The integer part is (? distributive ?)

[icantadd]

the integer part is ... (?? distributive ??)

Homework Statement

Define the floor of a real number k where [k] is the least smallest integer from k.

I want to show that [a - b] = [a] -

Homework Equations

n/a

The Attempt at a Solution

[1.2 - 5.7] = [-3.8] = -4
[1.2] - [5.7] = 1 - 5 = -4

I am not sure how to go about generalizing the observation above to all numbers.

 

[CompuChip]

You don't:
[5.2 - 2.3] = [2.9] = 2
[5.2] - [2.3] = [5] - [2] = 3

Sorry

 

[icantadd]

Yes, thank you CompuChip. However I stated the problem wrong. What I am thinking is this:

Let a > 0. Then
[a - 0] = [a] - 0 and
[a - 1] = [a] - 1.

I only care about the values being subtracted from a of 0 and 1, nothing else. Then if b = 0 or 1, [a-b] = [a] - b. This should hold, correct? Demonstrating it works for 0 is easy. I don't know how to show it for 1, but here is my attempt:

a - 1 < a
Suppose a is an integer. Then [a - 1] = a - 1 = [a] - 1.
Suppose a is not an integer. I am not sure. It seems to work fine with all the tests I can give (a>0). I don't know how to write this.

 

[CompuChip]

Let x be any (positive, although it probably works for any) number. Then write x as n + f, with n an integer and f the fractional part (0 <= f < 1).

 

[icantadd]

Let x be any (positive, although it probably works for any) number. Then write x as n + f, with n an integer and f the fractional part (0 <= f < 1).

Yes, thank you. Work backwards, of course.