[EventHorizon]

First a ground rule, no imaginary numbers (as it makes the questions too easy). So x can be any real number, and f(x) will stay a real number.

1. Can you find a function f such that f(f(x)) = -x?

2. Can you find a function f such that f(f(x)) = 1/x?

[Guest]

First one is easy enough.

f(x)=-[x]

Too tired to think about #2 right now

[Guest]

I think we have one more solution for first one (it is similar to vacian's answer)

f(x0 = -|1/x|

[Guest]

I thought of one that works tho i doubt its the answer you wanted, but as is I can't think of any other viable answers

f(x)=sqrt(x²⁾

^{this works because the sqrt has a positive and negative answer.}

^{just as a note the |x| answers dont work for negative x's}

[Guest]

I thought of one that works tho i doubt its the answer you wanted, but as is I can't think of any other viable answers

f(x)=sqrt(x²⁾

^{this works because the sqrt has a positive and negative answer.}

^{just as a note the |x| answers dont work for negative x's}

It is -|x| not |x|

[Guest]

It is -|x| not |x|

What he is saying is f(f(x)) != -x for f(x) = -|x| when x<0

eg -||-1|| != -(-1)

Edited July 29, 2009 by Semper Rideo

[Guest]

What he is saying is f(f(-1)) != -(-1) for f(x) = -|x|

Ya...i am sorry... i overlooked it

[Guest]

These solutions works, but it is a little messy. There may be a nicer solution out there...

In what follows, [x] refers to the floor of x. That is, [x] is the greatest integer less than or equal to x.

In what follows, {x} refers to the ceiling of x. That is, {x} is the smallest integer greater than or equal to x.

The definition of the function involves several parts:

If x is not an integer:

f(x) = x + 1, if x > 0 and [x] is even;

f(x) = -x + 1, if x > 0 and [x] is odd;

f(x) = x - 1, if x < 0 and {x} is even;

f(x) = -x - 1, if x < 0 and {x} is odd.

If x is an integer:

f(x) = x + 1, if x > 0 and x is odd;

f(x) = -x + 1, if x > 0 and x is even;

f(x) = x - 1, if x < 0 and x is odd;

f(x) = -x - 1, if x <0 and x is even;

f(x) = 0, if x = 0.

The way this works is it breaks the real line into little pieces, and then groups them together in groups of four pieces, so that each piece is in a group of four with its own negative. For example, the interval (0,1) is grouped with the intervals (0,-1), (1,2), and (-1,-2). Then in each group of four, the function takes the intervals to each other in such a way that after two times through, x is taken to -x. The integers are grouped in similar groups of 4, except for 0, which is sent to itself, since -0 = 0.

If you are confused by the definition, just try plugging in values to see how it works. It is best to cycle through four times, so you see how the function moves points around. Try plugging in both whole numbers and decimals.

For example:

f(2.25) = 2.25 + 1 = 3.25, since 2.25 > 0 and [2.25] = 2 is even.

f(3.25) = -3.25 + 1 = -2.25, since 3.25 > 0 and [3.25] = 3 is odd.

f(-2.25) = -2.25 - 1 = -3.25, since -2.25 < 0 and {-2.25} = -2 is even.

f(-3.25) = -(-3.25) - 1 = 2.25, since -3.25 < 0 and {-3.25} = -3 is odd.

The solution for the second problem follows.

This works in a similar way to the first problem, but the intervals are different, since the operation we are interested in is different.

In what follows, [x] refers to the floor of x. That is, [x] is the greatest integer less than or equal to x.

The definition of the function involves several parts:

If x is positive, and x is not an integer, and 1/x is not an integer:

f(x) = x + 1, if x > 1 and [x] is odd;

f(x) = 1/(x-1), if x > 1 and [x] is even;

f(x) = 1/(1/x + 1), if x < 1 and [1/x] is odd;

f(x) = 1/x - 1, if x < 1 and [1/x] is even.

If x is positive, and x is an integer or 1/x is an integer:

f(x) = x + 1, if x > 1 and x is even;

f(x) = 1/(x-1), if x > 1 and x is odd;

f(x) = 1/(1/x + 1), if x < 1 and 1/x is even;

f(x) = 1/x - 1, if x < 1 and 1/x is odd;

f(x) = 1, if x = 1.

If x is negative:

f(x) = -f(-x)

(That is, f for negative numbers is defined in terms of their positive counterparts.)

The way this works is it breaks the real line into little pieces, and then groups them together in groups of four pieces, so that each piece is in a group of four with its own reciprocal. For example, the interval (3,4) is grouped with the intervals (1/4,1/3), (4,5), and (1/5,1/4). Then in each group of four, the function takes the intervals to each other in such a way that after two times through, x is taken to 1/x. The integers and fractions of the form 1/n (where n is an integer) are grouped in similar groups of 4, except for 1, which is sent to itself, since 1/1 = 1.

If you are confused by the definition, just try plugging in values to see how it works. It is best to cycle through four times, so you see how the function moves points around. Try plugging in both whole numbers and fractions.

For example:

f(5/4) = 5/4 + 1 = 9/4, since 5/4 > 1 and [5/4] = 1 is odd;

f(9/4) = 1/(9/4 - 1) = 4/5, since 9/4 > 1 and [9/4] = 2 is even;

f(4/5) = 1(1/(4/5) + 1) = 1/(5/4 + 1) = 4/9, since 4/5 < 1 and [1/(4/5)] = 1 is odd;

f(4/9) = 1/(4/9) - 1 = 9/4 - 1 = 5/4, since 4/9 < 1 and [1/(4/9)] = 2 is even.

[EventHorizon]

These solutions works, but it is a little messy. There may be a nicer solution out there...

In what follows, [x] refers to the floor of x. That is, [x] is the greatest integer less than or equal to x.

In what follows, {x} refers to the ceiling of x. That is, {x} is the smallest integer greater than or equal to x.

The definition of the function involves several parts:

If x is not an integer:

f(x) = x + 1, if x > 0 and [x] is even;

f(x) = -x + 1, if x > 0 and [x] is odd;

f(x) = x - 1, if x < 0 and {x} is even;

f(x) = -x - 1, if x < 0 and {x} is odd.

If x is an integer:

f(x) = x + 1, if x > 0 and x is odd;

f(x) = -x + 1, if x > 0 and x is even;

f(x) = x - 1, if x < 0 and x is odd;

f(x) = -x - 1, if x <0 and x is even;

f(x) = 0, if x = 0.

The way this works is it breaks the real line into little pieces, and then groups them together in groups of four pieces, so that each piece is in a group of four with its own negative. For example, the interval (0,1) is grouped with the intervals (0,-1), (1,2), and (-1,-2). Then in each group of four, the function takes the intervals to each other in such a way that after two times through, x is taken to -x. The integers are grouped in similar groups of 4, except for 0, which is sent to itself, since -0 = 0.

If you are confused by the definition, just try plugging in values to see how it works. It is best to cycle through four times, so you see how the function moves points around. Try plugging in both whole numbers and decimals.

For example:

f(2.25) = 2.25 + 1 = 3.25, since 2.25 > 0 and [2.25] = 2 is even.

f(3.25) = -3.25 + 1 = -2.25, since 3.25 > 0 and [3.25] = 3 is odd.

f(-2.25) = -2.25 - 1 = -3.25, since -2.25 < 0 and {-2.25} = -2 is even.

f(-3.25) = -(-3.25) - 1 = 2.25, since -3.25 < 0 and {-3.25} = -3 is odd.

The solution for the second problem follows.

This works in a similar way to the first problem, but the intervals are different, since the operation we are interested in is different.

In what follows, [x] refers to the floor of x. That is, [x] is the greatest integer less than or equal to x.

The definition of the function involves several parts:

If x is positive, and x is not an integer, and 1/x is not an integer:

f(x) = x + 1, if x > 1 and [x] is odd;

f(x) = 1/(x-1), if x > 1 and [x] is even;

f(x) = 1/(1/x + 1), if x < 1 and [1/x] is odd;

f(x) = 1/x - 1, if x < 1 and [1/x] is even.

If x is positive, and x is an integer or 1/x is an integer:

f(x) = x + 1, if x > 1 and x is even;

f(x) = 1/(x-1), if x > 1 and x is odd;

f(x) = 1/(1/x + 1), if x < 1 and 1/x is even;

f(x) = 1/x - 1, if x < 1 and 1/x is odd;

f(x) = 1, if x = 1.

If x is negative:

f(x) = -f(-x)

(That is, f for negative numbers is defined in terms of their positive counterparts.)

The way this works is it breaks the real line into little pieces, and then groups them together in groups of four pieces, so that each piece is in a group of four with its own reciprocal. For example, the interval (3,4) is grouped with the intervals (1/4,1/3), (4,5), and (1/5,1/4). Then in each group of four, the function takes the intervals to each other in such a way that after two times through, x is taken to 1/x. The integers and fractions of the form 1/n (where n is an integer) are grouped in similar groups of 4, except for 1, which is sent to itself, since 1/1 = 1.

If you are confused by the definition, just try plugging in values to see how it works. It is best to cycle through four times, so you see how the function moves points around. Try plugging in both whole numbers and fractions.

For example:

f(5/4) = 5/4 + 1 = 9/4, since 5/4 > 1 and [5/4] = 1 is odd;

f(9/4) = 1/(9/4 - 1) = 4/5, since 9/4 > 1 and [9/4] = 2 is even;

f(4/5) = 1(1/(4/5) + 1) = 1/(5/4 + 1) = 4/9, since 4/5 < 1 and [1/(4/5)] = 1 is odd;

f(4/9) = 1/(4/9) - 1 = 9/4 - 1 = 5/4, since 4/9 < 1 and [1/(4/9)] = 2 is even.

Good job, both seem right. Your answer to number 1 was the first one I came up with, then I was able to reduce the number of discontinuities to 4. Same kind of thing for number 2.