Uniqueness of fractional part of square roots.
Carl G.
non-square integer if one is given only the fractional part (digits to the
right of the decimal point) of its square root. Is there a simple proof
that the fractional parts of non-square positive integers are unique? It is
hard to find a proof on the Internet without knowing the right keywords.
Here is an example to help make the question clearer:
sqrt(2) = 1.4142135623...
Show that only 2 has this particular fractional part (.4142135623...). If
one was given the fractional part, one could determine that the integer was
2. Note that -2 also has this fractionl part, but the problem is for
positive non-square integers. The integers must be non-square since the
square root of square integers (1, 4, 9, 27, etc.) are integers, and the
fractional parts are all zero.
Carl G.
Ilan Mayer
wrote:
If two different non-square integers n1 and n2 have the same
fractional part a in their square roots, then (m1+a)^2 = n1 and
(m2+a)^2 = n2 where m1 and m2 are also integers (and m1 is different
from m2, otherwise n1 = n2).
This means that:
a^2+2*m1*a+m1^2 = n1
a^2+2*m2*a+m2^2 = n2
Subtracting one equation from the other yields an expression for a:
a = (n1-n2-m1^2+m2^2)/(2*(m1-m2))
This means that a is a rational number, but the fractional part a of a
non-square integer is an irrational number - a contradiction.
This means that two different non-square integers cannot have the same
fractional part in their square roots.
Please reply to ilan dot mayer at hotmail dot com
__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||
jonnie303
>
> If two different non-square integers n1 and n2 have the same
> fractional part a in their square roots, then (m1+a)^2 = n1 and
> (m2+a)^2 = n2 where m1 and m2 are also integers (and m1 is different
> from m2, otherwise n1 = n2).
>
> This means that:
> a^2+2*m1*a+m1^2 = n1
> a^2+2*m2*a+m2^2 = n2
>
> Subtracting one equation from the other yields an expression for a:
> a = (n1-n2-m1^2+m2^2)/(2*(m1-m2))
>
> This means that a is a rational number, but the fractional part a of a
> non-square integer is an irrational number - a contradiction.
>
> This means that two different non-square integers cannot have the same
> fractional part in their square roots.
>
> Please reply to ilan dot mayer at hotmail dot com
>
> __/\__
> \ /
> __/\\ //\__ Ilan Mayer
> \ /
> /__ __\ Toronto, Canada
> /__ __\
>
> - Show quoted text -
Nice proof.
2 further questions seem to arise:
(a) Given the fractional part of a square root, how can one determine
the unique positive integer whose square root it is the fractional
part of? For example, what is the lowest integer whose square root
contains the fractional part .84417...?
(b) What is the relationship between those integers whose square roots
share the same initial digits in their fractional part? For example,
what is the next integer whose square root contains .414...?
---------------
jonnie303
Sevenoaks, UK
Don Del Grande
Maybe you can help me out on this one:
Is there a proof that, given a set of N positive integers, none of
which are perfect squares, the sum of the square roots of all of the
numbers in the set is irrational? I can get it for N = 1, 2, and 3,
but 4 (and higher) stumps me.
-- Don
Dr Nick
> On Feb 22, 5:50 pm, Ilan Mayer <ilan_no_s...@hotmail.com> wrote:
>>
>> If two different non-square integers n1 and n2 have the same
>> fractional part a in their square roots, then (m1+a)^2 = n1 and
>> (m2+a)^2 = n2 where m1 and m2 are also integers (and m1 is different
>> from m2, otherwise n1 = n2).
>>
>> This means that:
>> a^2+2*m1*a+m1^2 = n1
>> a^2+2*m2*a+m2^2 = n2
>>
>> Subtracting one equation from the other yields an expression for a:
>> a = (n1-n2-m1^2+m2^2)/(2*(m1-m2))
>>
>> This means that a is a rational number, but the fractional part a of a
>> non-square integer is an irrational number - a contradiction.
>>
>> This means that two different non-square integers cannot have the same
>> fractional part in their square roots.
>>
>
> Nice proof.
> 2 further questions seem to arise:
>
> (a) Given the fractional part of a square root, how can one determine
> the unique positive integer whose square root it is the fractional
> part of? For example, what is the lowest integer whose square root
> contains the fractional part .84417...?
>
> (b) What is the relationship between those integers whose square roots
> share the same initial digits in their fractional part? For example,
> what is the next integer whose square root contains .414...?
For a) I'm not convinced you can easily do it (although I'm no
mathematician). Among other things, the proof doesn't say anything
about how distinct the two fractional parts must be. And you get to
this in your part b. Have a look at the square roots of 42195 and
167620 and compare with root 2 for example. So I suspect that there can
be numbers that have identical fractional roots to any precision you
chose. That doesn't prove that reversing it isn't easy, of course, but
does suggest it might not be.
--
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Plan trips, see photos, check facilities | http://canalplan.org.uk
Nick Wedd
<3-no...@temporary-address.org.uk> writes
Indeed there must be. There are infinitely many non-square integers,
and they all have to fit the fractional parts of their square roots
between 0 and 1. So for any epsilon however small, there must be two
integers the fractional parts of whose square roots differ by less than
epsilon.
Nick
> That doesn't prove that reversing it isn't easy, of course, but
>does suggest it might not be.
--
Nick Wedd ni...@maproom.co.uk
Mark Brader
> > (a) Given the fractional part of a square root, how can one determine
> > the unique positive integer whose square root it is the fractional
> > part of? For example, what is the lowest integer whose square root
> > contains the fractional part .84417...?
> >
> > (b) What is the relationship between those integers whose square roots
> > share the same initial digits in their fractional part? For example,
> > what is the next integer whose square root contains .414...?
> For a) I'm not convinced you can easily do it (although I'm no
> mathematician). Among other things, the proof doesn't say anything
> about how distinct the two fractional parts must be. And you get to
> this in your part b. Have a look at the square roots of 42195 and
> 167620 and compare with root 2 for example.
Better yet, compare them with each other!
> So I suspect that there can be numbers that have identical fractional
> roots to any precision you chose.
Yes, that's easy to prove by piggybacking on the original problem.
Say we want two irrational roots whose fractional parts have values
within x of each other (where x is some number between 0 and 1).
Then all we have to do is set n = ceiling(1/x)+1 and examine the
first n irrational roots. We know from the original problem their
fractional parts are all unequal, and we also know that they are all
strictly between 0 and 1. If none of them were within x of each
other, then the difference between the highest and lowest values
would have to be at least (n-1)*x, which is >= 1.
["Ceiling(x)" means x if x is an integer, and the next higher integer
if not.]
For example, two irrational roots whose fractional parts are within
0.1 of each other must exist exist within the first 11 irrational
roots, namely sqrt(2) through sqrt(14). (In fact it happens that
there are several such pairs in that range.)
--
Mark Brader, Toronto | "If you want a 20th century solution, the
m...@vex.net | obvious answer is helicopters!" -- Bob Scheurle
My text in this article is in the public domain.
Andrew B.
A third extra question: are the fractional parts of square roots dense
in the interval (0,1) ?
The Qurqirish Dragon
Intuition tells me "yes"
reasoning: let d_n be defined by d_n := sqrt(n) - sqrt (n-1), for n>1
A few statements, which all seem easy to show:
i) d_2 = 0.414.....
ii) d_n is a strictly decreasing sequence
iii) for sufficiently large n, d_n is arbitrarily small, so d_n
converges to 0
proof sketch for density of fractional parts of square roots:
Let e > 0 be any value, and let x be any real 0 < x < 1
By (iii), let N be an integer such that d_N < e
Let k be an integer such that k^2 > N.
Then the fractional part of sqrt (k^2) = 0, and the fractional part of
sqrt((k+1)^2) = 0.
The fractional parts of sqrt (k^2 +1), sqrt (k^2 +2),...,sqrt ((k
+1)^2) all differ by less than e. In particular, one of them (call it
y) is such that | y - x | < e/2 < e.
Therefore, given any e > 0, and any 0<x<1, there is an integer M such
that the fractional part of |sqrt(M) - x| <e. Indeed, since by the
same reasoning, there is at least one such integer in every interval
(k + j)^2 ... (k + j + 1)^2, there are infinitely many such fractional
parts, and so the fractional parts of sqrt(n) are indeed dense in (0,1)
The Qurqirish Dragon
> On Feb 25, 4:37 am, "Andrew B." <bull...@gmail.com> wrote:
>
>
>
> > On Feb 22, 9:51 pm, jonnie303 <john.gri...@btinternet.com> wrote:
>
> > A third extra question: are the fractional parts of square roots dense
> > in the interval (0,1) ?
>
> Intuition tells me "yes"
> reasoning: let d_n be defined by d_n := sqrt(n) - sqrt (n-1), for n>1
> A few statements, which all seem easy to show:
> i) d_2 = 0.414.....
> ii) d_n is a strictly decreasing sequence
> iii) for sufficiently large n, d_n is arbitrarily small, so d_n
> converges to 0
>
Proof of (iii), in case you want it:
Let e>0 be a value for which you want d_n to be small.
Let n = ceiling (1/e), where ceiling (x) is the smallest integer
larger than x.
There are 2n+1 values between sqrt (n^2) and sqrt ((n+1)^2). Since
sqrt (n^2) and sqrt ((n+1)^2) are consecutive integers, at least two
of the square-roots must differ by less than 1/(2n) < e/2 <e, so d_k <
e for some k. By (ii), d_j < e for all j>k.
Thus, d_n < e for any e, for sufficiently large n, and (iii) is proven.
Andrew B.
Seems good to me.
Proof of (ii) is simple algebra:
x^2 > x^2-1 <=>
x > sqrt(x^2-1) <=>
2x > 2 sqrt(x^2-1) <=>
4x > 2 sqrt[(x+1)(x-1)] + (x+1) + (x-1) <=>
2 sqrt(x) > sqrt(x+1) + sqrt(x-1) <=>
sqrt(x+1) - sqrt(x) < sqrt(x) - sqrt(x-1).
mike
says...
> In message <87r5ax2...@temporary-address.org.uk>, Dr Nick
> <3-no...@temporary-address.org.uk> writes
> >jonnie303 <john....@btinternet.com> writes:
> >
> >For a) I'm not convinced you can easily do it (although I'm no
> >mathematician). Among other things, the proof doesn't say anything
> >about how distinct the two fractional parts must be. And you get to
> >this in your part b. Have a look at the square roots of 42195 and
> >167620 and compare with root 2 for example. So I suspect that there can
> >be numbers that have identical fractional roots to any precision you
> >chose.
>
> Indeed there must be. There are infinitely many non-square integers,
> and they all have to fit the fractional parts of their square roots
> between 0 and 1.
The comment above is true...
> So for any epsilon however small, there must be two
> integers the fractional parts of whose square roots differ by less than
> epsilon.
...and this bit is true as well, except for the "So ". In other words,
infinitely many fractional parts can fit between 0 and 1 (as they
actually do) with finite gaps between some of them (as they actually
don't, but which would have to be proven).
--
Mike
Brian Tung
> > Indeed there must be. There are infinitely many non-square integers,
> > and they all have to fit the fractional parts of their square roots
> > between 0 and 1.
>
> The comment above is true...
>
> > So for any epsilon however small, there must be two
> > integers the fractional parts of whose square roots differ by less than
> > epsilon.
>
> ...and this bit is true as well, except for the "So ". In other words,
> infinitely many fractional parts can fit between 0 and 1 (as they
> actually do) with finite gaps between some of them (as they actually
> don't, but which would have to be proven).
But the second comment *is* true because of the first. An extension of
the pigeonhole principle establishes that because of the first comment,
although there could be some gaps of positive length, for any e > 0,
there must be a gap smaller than e.
--
Brian Tung <brian....@gmail.com>
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Dan Hoey
[...]
> Intuition tells me "yes"
> reasoning: let d_n be defined by d_n := sqrt(n) - sqrt (n-1), for n>1
> iii) for sufficiently large n, d_n is arbitrarily small, so d_n
> converges to 0
Since sqrt'(x)=1/sqrt(4x)<1/sqrt(4n-4) for x in [n-1,n], it is
clear that d_n <= 1/sqrt(4n-4). It should be easy to prove
this without calculus, but I haven't managed it.
Dan Hoey
Brian Tung
> Since sqrt'(x)=1/sqrt(4x)<1/sqrt(4n-4) for x in [n-1,n], it is
> clear that d_n <= 1/sqrt(4n-4). It should be easy to prove
> this without calculus, but I haven't managed it.
sqrt(n)-sqrt(n-1) = [sqrt(n)-sqrt(n-1)][sqrt(n)+sqrt(n-1)]/[sqrt(n)+sqrt(n-1)]
= [n-(n-1)]/[sqrt(n)+sqrt(n-1)]
= 1/[sqrt(n)+sqrt(n-1)]
< 1/[sqrt(n-1)+sqrt(n-1)]
= 1/sqrt(4n-4)