-0.000 038 141 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 038 141(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 038 141(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 038 141| = 0.000 038 141


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 038 141.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 038 141 × 2 = 0 + 0.000 076 282;
  • 2) 0.000 076 282 × 2 = 0 + 0.000 152 564;
  • 3) 0.000 152 564 × 2 = 0 + 0.000 305 128;
  • 4) 0.000 305 128 × 2 = 0 + 0.000 610 256;
  • 5) 0.000 610 256 × 2 = 0 + 0.001 220 512;
  • 6) 0.001 220 512 × 2 = 0 + 0.002 441 024;
  • 7) 0.002 441 024 × 2 = 0 + 0.004 882 048;
  • 8) 0.004 882 048 × 2 = 0 + 0.009 764 096;
  • 9) 0.009 764 096 × 2 = 0 + 0.019 528 192;
  • 10) 0.019 528 192 × 2 = 0 + 0.039 056 384;
  • 11) 0.039 056 384 × 2 = 0 + 0.078 112 768;
  • 12) 0.078 112 768 × 2 = 0 + 0.156 225 536;
  • 13) 0.156 225 536 × 2 = 0 + 0.312 451 072;
  • 14) 0.312 451 072 × 2 = 0 + 0.624 902 144;
  • 15) 0.624 902 144 × 2 = 1 + 0.249 804 288;
  • 16) 0.249 804 288 × 2 = 0 + 0.499 608 576;
  • 17) 0.499 608 576 × 2 = 0 + 0.999 217 152;
  • 18) 0.999 217 152 × 2 = 1 + 0.998 434 304;
  • 19) 0.998 434 304 × 2 = 1 + 0.996 868 608;
  • 20) 0.996 868 608 × 2 = 1 + 0.993 737 216;
  • 21) 0.993 737 216 × 2 = 1 + 0.987 474 432;
  • 22) 0.987 474 432 × 2 = 1 + 0.974 948 864;
  • 23) 0.974 948 864 × 2 = 1 + 0.949 897 728;
  • 24) 0.949 897 728 × 2 = 1 + 0.899 795 456;
  • 25) 0.899 795 456 × 2 = 1 + 0.799 590 912;
  • 26) 0.799 590 912 × 2 = 1 + 0.599 181 824;
  • 27) 0.599 181 824 × 2 = 1 + 0.198 363 648;
  • 28) 0.198 363 648 × 2 = 0 + 0.396 727 296;
  • 29) 0.396 727 296 × 2 = 0 + 0.793 454 592;
  • 30) 0.793 454 592 × 2 = 1 + 0.586 909 184;
  • 31) 0.586 909 184 × 2 = 1 + 0.173 818 368;
  • 32) 0.173 818 368 × 2 = 0 + 0.347 636 736;
  • 33) 0.347 636 736 × 2 = 0 + 0.695 273 472;
  • 34) 0.695 273 472 × 2 = 1 + 0.390 546 944;
  • 35) 0.390 546 944 × 2 = 0 + 0.781 093 888;
  • 36) 0.781 093 888 × 2 = 1 + 0.562 187 776;
  • 37) 0.562 187 776 × 2 = 1 + 0.124 375 552;
  • 38) 0.124 375 552 × 2 = 0 + 0.248 751 104;
  • 39) 0.248 751 104 × 2 = 0 + 0.497 502 208;
  • 40) 0.497 502 208 × 2 = 0 + 0.995 004 416;
  • 41) 0.995 004 416 × 2 = 1 + 0.990 008 832;
  • 42) 0.990 008 832 × 2 = 1 + 0.980 017 664;
  • 43) 0.980 017 664 × 2 = 1 + 0.960 035 328;
  • 44) 0.960 035 328 × 2 = 1 + 0.920 070 656;
  • 45) 0.920 070 656 × 2 = 1 + 0.840 141 312;
  • 46) 0.840 141 312 × 2 = 1 + 0.680 282 624;
  • 47) 0.680 282 624 × 2 = 1 + 0.360 565 248;
  • 48) 0.360 565 248 × 2 = 0 + 0.721 130 496;
  • 49) 0.721 130 496 × 2 = 1 + 0.442 260 992;
  • 50) 0.442 260 992 × 2 = 0 + 0.884 521 984;
  • 51) 0.884 521 984 × 2 = 1 + 0.769 043 968;
  • 52) 0.769 043 968 × 2 = 1 + 0.538 087 936;
  • 53) 0.538 087 936 × 2 = 1 + 0.076 175 872;
  • 54) 0.076 175 872 × 2 = 0 + 0.152 351 744;
  • 55) 0.152 351 744 × 2 = 0 + 0.304 703 488;
  • 56) 0.304 703 488 × 2 = 0 + 0.609 406 976;
  • 57) 0.609 406 976 × 2 = 1 + 0.218 813 952;
  • 58) 0.218 813 952 × 2 = 0 + 0.437 627 904;
  • 59) 0.437 627 904 × 2 = 0 + 0.875 255 808;
  • 60) 0.875 255 808 × 2 = 1 + 0.750 511 616;
  • 61) 0.750 511 616 × 2 = 1 + 0.501 023 232;
  • 62) 0.501 023 232 × 2 = 1 + 0.002 046 464;
  • 63) 0.002 046 464 × 2 = 0 + 0.004 092 928;
  • 64) 0.004 092 928 × 2 = 0 + 0.008 185 856;
  • 65) 0.008 185 856 × 2 = 0 + 0.016 371 712;
  • 66) 0.016 371 712 × 2 = 0 + 0.032 743 424;
  • 67) 0.032 743 424 × 2 = 0 + 0.065 486 848;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 038 141(10) =

0.0000 0000 0000 0010 0111 1111 1110 0110 0101 1000 1111 1110 1011 1000 1001 1100 000(2)

6. Positive number before normalization:

0.000 038 141(10) =

0.0000 0000 0000 0010 0111 1111 1110 0110 0101 1000 1111 1110 1011 1000 1001 1100 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 038 141(10) =

0.0000 0000 0000 0010 0111 1111 1110 0110 0101 1000 1111 1110 1011 1000 1001 1100 000(2) =

0.0000 0000 0000 0010 0111 1111 1110 0110 0101 1000 1111 1110 1011 1000 1001 1100 000(2) × 20 =

1.0011 1111 1111 0011 0010 1100 0111 1111 0101 1100 0100 1110 0000(2) × 2-15


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -15

Mantissa (not normalized):
1.0011 1111 1111 0011 0010 1100 0111 1111 0101 1100 0100 1110 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1008(10) =

011 1111 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0011 1111 1111 0011 0010 1100 0111 1111 0101 1100 0100 1110 0000 =

0011 1111 1111 0011 0010 1100 0111 1111 0101 1100 0100 1110 0000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0000

Mantissa (52 bits) =
0011 1111 1111 0011 0010 1100 0111 1111 0101 1100 0100 1110 0000


Decimal number -0.000 038 141 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0011 1111 1111 0011 0010 1100 0111 1111 0101 1100 0100 1110 0000

» Convert decimal -0.000 038 147 to 64 bit double precision IEEE 754 binary floating point representation standard

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100