Re: [math-fun] "I don't find this proof beautiful" quote?
On 2018-05-29 09:59, Mike Stay wrote:
My favorite irrationality proof is one I heard from John Baez. Suppose the cube root of two were not irrational; then there would be two positive integers p, q such that p/q = ∛2. Multiplying both sides by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in the positive integers by Fermat's Last Theorem!
I just read this aloud to Rohan, who remarked "You can also use that for ∛9.", barely looking up from his video game. --rwg
:D https://cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf On Wed, May 30, 2018 at 2:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2018-05-29 09:59, Mike Stay wrote:
My favorite irrationality proof is one I heard from John Baez. Suppose the cube root of two were not irrational; then there would be two positive integers p, q such that p/q = ∛2. Multiplying both sides by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in the positive integers by Fermat's Last Theorem!
I just read this aloud to Rohan, who remarked "You can also use that for ∛9.", barely looking up from his video game. --rwg _______________________________________________ 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.math.ucr.edu/~mike http://reperiendi.wordpress.com
Is there any progress on the sum-of-three-cubes problem? I think 33 is the smallest number for which the answer is not known. On Wed, May 30, 2018 at 5:29 PM, Mike Stay <metaweta@gmail.com> wrote:
:D https://cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf
On Wed, May 30, 2018 at 2:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2018-05-29 09:59, Mike Stay wrote:
My favorite irrationality proof is one I heard from John Baez. Suppose the cube root of two were not irrational; then there would be two positive integers p, q such that p/q = ∛2. Multiplying both sides by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in the positive integers by Fermat's Last Theorem!
I just read this aloud to Rohan, who remarked "You can also use that for ∛9.", barely looking up from his video game. --rwg _______________________________________________ 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.math.ucr.edu/~mike http://reperiendi.wordpress.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Puzzle: How far can we take this idea? It obviously applies to cbrt7 as well, and to 2^(1/n) for n>=3. We can also go from "cbrt9 is irrational" to "cbrt3 is irrational". Rich --------------- Quoting Bill Gosper <billgosper@gmail.com>:
On 2018-05-29 09:59, Mike Stay wrote:
My favorite irrationality proof is one I heard from John Baez. Suppose the cube root of two were not irrational; then there would be two positive integers p, q such that p/q = ?2. Multiplying both sides by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in the positive integers by Fermat's Last Theorem!
I just read this aloud to Rohan, who remarked "You can also use that for ?9.", barely looking up from his video game. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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