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

[QUOTE=MattcAnderson;550906]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.[/QUOTE]I choose k = 0

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

[QUOTE=MattcAnderson;550910]
Let S1 = sqrt(2) + sqrt(3)
I am pretty sure that S1 is irrational. However, I do not know how to prove this.
[/QUOTE]
Did you try to google for an answer?

[QUOTE=MattcAnderson;550910]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.[/QUOTE]For k = 0: S = sqrt(0) = 0

Or do I misunderstand?

A google search of
"Is square root of 2 plus square root of 3 rational"
helped me find a webpage at [URL="http://proofsfromthebook.com/2012/12/24/root-of-2-plus-root-of-3-irrational/"]proofsfromthebook.com[/URL]

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

There is a recent Mathologer video about this. Quite nice to watch, actually (don't look to his eyes! :razz:)

Since 1 is the only squarefree positive integer that is also a square, you could specify that a[sub]i[/sub] is squarefree, and a[sub]i[/sub] > 1 for every i from 0 to k.

Assertion: If no non-empty product of the a[sub]i[/sub] is a square, then the numbers 1, sqrt(a[sub]0[/sub]), sqrt(a[sub]1[/sub]), ... sqrt(a[sub]k[/sub]) are linearly independent over Q (field of rational numbers). [This assertion implies your result.]

Sketch of proof:

Let p[sub]1[/sub], p[sub]2[/sub], ..., p[sub]r[/sub] be distinct prime numbers such that a[sub]i[/sub] divides the product p[sub]1[/sub]p[sub]2[/sub]...p[sub]r[/sub] for i = 0 to k.

The extension K = Q(sqrt(p[sub]1[/sub]), sqrt(p[sub]2[/sub]), ... sqrt(p[sub]r[/sub])) is of degree 2[sup]r[/sup] over Q.

For each subset s of S = {1, 2, ...r} let P[sub]s[/sub] be the product of the primes p[sub]i[/sub], i in s. Then there are 2[sup]r[/sup] products P[sub]s[/sub]. These products include 1 (empty product) and all the a[sub]i[/sub]. Every element of K can be expressed as a Q-linear combination of these 2[sup]r[/sup] 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.

[QUOTE=retina;550912]For k = 0: S = sqrt(0) = 0

Or do I misunderstand?[/QUOTE]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".

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.

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