20200718, 07:15  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}·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 
20200718, 07:36  #2 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3·23·83 Posts 

20200718, 07:58  #3 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}×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 
20200718, 08:14  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
5^{3}·73 Posts 

20200718, 08:42  #5 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1011001011111_{2} Posts 

20200718, 09:04  #6 
"Matthew Anderson"
Dec 2010
Oregon, USA
254_{16} 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 20200718 at 09:06 Reason: added some set notation for clarity 
20200718, 09:31  #7 
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! )

20200718, 11:47  #8 
Feb 2017
Nowhere
2^{2}·5·173 Posts 
Since 1 is the only squarefree positive integer that is also a square, you could specify that a_{i} is squarefree, and a_{i} > 1 for every i from 0 to k.
Assertion: If no nonempty product of the a_{i} is a square, then the numbers 1, sqrt(a_{0}), sqrt(a_{1}), ... sqrt(a_{k}) are linearly independent over Q (field of rational numbers). [This assertion implies your result.] Sketch of proof: Let p_{1}, p_{2}, ..., p_{r} be distinct prime numbers such that a_{i} divides the product p_{1}p_{2}...p_{r} for i = 0 to k. The extension K = Q(sqrt(p_{1}), sqrt(p_{2}), ... sqrt(p_{r})) is of degree 2^{r} over Q. For each subset s of S = {1, 2, ...r} let P_{s} be the product of the primes p_{i}, i in s. Then there are 2^{r} products P_{s}. These products include 1 (empty product) and all the a_{i}. Every element of K can be expressed as a Qlinear combination of these 2^{r} products. They are, therefore, a Qbasis for K, hence linearly independent over Q. Thus any nonempty subset is also linearly independent over Q. Last fiddled with by Dr Sardonicus on 20200718 at 11:49 Reason: xifnix ostpy 
20200718, 16:19  #9 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2×3^{2}×5×7^{2} Posts 
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 20200718 at 16:23 
20200720, 03:53  #10 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}×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. 
20200720, 05:09  #11 
"Matthew Anderson"
Dec 2010
Oregon, USA
596_{10} 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 
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