mod function return wrong answer?

How can I make sure mod(a,b) returns value less than b, without explicitly calling min( mod(a,b), b )?

mod( -eps, 1 ) = 1 when eps ~ 10^-17.

Numerical accuracy is not an issue, but I need mod( a, 1 ) be less than 1.

[Sindar]

What value would you like it to take?

mod(a,m) is expressed as:

b = a - m.*floor(a./m);

so, you have b = -eps - (1)*floor(-eps) = -eps - (-1) = 1

rem(-eps,1) will return -eps

[madhan ravi]

Sindar I suggest moving your comment as an answer

[Hiroaki Yamamoto]

I expected mod( a, m ) < m (less than m),

but numerically mod(a, m) <= m (less than or equal to m) is the result when a = -1e-17 and m = 1.

If, internally, 1 - 10^-17 = 1 is calculated and returned, I think 0 is the mathematically correct return value rather than 1.

Answer 1

[madhan ravi]

>> mod( -eps, 1 ) < 1

ans =

    logical

     1

[Hiroaki Yamamoto]

I get different answers as follows, which indicare mod(a, m) can be equal to m, not less than m.

>> mod(-1e-17,1) < 1

ans =

logical

0

>> mod(-1e-17,1) == 1

ans =

logical

1

[madhan ravi]

 >> sprintf('%.32f', mod(-1e-17, 1))

 ans =

    '1.00000000000000000000000000000000'

 >> sprintf('%.32f', mod(-eps, 1))

 ans =

    '0.99999999999999977795539507496869'

[Hiroaki Yamamoto]

My question is, mathematically which of the following is correct or intended to be implemented.

1) mod(a, m) < m

or

2) mod(a, m) <= m.

Practically, I found mod(a,m) = m can happen and I do explicit check and avoid problems caused by the case mod(a,m) = m.

[madhan ravi]

Practically speaking it depends on what m is.