Thread: A probability problem with roots

1. A probability problem with roots

What percentage of $\sqrt{n}$ where n belongs to Z are rational?

a) 1<(or the same)n<(or the same)2000?

(I am sorry, I do not know how to do the little line things).

Thank you!

2. Originally Posted by Coach

What percentage of $\sqrt{n}$ where n belongs to Z are rational?

a) 1<(or the same)n<(or the same)2000?

(I am sorry, I do not know how to do the little line things).

Thank you!
$\sqrt{n},\ n \in \bold{N}$ is rational if an only if $n$ is a perfect square.

So the question is what fraction of integers $1 \le n \le 2000$
are perfect squares (there are 44 perfect squares less than 2000, counting
not counting 0, can you prove that?).

RonL

3. do you mean Z or Q ?
members of Z are signed natural numbers
i mean if a belongs to Z -a or a belongs to N
as only positive numbers got a square root you may be had made a little confusion or mistake!

anyway there is a litle thing wrong in speaking of percentage of an infinite set
some will tell and make you the demonstration that they are as much even (or odd) numbers than natural numbers by making a bijection between those two set
though i believe they are fifty percent of the numbers that are odd (or even )

if i where you i'll tried to find if sqr(n) rational => n=m^2 (so n is a perfect square)
I'm not sure of that but i would bet
you'll easyly find the demonstration on this forum or in the net that sqr(2) is irrational maybe (i'd bet but i would'nt cut my head off) this demonstration could be addapted to others numbers which are not perfect square??

so if i were true for every numbers A they will be only entire-part(sqr(A)) numbers <=A that are perfect square so if i trust my feeling for wich sqr(them) are rational
so for a given number A you could find the 'percentage' but for entire N it will be zero at the limit (if suc a limit have a sence in term of percentage) and the half of zero for Z!

4. Originally Posted by SkyWatcher
....
anyway there is a litle thing wrong in speaking of percentage of an infinite set...
we're not dealing with an infinite set. only the integers between 0 and 2000 (inclusive)

5. Originally Posted by Jhevon
we're not dealing with an infinite set. only the integers between 0 and 2000 (inclusive)
Actually, the orginal question is about integers from 1 to 2000; $1 \le n \le 2000$.

6. Originally Posted by Plato
Actually, the orginal question is about integers from 1 to 2000; $1 \le n \le 2000$.
right. it's still not infinite though

7. Originally Posted by Plato
Actually, the orginal question is about integers from 1 to 2000; $1 \le n \le 2000$.
I must say that I thought it said 0, but on reexamination it says 1, I must
have read the original question in a parallel universe

RonL

8. Originally Posted by SkyWatcher
if i where you i'll tried to find if sqr(n) rational => n=m^2 (so n is a perfect square)
I'm not sure of that but i would bet
you'll easyly find the demonstration on this forum or in the net that sqr(2) is irrational maybe (i'd bet but i would'nt cut my head off) this demonstration could be addapted to others numbers which are not perfect square??
It is standard result in elementary number theory that for any natural number n sqrt(n)
is irrational except when n is a perfect square. The proof is a easy extension of the
proof that sqrt(2) is irrational.

RonL

9. Originally Posted by Jhevon
we're not dealing with an infinite set. only the integers between 0 and 2000 (inclusive)
it's coach who is dealing whith that stuff it's him to tell!
and according to the second phrase of is post the question was posed!
though there is a number a) in the third line dealing whith numbers only in the interval you mentionned the formulation was not to very clear: to "(or the same)" in the same phrase i suposed the first line was an expression of a mathèmatical question (refered in the second question (third line) by "the same")

10. Originally Posted by CaptainBlack
It is standard result in elementary number theory that for any natural number n sqrt(n)
is irrational except when n is a perfect square. The proof is a easy extension of the
proof that sqrt(2) is irrational.

RonL
i would have bet that !
but as the question is about how many sqrt(n) are rational I would'nt see were would be the plot to ask such a quetion if the elementary result was known because the question will be then much more elementary than the theorème involved and would had not much to do about rational or irational but how many perfect square express in percentage can you find below a certain number (in the question were dealing) which is a very elementary question when your supose to deal whith irationality!
but of course i'm not very aware of what's going in college nowadays

11. Originally Posted by SkyWatcher
i would have bet that !
but as the question is about how many sqrt(n) are rational I would'nt see were would be the plot to ask such a quetion if the elementary result was known because the question will be then much more elementary than the theorème involved and would had not much to do about rational or irational but how many perfect square express in percentage can you find below a certain number (in the question were dealing) which is a very elementary question when your supose to deal whith irationality!
but of course i'm not very aware of what's going in college nowadays
If you know the rational root theorem we have: let y=sqrt(n), then:

y^2-n=0

and if this has rational roots they are amoung the factors (positive or negative) of n.
So we have sqrt(n) is an integer, and [sqrt(n)]^2=n, that is n is a perfect square.

RonL

12. Originally Posted by SkyWatcher
it's coach who is dealing whith that stuff it's him to tell!
and according to the second phrase of is post the question was posed!
though there is a number a) in the third line dealing whith numbers only in the interval you mentionned the formulation was not to very clear: to "(or the same)" in the same phrase i suposed the first line was an expression of a mathèmatical question (refered in the second question (third line) by "the same")

I am sorry I confused you.

The question is actually from my math book, and so that is where the a comes from.

The answer my math book gives is 2,2%.