In a column, write the integer part of the integer multiples of . Then, in the second column, write the missing integers in increasing order. In the third column, write the difference between the second and first columns. You get :
1 3 2
2 6 4
4 10 6
5 13 8
7 17 10
8 20 12
9 23 14
11 27 16
12 30 18
14 34 20
15 37 22
16 40 24
18 44 26
19 47 28
21 51 30
22 54 32
24 58 34
25 61 36
26 64 38
28 68 40
29 71 42
Why does the third column consist of the even integers?
Hmmm...do you have a proof that the third column is in fact the set of positive even integers? Any easy argument would seem to require that the first column follows a pattern - at first glance it is unclear to me if this is true.
Yes, it's true, and I'm challenging you to find the proof!
Originally Posted by DrSteve
HI , i think you are asking us to show that the form represents all the missing integers . I am now going to prove this fact .
Before the proof , let me call the integers which cannot be expressed as happy integers . It is to avoid typing so much !
Observation 1 :
Let be a positive integer , then is happy if and only if .
Proof : It is easy to check that is happy if and only if there is no real number of the form lying between and , since the common difference of the sequence is , we can see that there must be such a number lying between and . In other words , there exists such that or . Therefore ,
Now i am going to show all the integers of the form are happy :
Since we have
Therefore , is happy for all positive integers .
Observation 2 :
If is happy , the next happy integer is greater than .
( since )
so is not happy .
( since )
so is not happy !
We can see that is the first happy integer which is of this form . Moreover , . The conclusion is : there doesn't exist a happy integer lying between and exclusively as from the second observation , the separation between them will be greater than 4 which is impossible .
Therefore , represents all the happy integers ! (Happy)
Good job simplependulum, you rock as always!
This fact is a consequence of the following more general and quite surprising theorem (a theorem which, obviously, you weren't expected to know off the top of your head) : the sets form a partition of the integers if and only if is irrational and .
So with , we find that , which solves the problem!