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Problem #1

Silly Druid

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Here is my first problem for you all to solve. I figured it out myself, and it's pretty easy if you know how to approach it.

It's a well-known fact (and easy to prove) that all square roots of non-square natural numbers are irrational. But knowing that, can you prove that the sum of any two such square roots is also irrational?

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It has been a while, I'm more applied maths guy and less proofing. But let's see if I can still do this :>


Let's put down what we know from the problem

image.png    (1)


Let's try proof by contradiction on this, so let's assume such roots would add up to a rational number p/q

image.png     (2)

With pq in Z and q ≠ 0

Any fraction/rational number you can split up in two smaller fraction which would add up the original fraction, like this

image.png    (3)

with of course

image.png    (4)


let's define these two new variables δ, ε regarding the difference of these fractions and those roots

image.png    (5)

Now let's think: δ, ε must be both unequal to zero. Because if these would be zero then square roots of n and m would be both rational numbers and would contradict the problem.

image.png     (6)

Now let's add these two equations/ definitions together

image.png    (7)

since all these fractions p_i/q_i are the same, we can use the earlier equation and add the sub fraction together to the original one

image.png    (8)

and this contradicts 1, since δ, ε ≠ 0.

Or in simple words: either  those roots of n and m are both rational numbers (δ, ε = 0) or they dont add up to a rational number (δ, ε ≠ 0).  When both conditions are applied then this leads to the contradiction of δ, ε = 0 and δ, ε ≠ 0 at the same time

So, disproved which means two rational numbers don't add up to a rational number

 My thoughts

I think this proof is a bit too unspecific to your problem/ I think there maybe a more direct one since this is very general way

This proof relies on the proof that a irrational number cannot be written as a fraction with p/q with p,q in Z and q unequal to 0. So I had the feeling while reading the problem that some things are known.

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@Astralshy About your proof: I don't fully understand it. Because p1/q1 + p2/q2 = sqrt(n) + sqrt(m), δ and ε must add up to 0 (one is positive and one is negative) so there is no contradiction, unless I'm missing something here (also image #4 doesn't work for me).

Here is my proof. Sorry for not using math notation, but I think it should be readable.


Let's say n and m are natural numbers that are not squares of other natural numbers (so no 0, 1, 4, 9, 16, etc.)

Now let's label the sum of their square roots as s:

sqrt (n) + sqrt (m) = s

Now, let's move sqrt (n) to the other side:

sqrt (m) = s - sqrt (n)

Square both sides:

m = s^2 - 2s sqrt (n) + n

Solve for sqrt (n):

2s sqrt (n)  = s^2 - m + n

sqrt (n) = (s^2 - m + n) / 2s

Now, if we assume that s is a rational number, then (s^2 - m + n) / 2s is also a rational number. And we know that sqrt (n) is irrational, so we have a contradiction. Hence, s must be irrational. QED

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