-0.000 282 005 879 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 879 4(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
-0.000 282 005 879 4(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:

|-0.000 282 005 879 4| = 0.000 282 005 879 4


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: 0.000 282 005 879 4.

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) 0.000 282 005 879 4 × 2 = 0 + 0.000 564 011 758 8;
  • 2) 0.000 564 011 758 8 × 2 = 0 + 0.001 128 023 517 6;
  • 3) 0.001 128 023 517 6 × 2 = 0 + 0.002 256 047 035 2;
  • 4) 0.002 256 047 035 2 × 2 = 0 + 0.004 512 094 070 4;
  • 5) 0.004 512 094 070 4 × 2 = 0 + 0.009 024 188 140 8;
  • 6) 0.009 024 188 140 8 × 2 = 0 + 0.018 048 376 281 6;
  • 7) 0.018 048 376 281 6 × 2 = 0 + 0.036 096 752 563 2;
  • 8) 0.036 096 752 563 2 × 2 = 0 + 0.072 193 505 126 4;
  • 9) 0.072 193 505 126 4 × 2 = 0 + 0.144 387 010 252 8;
  • 10) 0.144 387 010 252 8 × 2 = 0 + 0.288 774 020 505 6;
  • 11) 0.288 774 020 505 6 × 2 = 0 + 0.577 548 041 011 2;
  • 12) 0.577 548 041 011 2 × 2 = 1 + 0.155 096 082 022 4;
  • 13) 0.155 096 082 022 4 × 2 = 0 + 0.310 192 164 044 8;
  • 14) 0.310 192 164 044 8 × 2 = 0 + 0.620 384 328 089 6;
  • 15) 0.620 384 328 089 6 × 2 = 1 + 0.240 768 656 179 2;
  • 16) 0.240 768 656 179 2 × 2 = 0 + 0.481 537 312 358 4;
  • 17) 0.481 537 312 358 4 × 2 = 0 + 0.963 074 624 716 8;
  • 18) 0.963 074 624 716 8 × 2 = 1 + 0.926 149 249 433 6;
  • 19) 0.926 149 249 433 6 × 2 = 1 + 0.852 298 498 867 2;
  • 20) 0.852 298 498 867 2 × 2 = 1 + 0.704 596 997 734 4;
  • 21) 0.704 596 997 734 4 × 2 = 1 + 0.409 193 995 468 8;
  • 22) 0.409 193 995 468 8 × 2 = 0 + 0.818 387 990 937 6;
  • 23) 0.818 387 990 937 6 × 2 = 1 + 0.636 775 981 875 2;
  • 24) 0.636 775 981 875 2 × 2 = 1 + 0.273 551 963 750 4;
  • 25) 0.273 551 963 750 4 × 2 = 0 + 0.547 103 927 500 8;
  • 26) 0.547 103 927 500 8 × 2 = 1 + 0.094 207 855 001 6;
  • 27) 0.094 207 855 001 6 × 2 = 0 + 0.188 415 710 003 2;
  • 28) 0.188 415 710 003 2 × 2 = 0 + 0.376 831 420 006 4;
  • 29) 0.376 831 420 006 4 × 2 = 0 + 0.753 662 840 012 8;
  • 30) 0.753 662 840 012 8 × 2 = 1 + 0.507 325 680 025 6;
  • 31) 0.507 325 680 025 6 × 2 = 1 + 0.014 651 360 051 2;
  • 32) 0.014 651 360 051 2 × 2 = 0 + 0.029 302 720 102 4;
  • 33) 0.029 302 720 102 4 × 2 = 0 + 0.058 605 440 204 8;
  • 34) 0.058 605 440 204 8 × 2 = 0 + 0.117 210 880 409 6;
  • 35) 0.117 210 880 409 6 × 2 = 0 + 0.234 421 760 819 2;
  • 36) 0.234 421 760 819 2 × 2 = 0 + 0.468 843 521 638 4;
  • 37) 0.468 843 521 638 4 × 2 = 0 + 0.937 687 043 276 8;
  • 38) 0.937 687 043 276 8 × 2 = 1 + 0.875 374 086 553 6;
  • 39) 0.875 374 086 553 6 × 2 = 1 + 0.750 748 173 107 2;
  • 40) 0.750 748 173 107 2 × 2 = 1 + 0.501 496 346 214 4;
  • 41) 0.501 496 346 214 4 × 2 = 1 + 0.002 992 692 428 8;
  • 42) 0.002 992 692 428 8 × 2 = 0 + 0.005 985 384 857 6;
  • 43) 0.005 985 384 857 6 × 2 = 0 + 0.011 970 769 715 2;
  • 44) 0.011 970 769 715 2 × 2 = 0 + 0.023 941 539 430 4;
  • 45) 0.023 941 539 430 4 × 2 = 0 + 0.047 883 078 860 8;
  • 46) 0.047 883 078 860 8 × 2 = 0 + 0.095 766 157 721 6;
  • 47) 0.095 766 157 721 6 × 2 = 0 + 0.191 532 315 443 2;
  • 48) 0.191 532 315 443 2 × 2 = 0 + 0.383 064 630 886 4;
  • 49) 0.383 064 630 886 4 × 2 = 0 + 0.766 129 261 772 8;
  • 50) 0.766 129 261 772 8 × 2 = 1 + 0.532 258 523 545 6;
  • 51) 0.532 258 523 545 6 × 2 = 1 + 0.064 517 047 091 2;
  • 52) 0.064 517 047 091 2 × 2 = 0 + 0.129 034 094 182 4;
  • 53) 0.129 034 094 182 4 × 2 = 0 + 0.258 068 188 364 8;
  • 54) 0.258 068 188 364 8 × 2 = 0 + 0.516 136 376 729 6;
  • 55) 0.516 136 376 729 6 × 2 = 1 + 0.032 272 753 459 2;
  • 56) 0.032 272 753 459 2 × 2 = 0 + 0.064 545 506 918 4;
  • 57) 0.064 545 506 918 4 × 2 = 0 + 0.129 091 013 836 8;
  • 58) 0.129 091 013 836 8 × 2 = 0 + 0.258 182 027 673 6;
  • 59) 0.258 182 027 673 6 × 2 = 0 + 0.516 364 055 347 2;
  • 60) 0.516 364 055 347 2 × 2 = 1 + 0.032 728 110 694 4;
  • 61) 0.032 728 110 694 4 × 2 = 0 + 0.065 456 221 388 8;
  • 62) 0.065 456 221 388 8 × 2 = 0 + 0.130 912 442 777 6;
  • 63) 0.130 912 442 777 6 × 2 = 0 + 0.261 824 885 555 2;
  • 64) 0.261 824 885 555 2 × 2 = 0 + 0.523 649 771 110 4;

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:


0.000 282 005 879 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000(2)

6. Positive number before normalization:

0.000 282 005 879 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 879 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000 =

0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000


Decimal number -0.000 282 005 879 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0000 0111 1000 0000 0110 0010 0001 0000

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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