0.1 float is greater than 0.1 double. I expected it to be false [duplicate]

I'd say the answer depends on the rounding mode when converting the double to float. float has 24 binary bits of precision, and double has 53. In binary, 0.1 is:

0.1₁₀ = 0.0001100110011001100110011001100110011001100110011…₂             ^        ^         ^   ^              1       10        20  24

So if we round up at the 24th digit, we'll get

0.1₁₀ ~ 0.000110011001100110011001101

which is greater than the exact value and the more precise approximation at 53 digits.

The number 0.1 will be rounded to the closest floating-point representation with the given precision. This approximation might be either greater than or less than 0.1, so without looking at the actual values, you can't predict whether the single precision or double precision approximation is greater.

Here's what the double precision value gets rounded to (using a Python interpreter):

>>> "%.55f" % 0.1'0.1000000000000000055511151231257827021181583404541015625'

And here's the single precision value:

>>> "%.55f" % numpy.float32("0.1")'0.1000000014901161193847656250000000000000000000000000000'

So you can see that the single precision approximation is greater.

If you convert .1 to binary you get:

0.000110011001100110011001100110011001100110011001100...

repeating forever

Mapping to data types, you get:

float(.1)  = %.00011001100110011001101                                     ^--- note roundingdouble(.1) = %.0001100110011001100110011001100110011001100110011010

Convert that to base 10:

float(.1)  = .10000002384185791015625double(.1) = .100000000000000088817841970012523233890533447265625

This was taken from an article written by Bruce Dawson. it can be found here:
Doubles are not floats, so don’t compare them