Sum of 2 normal numbers still normal?


by Liji.h
Tags: normal, numbers
Liji.h
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#1
Oct3-10, 10:33 AM
P: 8
Recently when I talked to a professor specialized in number theory, I learned about normal numbers.

So the other day I asked myself, is the sum of two normal numbers still normal? I quickly realized if you take [tex]\sqrt{2}[/tex] and [tex]-\sqrt{2}[/tex] you get 0 so the answer is obviously no.

But then I started to wonder, what if we know the sum is an irrational number? is it then normal?

I am intuitively guessing that the answer is yes...

Thanks in advance for any input!
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hamster143
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#2
Oct4-10, 04:54 AM
P: 986
This seems like a hard problem. But you can construct a strong hint that that it's false for numbers normal in any particular base (say, base 10).

First of all, if x is normal, then so is [itex]\lceil x \rceil - x[/itex], where [itex]\lceil x \rceil[/itex] is the first integer larger than x. (It's decimal representaton consists entirely of 9's.) But their sum is clearly not normal.

Now, all you have to do to disprove it, is to replace enough digits in [itex]\lceil x \rceil[/itex] to make it irrational, without making either of the two summands non-normal. It should be achievable, for example, by replacing the '9' at 10^n'th position with a '8' iff the colocated digit in x is <=8.
disregardthat
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#3
Oct4-10, 08:02 AM
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Pick a normal number x. For each interval [tex][2^k,2^{k+1})[/tex], reduce one of the non-zero digits between the [tex]2^k[/tex] (included) and [tex]2^{k+1}[/tex]'th (not included) digit by 1. If all digits in this interval is 0 do nothing. Since it is normal you can do this infinitely many times. Let the resulting number be y. It shouldn't be too hard to prove that y is also normal, since we change x so "rarely" along its expansion ("exponentially"). However, x-y is obviously irrational, but certainly not normal (it only contains 1's and 0's).

CRGreathouse
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#4
Oct4-10, 09:13 AM
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Sum of 2 normal numbers still normal?


Quote Quote by Liji.h View Post
But then I started to wonder, what if we know the sum is an irrational number? is it then normal?
Suppose x is a transcendental (I assume this is what you meant when you wrote "irrational") normal number. Then y_b is normal, where y_b is x with its base-b digits d replaced by b - d (or 0, if d = 0). But if b > 2, x + y_b is clearly not normal since all of its base-b digits are 0 or 1. It remains to be proven that the sum is transcendental, but this seems clear (even if difficult). At the least it should be provable by density that such sums exist.
Liji.h
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#5
Oct4-10, 10:12 AM
P: 8
thanks for the replies.

i noticed that all the counter-examples are irrationals that are constructed in a particular way.

now let me make the question more specific.

What if the numbers that we are adding up are the integer multiples of square roots of two non-perfect-square integers?

for the sake of this question, we shall assume that the root of a non-perfect-square integer is a normal number.

the reason i find this question fascinating is because, the non-constructed normal numbers that we see everyday, like [tex]\sqrt{2}[/tex], e and [tex]\pi[/tex] look very "random", so to me, they are normal in every way possible. however, if you construct a normal number, it may not be normal in another base, also constructed normal numbers have absolutely no "relationship" to an integer, where as, ([tex]\sqrt{2}[/tex])^2=2, Lim (1+1/n)^n=e, [tex]\pi[/tex]/2=[tex]\frac{22446688...}{13355779}[/tex] etc.
CRGreathouse
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#6
Oct4-10, 02:16 PM
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I'm sure this is too hard to solve in general, since deciding if [tex]\sqrt2+\sqrt2[/tex] is normal would presumably determine whether [tex]\sqrt2[/tex] was normal.
hamster143
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#7
Oct4-10, 05:18 PM
P: 986
Quote Quote by Jarle View Post
Pick a normal number x. For each interval [tex][2^k,2^{k+1})[/tex], reduce one of the non-zero digits between the [tex]2^k[/tex] (included) and [tex]2^{k+1}[/tex]'th (not included) digit by 1. If all digits in this interval is 0 do nothing. Since it is normal you can do this infinitely many times. Let the resulting number be y. It shouldn't be too hard to prove that y is also normal, since we change x so "rarely" along its expansion ("exponentially"). However, x-y is obviously irrational, but certainly not normal (it only contains 1's and 0's).
Are you sure that you can prove that y is still normal in all bases (as opposed to the one base where you did the reducing)?
disregardthat
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#8
Oct5-10, 10:27 AM
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Quote Quote by hamster143 View Post
Are you sure that you can prove that y is still normal in all bases (as opposed to the one base where you did the reducing)?
Good point, I overlooked that. Each reduction is subtracting a number on the form 0.00...0010000... in base 10, and it will maintain its form in every base. If we could somehow ensure that this wouldn't change the number too much, it might work. Subtraction 0.00...0100... from x would change all consecutive sequences of 0's in x ending in the same decimal place (or in any other base). I don't see how we can be sure to maintain normality, too many 0's could be destroyed in the process in some base.
hamster143
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#9
Oct5-10, 05:34 PM
P: 986
According to wolframalpha, 0.001000000000 in base 10 is 0.0(15343a0b62a68781b059) in base 12. So, subtracting 0.00100000000 will not change the number too much in base 10, but it will seriously wreck it in base 12.


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