How to use Floor command

I have some wrong thing with floor command. This problem is that

Step 1: Initial step

t=[0.90 1.00 1.30 1.80 1.90];
step = (t(5)-t(1))/100;
time = t(1):step:10;

Step 2: Check floor command

time(181)

ans =

    2.7000
  floor((time(181)-0.9)/(2*0.9))

ans =

     0
*But i realize that*

floor((2.7-0.9)/(2*0.9))

ans =

     1

What problem was happen in this case?

Question 2: Logic condition

t_ = 4.5 - 4*0.9

t_ =

    0.9000
  t_ >= 0.9

ans =

     0
But

0.9>=0.9

ans =

     1

Answer 1

[Matt Fig]

You have fallen into thinking that you are using infinite precision arithmetic when you are only using double precision. This is such a common mistake that it has its own FAQ.

Here is a little more about is going on:

t = [0.90 1.00 1.30 1.80 1.90];
step = (t(5)-t(1))/100;
time = t(1):step:10;
A = time(181)-0.9;
B = 2*0.9; 
C = A/B;
fprintf('%16.16f\n',time(181),A,B,C)

2.6999999999999997
1.7999999999999998
1.8000000000000000
0.9999999999999999

As for the second question: same problem:

fprintf('%16.16f\n',4.5 - 4*0.9)
0.8999999999999999

Answer 2

[Matt J]

You're not accounting for floating point errors.

4.5 - 4*0.9

will not give you precisely 9/10, but rather something accurate to many decimal places.

Answer 3

[Honglei Chen]

This has nothing to do with floor, it's part of the floating point computation. See the link below

http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

Answer 4

[trung trinh]

Thank a lot :)