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

10110010111112 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 25416 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 2×11×397 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 59610 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

 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

All times are UTC. The time now is 10:42.

Mon Sep 21 10:42:48 UTC 2020 up 11 days, 7:53, 0 users, load averages: 1.70, 1.81, 1.67