User:BZAW31559/sandbox/floatingpointimprecision

For more information, see Floating-point arithmetic.

0.1 + 0.2 = ? in single precision

$0.1={\frac {\mathrm {round} (0.1\times 2^{27})}{2^{27}}}=0.100000001490116119384765625\approx 0.1$

$0.2={\frac {\mathrm {round} (0.2\times 2^{26})}{2^{26}}}=0.20000000298023223876953125\approx 0.2$

$0.1+0.2={\frac {\mathrm {round} (0.1\times 2^{27})}{2^{27}}}+{\frac {\mathrm {round} (0.2\times 2^{26})}{2^{26}}}=0.300000004470348358154296875\approx 0.3$

2^-24 in decimal32 loses the number of digits?

2^-24 in decimal32 loses the number of digits?

Yes, the decimal32 floating-point format only supports 7 decimal digits, but the half precision of this value is correct and can fit this value exactly. –BZAW31559 (talk) 07:30, 4 May 2019 (UTC)

See math: $2^{-24}={\frac {\mathrm {round} (2^{-24}\times 10^{14})}{10^{14}}}=0.00000005960464\approx 0.000000059604644775390625$

4-bit minifloat

Form type	0 0 00	0 0 01	0 0 10	0 0 11	0 1 00	0 1 01	0 1 10	0 1 11	1 0 00	1 0 01	1 0 10	1 0 11	1 1 00	1 1 01	1 1 10	1 1 11
Signed decimal	0	0.25	0.5	0.75	1	1.25	1.5	1.75	−0	−0.25	−0.5	−0.75	−1	−1.25	−1.5	−1.75
Unsigned decimal	0	0.25	0.5	0.75	1	1.25	1.5	1.75	2	2.5	3	3.5	4	5	6	7
Floating imaginary	0	0.25	0.5	0.75	1	Inf	NaN	NaN	–0	–0.25	–0.5	–0.75	–1	–Inf	NaN	NaN
Signed integer	0	1	2	3	4	5	6	7	−0	−1	−2	−3	−4	−5	−6	−7
Unsigned integer	0	1	2	3	4	5	6	7	8	10	12	14	16	20	24	28
Fixed quarter imaginary	0	1	2	3	2i	4i	6i	1+2i	1+4i	1+6i	2+2i	2+4i	2+6i	3+2i	3+4i	3+6i
Floating quarter imaginary	0	1	2	3	2i	4i	6i	−4	−8	−12	−8i	−16i	−24i	16	32i	NaN

2-bit data type

There are no significand or mantissa bits in this format. Only sign bit and combination bit.

Form type	0b00	0b01	0b10	0b11
Integer	0	1	2	3
Inverse boolean	0	1	–0	–1
Inverse imaginary	0	1	–1	NaN
Unsigned imaginary	0	1	Infinity	NaN
Quarter imaginary	0	1	2i	1+2i
Deprecated imaginary	0	1	2i	NaN