Question

There is a sequence {x_n} of non-square numbers (example: 2,3,5,6,7,8,10,...). If m^2<=x_n<=(m+1)^2, then prove that m=[n+0.5], where [] is the greatest integer function.

Answer 1

Can we proceed by induction here. I used induction on m and n. (clearly m and n are positive integers)
the proof goes as:
Base step: for n=1, 1<2<4 u get m=1
Induction hypothesis: assume p=[q+0.5] to be true. add 1 on both sides and u get p+1=[q+0.5]+1=[g+1+0.5].
But this implies that the equation to be proved holds true for p+1 when it holds true for p.
So it holds true for all n, using principle of mathematical induction.
Is this correct? thnx. :)

Answer 2

is there any other method than induction... it will be gr8 if u come with one.. :)
also please correct me if i am wrong anywhere! :)

Answer 3

I don't think this is true, because there is no given relation between x_n and n other than that the sequence contains no squares...

Answer 4

yup, nowhereman thats the place where this question (and the proof) goes weird...there is no place where i m using any thing about squares!

Answer 5

Already your induction start won't work. You can not choose an arbitrary m. The proposition is that for any m, if the inequality holds, then m=[n+0.5], but you don't know what \(x_1\) is! So [n+0.5] will always be 1 but m could be any number.

Answer 6

can u tell me what is n exactly here?

Answer 7

if it is an integer then [n+.5] has no meaning...

Answer 8

k...nowhereman..u may be correct..
i am wrong in the induction hypothesis part...
but we can have that the equation holds true for all such q i.e. n
(its easy to prove and on similar lines just like the one i gave)

Answer 9

then the proof seems correct? (but nevertheless weird since no use of the fact that they are all non-squares)

Answer 10

No, the proposition is false. You can easily construct a sequence of non-square numbers where \(m^2 \leq x_n \leq (m+1)^2\) but \(m \neq n = [n+0.5]\).

Answer 11

so how will u prove?

Answer 12

You can not prove something which is wrong...

Answer 13

the question is right! how do u say its wrong?

Answer 14

take the sequence \(100n^2+1\) It certainly contains no squares. But the first element is \(101\) So for \(m=10\) you have \(m^2 = 100 \leq 101 = x_1 \leq 121 = (m+1)^2\) but you also have \(m = 10 \neq 1 = [1 + 0.5] = [n+0.5]\).

Answer 15

u must have the sequence starting with 2, i.e. x_1=2,x_2=3 and so on...not from any arbitrary non-square number....

Answer 16

u misinterpreted the question, nowhereman....

Answer 17

or maybe i should have explicitly told...sorry :)

Answer 18

u have the sequence {x_n} with x_1=2, x_2=3, x_3=5, ...

Answer 19

now having known this, is my proof correct?

Answer 20

Wait, so should the proposition be "Let \((x_n)_{n=1}^∞\) be the ascending sequence of all non-square integers."?

Answer 21

yep...and the sequence satisfies the given inequality

Answer 22

Then you really should have wrote so in the beginning. Mathematics is all about exactness. And "a sequence" and "example" say something completely different from what you meant...

Answer 23

i m sorry..i told u.. :) so my proof correct?

Answer 24

It still doesn't work out. Let's look at the second element in the sequence. There you have \(n=2\) and \(x_2 = 3\) so for \(m=1\) you have the inequality, but \(m = 1 \neq 2 = [n+0.5]\)