No Divisors Between an Integer and Twice it

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In summary, the conversation discusses a problem where ##x## is a natural number between ##n## and ##2n##, and the question is whether ##x## can divide ##2n##. The attempt at a solution involves assuming that ##x## divides ##2n## and using algebraic manipulations to reach a contradiction. It is then suggested to substitute ##x## with ##\frac{2n}{k}## and divide by ##n## to make the argument more rigorous.
  • #1
Bashyboy
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Homework Statement


Suppose that ##x \in \mathbb{N}## is such that ##n < x < 2n##, where ##n## is another natural number. I want to conclude that it is impossible for ##x## to divide ##2n##. Also, is there a way to naturally generalize this to all integers?

Homework Equations

The Attempt at a Solution


Since ##x \in (n,2n)##< then ##x = n + \ell## for some natural number ##\ell \in (0,n)##. Now, suppose the contrary, that ##x## divides ##2n##. Then ##2n = k(n + \ell)## for some ##k \in \mathbb{Z}##. Immediately we can say that ##k## must be a positive integer. Rearranging the equation gives

##n(2-k) = k(n+\ell)##

If ##k## is either ##1## or ##2##, then we get ##n = \ell## or ##2 \ell = 0##, both of which are contradictions. If ##k > 2##, then the LHS will be negative but the RHS will be positive, a contradiction.

How do I make this a little more rigorous?
 
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  • #2
Bashyboy said:

Homework Statement


Suppose that ##x \in \mathbb{N}## is such that ##n < x < 2n##, where ##n## is another natural number. I want to conclude that it is impossible for ##x## to divide ##2n##. Also, is there a way to naturally generalize this to all integers?

Homework Equations

The Attempt at a Solution


Since ##x \in (n,2n)##< then ##x = n + \ell## for some natural number ##\ell \in (0,n)##. Now, suppose the contrary, that ##x## divides ##2n##. Then ##2n = k(n + \ell)## for some ##k \in \mathbb{Z}##. Immediately we can say that ##k## must be a positive integer. Rearranging the equation gives

##n(2-k) = k(n+\ell)##

If ##k## is either ##1## or ##2##, then we get ##n = \ell## or ##2 \ell = 0##, both of which are contradictions. If ##k > 2##, then the LHS will be negative but the RHS will be positive, a contradiction.

How do I make this a little more rigorous?
##2n = k(n + \ell) \nRightarrow n(2-k) = k(n+\ell)## and I don't see why you may assume ##k \in \{1,2\}##.
However, I don't think ##\ell## is needed at all. What do you get by simply assuming ##xk=2n##?
 
  • #3
fresh_42 said:
##2n = k(n + \ell) \nRightarrow n(2-k) = k(n+\ell)## and I don't see why you may assume ##k \in \{1,2\}##.
However, I don't think ##\ell## is needed at all. What do you get by simply assuming ##xk=2n##?

Whoops! Rearranging the equation should actually give ##n(2-k) = k \ell##. And I am not not assuming that ##k## always takes the values ##1## and ##2##: I was just considering special cases.

To answer your question, I don't know what ##xk=2n## implies.
 
  • #4
Bashyboy said:
Whoops! Rearranging the equation should actually give ##n(2-k) = k \ell##. And I am not not assuming that ##k## always takes the values ##1## and ##2##: I was just considering special cases.

To answer your question, I don't know what ##xk=2n## implies.
I meant, if you substituted ##x## by ##\frac{2n}{k}## in your assumption ##n<x<2n##, you could divide the whole thing by ##n## and see where it get's you to.
 

FAQ: No Divisors Between an Integer and Twice it

1. What does "No Divisors Between an Integer and Twice it" mean?

"No Divisors Between an Integer and Twice it" refers to a mathematical concept where there are no whole numbers that can evenly divide both an integer and twice that integer (2x). In other words, there are no numbers that can be multiplied by 2 and still result in the original integer.

2. Can any integer have no divisors between it and twice it?

No, not all integers have no divisors between it and twice it. For example, the integer 2 has a divisor (1) between it and twice it (4). However, there are some integers that do have this property, such as 3, 5, and 7.

3. How is "No Divisors Between an Integer and Twice it" used in mathematics?

This concept is used in various areas of mathematics, such as number theory and algebra. It can also be used to solve problems involving divisibility and factorization.

4. Are there any real-world applications of "No Divisors Between an Integer and Twice it"?

Yes, this concept has practical applications in fields such as cryptography, where it is used to create secure encryption algorithms. It can also be used in computer science for optimizing algorithms and data structures.

5. Can "No Divisors Between an Integer and Twice it" be extended to other multiples?

Yes, the concept of "No Divisors Between an Integer and Twice it" can be extended to other multiples, such as 3x, 4x, and so on. However, the property may not hold for all integers and their multiples, as it depends on the specific numbers involved.

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