Function‘round’. Numerical calculation question

Problem recurrence：

>> round(((6/40000)*10000))

ans =

     1

but，

>> round(1.500000000000000)

ans =

     2

Why is 1.5 such a value appearing when rounding off?The same value calculation results are different.

Answer 1

[Bruno Luong]

Welcome to the work of floating point arithmetic

>> (6/40000)*10000-1.5

ans =

  -2.2204e-16

>> 

So actually the calculation of (6/40000)*10000 returns value stricly less than 1.5=3/2. Therefore what you have observed with round is explained.

[yang li]

Thank you for your answer, I need to look at the knowledge of floating point calculations.

[Walter Roberson]

When you think of 1/10 in binary it can help to think about 1/7 in decimal, in that 1/7 decimal is an infinitely repeating multidigit pattern . If you truncate to any finite number of digits and then multiply back by 7 then you will not get exactly 1. Exactly the same mathematics reasons that lead 1/7 decimal to be infinite lead 1/10 to be infinite repeating multiple digits in binary. It is not a "quality of implementation" issue but rather a fundamental limitation of using any finite number of digits in a "base" numeric representation instead of using rational numbers as the representation .

Answer 2

[Walter Roberson]

1/10 is not exactly representable in binary floating point so you would need to calculate carefully to figure what is being computed.

But algebraically the first of those would be round(2/4) which would be round(0.5) which would round up to 1 and likewise round(1.5) rounds up to 2 so the results you see happen to agree with the algebraic results . Other combinations that are algebraically the same could come out differently because of 1/10 not being exactly representable.

[yang li]

I am sorry that my code part is wrong. The problem is as follows.

>> round((6/40000)*10000)

ans =

     1

and

     >> round(1.500000000000000)

ans =

     2

I use matlab to calculate directly, but the same value gives different results.

Answer 3

[KSSV]

if your number is less than 0.5.._round_ acts like floor function. For number greater than or equla to 0.5 it acts like ceil. YOu may check it yourself. Read about ceil and _floor_

x = 0:0.1:1
floor(x)
ceil(x)
round(x)

[yang li]

I am sorry that my code part is wrong. The problem is as follows.

>> round((6/40000)*10000)

ans =

     1

and

     >> round(1.500000000000000)

ans =

     2

I use matlab to calculate directly, but the same value gives different results.