[Guest]

Determine all possible value(s) of a real number N, such that N satisfies this equation:

(N) = N*<N>

Note: <x> is the least integer ≥ x, and, (x) = <x> - x.

[Pickett]

Determine all possible value(s) of a real number N, such that N satisfies this equation:

(N) = N*<N>

Note: <x> is the least integer ≥ x, and, (x) = <x> - x.

N = 0:

Basically the left side becomes <0> - 0...which is 0, and the right side is 0 * 0...which is also 0...so N = 0 works

N = 0.5:

The left side becomes 1 - 0.5...which is 0.5 and the right side is 0.5 * 1...which is also 0.5...so N = 0.5 works

...and I'm just now looking at this, but I'm thinking those may be only answers...Non-zero integers do not work since they would cause the left side to be zero always, but the right side to be a positive number...still checking and working on a proof...

Edited June 25, 2009 by Pickett

[Pickett]

Ok, so, let's divide up the real numbers into 5 groups and prove values in each group:

N < -1

If N was less than -1, then the right hand side of the equation would yield a positive number greater than 1 ALWAYS. The left hand side of the equation would always be a negative number somewhere between 0 and -1...therefore there are no solutions for N < -1.

-1 <= N < 0

If N was in this range, we get the right hand side of the equation to be 0 (since <N> becomes 0), but the left hand side would have to be 0 - N...so the solution would have to satisfy 0 - N = 0...which does not exist in this range.

0

From the above proof, we see that 0 - N = 0 is a solution...so when N = 0, we have a solution.

0 < N <= 1

IF N was in this range, the right hand side is N (because <N> becomes 1), and the left hand side would be 1-N...so the solution would have to satisfy 1 - N = N...therefore, the only solution in this range is 0.5

N > 1

This is the same argument as the N < -1...RHS becomes positive number greater than 1, LHS always is less than 1...so no solutions exist in this range.

Therefore, N = 0 and N = 0.5 are the only solutions that satisfy this equation. QED

[unreality]

Determine all possible value(s) of a real number N, such that N satisfies this equation:

(N) = N*<N>

Note: <x> is the least integer ≥ x, and, (x) = <x> - x.

if x is an integer, (x) = 0 and x<x> = x^2, so if x is an integer then x=0

it's most helpful to treat <x> as x+f, where f=[0,1) and multiple f's in an equation can all be different things.

thus (x) = <x> - x = x+f - x = f

so f on the left side, always

on the right we have x(x+f)

in other words, x multiplied by a fraction above itself must equal a fraction between 0 and 1 (or 0, but we've already covered the x=0 only-integer option)

for x(x+f) to equal f, it must be between 0 and 1 itself... or between -1 and 0

thus all answers are between -1 and 1

let's just consider the first part, x being between 0 and 1...

all <x> leads to 1, therefore the equation becomes:

1-x = x*1

that is, 1-x = x or 1=2x or x=.5

Thus the ONLY positive solutions are 0 and .5

For negative solutions, they must be between 0 and -1 as we showed earlier... thus <x> is always 0, therefore the equation becomes:

0-x = x*0

of which only 0 works.

so I'm pretty sure only 0 and 0.5 work

[EventHorizon]

N = 0:

Basically the left side becomes <0> - 0...which is 0, and the right side is 0 * 0...which is also 0...so N = 0 works

N = 0.5:

The left side becomes 1 - 0.5...which is 0.5 and the right side is 0.5 * 1...which is also 0.5...so N = 0.5 works

...and I'm just now looking at this, but I'm thinking those may be only answers...Non-zero integers do not work since they would cause the left side to be zero always, but the right side to be a positive number...still checking and working on a proof...

(N) = <N>*N

<N>-N = <N>*N

<N> = <N>*N+N

<N> = [<N>+1]*N

<N>/[<N>+1] = N

If N <= -1, plugging N into the left side of the equation gives a non-negative number.

If N is in (-1,0), LHS of the equation is 0/1 = 0 != N.

If N is 0, LHS of the equation is 0/1 = 0 = N. **It works!**

If N is in (0,.5), LHS = 1/2 = .5 != N.

If N is .5, LHS of the equation is 1/2 = .5 = N. **It works!**

If N is in (.5,1], LHS = 1/2 = .5 != N.

If N is in (1,infinity), obviously the left hand side of the equation cannot be greater than 1 because the denominator is one greater than the numerator, so the equation would fail again.

Therefore, the only answers are 0 and .5

[Guest]

if x is an integer, (x) = 0 and x<x> = x^2, so if x is an integer then x=0

it's most helpful to treat <x> as x+f, where f=[0,1) and multiple f's in an equation can all be different things.

thus (x) = <x> - x = x+f - x = f

so f on the left side, always

on the right we have x(x+f)

in other words, x multiplied by a fraction above itself must equal a fraction between 0 and 1 (or 0, but we've already covered the x=0 only-integer option)

for x(x+f) to equal f, it must be between 0 and 1 itself... or between -1 and 0

thus all answers are between -1 and 1

let's just consider the first part, x being between 0 and 1...

all <x> leads to 1, therefore the equation becomes:

1-x = x*1

that is, 1-x = x or 1=2x or x=.5

Thus the ONLY positive solutions are 0 and .5

For negative solutions, they must be between 0 and -1 as we showed earlier... thus <x> is always 0, therefore the equation becomes:

0-x = x*0

of which only 0 works.

so I'm pretty sure only 0 and 0.5 work

I agree the ONLY solutions are 0 and 0.5. Here is my proof:

I will be using 'r' (as readl) to mean N, and 'd' (as decimal) to be <N>

We know that (N) = r*d

and (N) = d-r

==> r*d= d-r

If r is also a decimal ==> r = d => d-r = 0 => d*r =0 => r=d=0 First Solution

Otherwise r < d => 0 < d-r < 1 ==> 0 < d*r < 1 ==> 0 < r < 1 ==> d=1

Now since r*d = d -r therefore r = 1-r ==> r = 0.5 Second solution

Edited June 26, 2009 by bonanova
spoiler added