
#1
Oct310, 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! 



#2
Oct410, 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 nonnormal. 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. 



#3
Oct410, 08:02 AM

Sci Advisor
P: 1,707

Pick a normal number x. For each interval [tex][2^k,2^{k+1})[/tex], reduce one of the nonzero 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, xy is obviously irrational, but certainly not normal (it only contains 1's and 0's).




#4
Oct410, 09:13 AM

Sci Advisor
HW Helper
P: 3,680

Sum of 2 normal numbers still normal? 



#5
Oct410, 10:12 AM

P: 8

thanks for the replies.
i noticed that all the counterexamples 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 nonperfectsquare integers? for the sake of this question, we shall assume that the root of a nonperfectsquare integer is a normal number. the reason i find this question fascinating is because, the nonconstructed 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. 



#6
Oct410, 02:16 PM

Sci Advisor
HW Helper
P: 3,680

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.




#7
Oct410, 05:18 PM

P: 986





#8
Oct510, 10:27 AM

Sci Advisor
P: 1,707





#9
Oct510, 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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