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Old 2020-07-18, 07:15   #1
MattcAnderson
 
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"Matthew Anderson"
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Smile 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
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Old 2020-07-18, 07:36   #2
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Quote:
Originally Posted by MattcAnderson View Post
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
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Old 2020-07-18, 07:58   #3
MattcAnderson
 
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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
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Old 2020-07-18, 08:14   #4
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Quote:
Originally Posted by MattcAnderson View Post
Let S1 = sqrt(2) + sqrt(3)
I am pretty sure that S1 is irrational. However, I do not know how to prove this.
Did you try to google for an answer?
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Old 2020-07-18, 08:42   #5
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Quote:
Originally Posted by MattcAnderson View Post
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?
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Old 2020-07-18, 09:04   #6
MattcAnderson
 
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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
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Old 2020-07-18, 09:31   #7
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There is a recent Mathologer video about this. Quite nice to watch, actually (don't look to his eyes! )
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Old 2020-07-18, 11:47   #8
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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
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Old 2020-07-18, 16:19   #9
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Quote:
Originally Posted by retina View Post
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
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Old 2020-07-20, 03:53   #10
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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.
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Old 2020-07-20, 05:09   #11
MattcAnderson
 
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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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