
Let us say that, if Theorem 1 (T1) can be used to prove Theorem 2 (T2), then T1 --> T2. Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children? Are there cases of longer loops T1 --> ... --> Tn --> T1 which cannot, or cannot naturally, be shortened ? Schoolchildren-compatable preferred again !
What is a school child? Victor miller told me he read Hogben's Mathematics for the Million at an age that put me to shame, five or eight or eleven or something like that. (Perhaps he will chime in and correct.)
When I was five or six I learned that there was a "Little Golden Book of Mathematics" and also a "World of Mathematics". I saved my allowance for months, waiting for a family trip to San Francisco where there were large bookstores that had these treasures. Dressed properly for The City, wearing a new dress, coat, Mary Jane shoes, and gloves, I handed over my money at City of Paris, thrilled with the purchases. It seems almost unbelieveable to think that there was a time when information was so hard to come by, yet the acquisition of the books was sublimely wonderful, as magical as any of fairy tales that were available in my otherwise favorite books --- the Bookhouse books.
Ignoring the school child requirement, I think you find constructions of that sort in Halmos Naive Set Theory in regard to a variety of things equivalent to the axiom of choice. It also seems familiar from algebra --- ``the following definitions of a group are all equivalent'' ---- and perhaps topology.
I doubt the shortest-chain requirement can be formalized. If T1 --> T2 --> T3, you can usually rewrite the proof to avoid explicit statement of T2. This comes up at ``problem-set level.'' I recall being asked to proved something without the mean-value theorem: no one in class could do it but we sure found some ways of disguising it. I suspect it also occurs at ``Fields-Medal level.'' There is a paper by someone Reed in the '50 proving (if I recall it right) that differentiable equals analytic without the Cauchy integral theorem. This may be a case where it took a century to squeeze out T2.
There is another Minsky thing that works like this. Minsky said that the way to handle trigonometric-identities problems on exams is to write the left hand side at the top and the right hand side at the bottom. You then work forward from the top and backward from the bottom and put an equal sign in the middle. Since it is not clear what the rules of inference are in trigonometric identities, you are most likely to be graded correct. (And you get untll the exam is returned to figure out an explanation of why the middle equal is obvious.) The point is that if you put a trigonometric identities proof in your form, who can say how many intermediate Ts there are.
In high school I remember one of my classmates in analytic geometry trying to prove a theorm at the blackboard. He was animated and enthusiastic, drawing on almost everything in three years of high school math, covering several feet of board space with geometric diagrams and trigonmetric equations and algebra. It was definitely entertaining, though I suspect his point may have been to use up the class period so thoroughly as to shut out the easy going and boring teacher. I waited until the very end to point out his error in the middle of the derivation. A good time was had by all. Hilarie
I read Mathematics for the Million at age 6 (or maybe 7). But I don't think that's the target audience they had in mind 😄 Victor Sent from my iPhone On Mar 23, 2015, at 12:49, Whitfield Diffie <whitfield.diffie@gmail.com> wrote:
Let us say that, if Theorem 1 (T1) can be used to prove Theorem 2 (T2), then T1 --> T2. Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children? Are there cases of longer loops T1 --> ... --> Tn --> T1 which cannot, or cannot naturally, be shortened ? Schoolchildren-compatable preferred again !
What is a school child? Victor miller told me he read Hogben's Mathematics for the Million at an age that put me to shame, five or eight or eleven or something like that. (Perhaps he will chime in and correct.)
Ignoring the school child requirement, I think you find constructions of that sort in Halmos Naive Set Theory in regard to a variety of things equivalent to the axiom of choice. It also seems familiar from algebra --- ``the following definitions of a group are all equivalent'' ---- and perhaps topology.
I doubt the shortest-chain requirement can be formalized. If T1 --> T2 --> T3, you can usually rewrite the proof to avoid explicit statement of T2. This comes up at ``problem-set level.'' I recall being asked to proved something without the mean-value theorem: no one in class could do it but we sure found some ways of disguising it. I suspect it also occurs at ``Fields-Medal level.'' There is a paper by someone Reed in the '50 proving (if I recall it right) that differentiable equals analytic without the Cauchy integral theorem. This may be a case where it took a century to squeeze out T2.
There is another Minsky thing that works like this. Minsky said that the way to handle trigonometric-identities problems on exams is to write the left hand side at the top and the right hand side at the bottom. You then work forward from the top and backward from the bottom and put an equal sign in the middle. Since it is not clear what the rules of inference are in trigonometric identities, you are most likely to be graded correct. (And you get untll the exam is returned to figure out an explanation of why the middle equal is obvious.) The point is that if you put a trigonometric identities proof in your form, who can say how many intermediate Ts there are.
Whit
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There are a lot of theorem of the form The following are equivalent: Condition_1, ..., Condition_N. Occasionally the standard proof of such a theorem is of the form Condition_1 => Condition_2 => ... => Condition_N => Condition_1 . --Dan
On Mar 23, 2015, at 2:29 AM, Guy Haworth <g.haworth@reading.ac.uk> wrote:
Let us say that, if Theorem 1 (T1) can be used to prove Theorem 2 (T2), then T1 --> T2.
Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children?
Are there cases of longer loops T1 --> ... --> Tn --> T1 which cannot, or cannot naturally, be shortened ? Schoolchildren-compatable preferred again !
Thanks - Guy
Here's one example that can be proved in the format of Condition_1 => Condition_2 => ... => Condition_27 => Condition_1 : Theorem ------- Let A be an n×n matrix, and let T : R^n → R^n be the linear transformation defined by T(x) = Ax for x ∈ R^n. Then the following are equivalent: • A is invertible. • There exists an n × n matrix B such that BA = I_n. • There exists an n × n matrix C such that AC = I_n. • Every column of A is a pivot column. • Every column of rref(A) is a pivot column. • rref(A) has no zero rows. • rref(A)=I_n. • The only solution to Ax=0 is x=0. • For every b ∈ R^n, Ax = b has at least one solution x ∈ Rn. • For every b ∈ R^n, Ax=b has at most one solution x∈R^n. • rank A=n. • nullity A = 0. • Null(A) is the zero subspace of R^n. • The columns of A span R^n. • The columns of A are linearly independent. • The columns of A are a basis for R^n. • T is onto. • T is one-to-one. • T is invertible. • im(T) =R^n. • rank T = n. • nullity T = 0. • ker(T) is the zero subspace of R^n. • A is a product of elementary matrices. • det(A) is not equal to 0. • 0 is not an eigenvalue of A. • 0 is not an eigenvalue of T. --Dan
On Mar 23, 2015, at 10:56 AM, Dan Asimov <asimov@msri.org> wrote:
There are a lot of theorem of the form
The following are equivalent: Condition_1, ..., Condition_N.
Occasionally the standard proof of such a theorem is of the form
Condition_1 => Condition_2 => ... => Condition_N => Condition_1 .
--Dan
On Mar 23, 2015, at 2:29 AM, Guy Haworth <g.haworth@reading.ac.uk> wrote:
Let us say that, if Theorem 1 (T1) can be used to prove Theorem 2 (T2), then T1 --> T2.
Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children?
Are there cases of longer loops T1 --> ... --> Tn --> T1 which cannot, or cannot naturally, be shortened ? Schoolchildren-compatable preferred again !
Thanks - Guy
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="Guy Haworth" <g.haworth@reading.ac.uk> Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children?
Are these too trivial? A = B + C B = A - C C = A - B Might one take the first as given, prove the second as T1 then show it implies the third as T2, and vice versa?
How about Ko? http://senseis.xmp.net/?Ko On Mon, Mar 23, 2015 at 1:35 PM, Marc LeBrun <mlb@well.com> wrote:
="Guy Haworth" <g.haworth@reading.ac.uk> Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children?
Are these too trivial? A = B + C B = A - C C = A - B
Might one take the first as given, prove the second as T1 then show it implies the third as T2, and vice versa?
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Another very simple example would be a statement and its contrapositive. For instance: If Spot is a dog, then Spot is a mammal. If Spot is not a mammal, then Spot is not a dog. Or perhaps: If Spot is a dog, then Spot is not a bird. If Spot is a bird, then Spot is not a dog. Tom Mike Stay writes:
How about Ko? http://senseis.xmp.net/?Ko
On Mon, Mar 23, 2015 at 1:35 PM, Marc LeBrun <mlb@well.com> wrote:
="Guy Haworth" <g.haworth@reading.ac.uk> Are there simple examples of 'equivalence', T1 --> T2 and T2 --> T1 (setting up loop T1 --> T2 --> T1) that could be explained to school children?
Are these too trivial? A = B + C B = A - C C = A - B
Might one take the first as given, prove the second as T1 then show it implies the third as T2, and vice versa?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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participants (8)
-
Dan Asimov -
Guy Haworth -
Hilarie Orman -
Marc LeBrun -
Mike Stay -
Tom Karzes -
Victor S. Miller -
Whitfield Diffie