 mersenneforum.org > Math a question regarding irrational numbers
 Register FAQ Search Today's Posts Mark Forums Read 2020-07-18, 07:15 #1 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 22·149 Posts a question regarding irrational numbers Hi again all, Please look at the following conjecture. Note that in mathematics, conjecture means 'probably true'. Conjecture about irrational numbers Let k be an arbitrary chosen counting number. Consider S = sqrt(a0) + sqrt(a1) + ... + sqrt(ak). Where a0, a1, ... , ak are square free positive integers. Conjecture that S is therefore irrational. This might be an open question. For what its worth Matt   2020-07-18, 07:36   #2
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

3·23·83 Posts Quote:
 Originally Posted by MattcAnderson Let k be an arbitrary chosen counting number. Consider S = sqrt(a0) + sqrt(a1) + ... + sqrt(ak). Where a0, a1, ... , ak are square free positive integers. Conjecture that S is therefore irrational.
I choose k = 0   2020-07-18, 07:58 #3 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 22·149 Posts k=0 is an excellent choice. From the rational root theorem, we have that the square root of any positive integer that is not itself a perfect square, is irrational. Consider an example with k=1. Let S1 = sqrt(2) + sqrt(3) I am pretty sure that S1 is irrational. However, I do not know how to prove this. Regards, Matt   2020-07-18, 08:14   #4
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

53·73 Posts Quote:
 Originally Posted by MattcAnderson Let S1 = sqrt(2) + sqrt(3) I am pretty sure that S1 is irrational. However, I do not know how to prove this.   2020-07-18, 08:42   #5
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

131378 Posts Quote:
 Originally Posted by MattcAnderson k=0 is an excellent choice. From the rational root theorem, we have that the square root of any positive integer that is not itself a perfect square, is irrational.
For k = 0: S = sqrt(0) = 0

Or do I misunderstand?   2020-07-18, 09:04 #6 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 22×149 Posts A google search of "Is square root of 2 plus square root of 3 rational" helped me find a webpage at proofsfromthebook.com The restriction on the a_i is that the numbers are Positive and not squares. So zero is not allowed in this conjecture. The various a_i must be in the set {2,3,5,6,7,8,10, …} Regards, Matt Last fiddled with by MattcAnderson on 2020-07-18 at 09:06 Reason: added some set notation for clarity   2020-07-18, 09:31 #7 LaurV Romulan Interpreter   Jun 2011 Thailand 210368 Posts There is a recent Mathologer video about this. Quite nice to watch, actually (don't look to his eyes! )   2020-07-18, 11:47 #8 Dr Sardonicus   Feb 2017 Nowhere 22×5×173 Posts Since 1 is the only squarefree positive integer that is also a square, you could specify that ai is squarefree, and ai > 1 for every i from 0 to k. Assertion: If no non-empty product of the ai is a square, then the numbers 1, sqrt(a0), sqrt(a1), ... sqrt(ak) are linearly independent over Q (field of rational numbers). [This assertion implies your result.] Sketch of proof: Let p1, p2, ..., pr be distinct prime numbers such that ai divides the product p1p2...pr for i = 0 to k. The extension K = Q(sqrt(p1), sqrt(p2), ... sqrt(pr)) is of degree 2r over Q. For each subset s of S = {1, 2, ...r} let Ps be the product of the primes pi, i in s. Then there are 2r products Ps. These products include 1 (empty product) and all the ai. Every element of K can be expressed as a Q-linear combination of these 2r products. They are, therefore, a Q-basis for K, hence linearly independent over Q. Thus any non-empty subset is also linearly independent over Q. Last fiddled with by Dr Sardonicus on 2020-07-18 at 11:49 Reason: xifnix ostpy   2020-07-18, 16:19   #9
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

2×32×5×72 Posts Quote:
 Originally Posted by retina For k = 0: S = sqrt(0) = 0 Or do I misunderstand?
I took his description to be sum of k terms.
The sum of 0 terms would be 0.
S=sum over 0 to k of sqrt(k). Depending on whether you test before or after summing, you get 1 or 0 terms. But 0 as a term may be as disqualified as 1.
k=0 would seem to be excluded by "counting number".

Last fiddled with by kriesel on 2020-07-18 at 16:23   2020-07-20, 03:53 #10 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 22×149 Posts I learned something. The mathologger video about irrationals was helpful. Wolfram Alpha has a command called MinimumPolynomial(). You input an expression, which can have square roots, and it will give you the monic polynomial with minimal degree, that has a root that is the argument in the command. This with the integer root theorem shows the irrationality of the expression S = sqrt(a0) + sqrt(a1) + … + sqrt(ak) and require all ai >2, and integer and not a perfect square. Done.   2020-07-20, 05:09 #11 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 22·149 Posts oops, I meant to allow 2 for any ai with 1>= i >= k Also, a related conjecture is Let S2 = q + sqrt(r). where q is any rational number and r is and integer and not a square and greater than one. show that S2 is irrational. Regards, Matt  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post a1call Miscellaneous Math 16 2017-05-13 22:50 LaurV Math 11 2012-03-16 12:10 jasong Soap Box 8 2006-05-22 17:32 WraithX GMP-ECM 1 2006-03-19 22:16 ColdFury Software 14 2003-06-20 01:26

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