Sum of 2 normal numbers still normal?

In summary: So the idea is to find a way to reduce the number of digits in the base where destroying a lot of digits would ruin normality.In summary, the conversation revolved around the question of whether the sum of two normal numbers is still normal, particularly when the numbers are integer multiples of square roots of non-perfect-square integers. Various counter-examples were presented, and the difficulty of solving this problem in general was acknowledged. A method of constructing a normal number through reducing certain digits was proposed, but the issue of maintaining normality in all bases was raised. The conversation ended with a discussion on finding a way to reduce the number of digits in a base where destroying a lot of digits would not affect normality.
  • #1
Liji.h
8
0
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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  • #2
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.
 
  • #3
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).
 
  • #4
Liji.h said:
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.
 
  • #5
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.
 
  • #6
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
Jarle said:
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)?
 
  • #8
hamster143 said:
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.
 
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  • #9
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.
 

Related to Sum of 2 normal numbers still normal?

1. What is a normal number?

A normal number is a number that follows a specific distribution pattern, where the majority of values cluster around the mean (average) and follow the familiar bell-shaped curve called the normal distribution.

2. How do you define the sum of two normal numbers?

The sum of two normal numbers is simply the result of adding the two numbers together. For example, if we have two normal numbers, A and B, their sum would be A+B.

3. Is the sum of two normal numbers always a normal number?

Yes, the sum of two normal numbers is always a normal number. This is because the normal distribution is closed under addition, meaning that when you add two normal numbers, the resulting number will still follow the normal distribution.

4. Are there any exceptions to the rule that the sum of two normal numbers is still normal?

No, there are no exceptions to this rule. As mentioned before, the normal distribution is closed under addition, which means that the sum of any two normal numbers will always be a normal number.

5. How does the sum of two normal numbers affect the mean and standard deviation?

When two normal numbers are added, the mean of the resulting sum will be the sum of the individual means. The standard deviation of the resulting sum will be the square root of the sum of the squares of the individual standard deviations. In other words, the mean and standard deviation of the sum will be affected by the individual values of the two normal numbers being added.

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