• Dec 13th 2009, 02:35 AM
Also sprach Zarathustra
In front of you four statements that regard to functions f:R-->R.

1. Exist a>0 and exist x in R so that f(x+a)=f(x)

2. For every a>0, exist x in R so that f(x+a)=f(x)

3. Exist a>0 so that for every x in R so thatf(x+a)=f(x)

4. For every a>0 and every x in R so that f(x+a)=f(x)

Questions:

i. Characterize all functions from 1.
ii. Characterize all functions from 4.
iii. Find a function that is 1, but not 2,3,4
iv. Find a function that is 2, but not 3 and 4.
v. Find a function that is 3, but not 2 and 4.

Thanks!
• Dec 13th 2009, 05:22 AM
HallsofIvy
Quote:

Originally Posted by Also sprach Zarathustra
In front of you four statements that regard to functions f:R-->R.

1. Exist a>0 and exist x in R so that f(x+a)=f(x)

2. For every a>0, exist x in R so that f(x+a)=f(x)

3. Exist a>0 so that for every x in R so thatf(x+a)=f(x)

4. For every a>0 and every x in R so that f(x+a)=f(x)

Questions:

i. Characterize all functions from 1.

I take it you mean all functions that have property #1. Since a> 0 that simply means that there exist two distinct numbers, x and y, such that f(x)= f(y).

Quote:

ii. Characterize all functions from 4.
Let x and y be any two numbers. Then we can take a= |y-x| so that f(x)= f(x+a)= f(y).

Quote:

iii. Find a function that is 1, but not 2,3,4
There are many many easy examples. Try!

Quote:

iv. Find a function that is 2, but not 3 and 4.
v. Find a function that is 3, but not 2 and 4.

Thanks!
• Dec 13th 2009, 06:48 AM
Also sprach Zarathustra
I am not understanding this! What I need to show...? Can you please give more detailed answer?
• Dec 13th 2009, 07:46 AM
HallsofIvy
For (i) you simply have functions that are not "one to one".

For (i), as I said, you must have f(x)= f(y) for all x and y- a constant function.
• Dec 13th 2009, 08:37 AM
Also sprach Zarathustra
Yes, this (i) is easy, but what about (ii), (iii), (iv) and (v)?

Thank you for your time and knowledge!