-0.000 282 005 956 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 956 5(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
-0.000 282 005 956 5(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:

|-0.000 282 005 956 5| = 0.000 282 005 956 5


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: 0.000 282 005 956 5.

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) 0.000 282 005 956 5 × 2 = 0 + 0.000 564 011 913;
  • 2) 0.000 564 011 913 × 2 = 0 + 0.001 128 023 826;
  • 3) 0.001 128 023 826 × 2 = 0 + 0.002 256 047 652;
  • 4) 0.002 256 047 652 × 2 = 0 + 0.004 512 095 304;
  • 5) 0.004 512 095 304 × 2 = 0 + 0.009 024 190 608;
  • 6) 0.009 024 190 608 × 2 = 0 + 0.018 048 381 216;
  • 7) 0.018 048 381 216 × 2 = 0 + 0.036 096 762 432;
  • 8) 0.036 096 762 432 × 2 = 0 + 0.072 193 524 864;
  • 9) 0.072 193 524 864 × 2 = 0 + 0.144 387 049 728;
  • 10) 0.144 387 049 728 × 2 = 0 + 0.288 774 099 456;
  • 11) 0.288 774 099 456 × 2 = 0 + 0.577 548 198 912;
  • 12) 0.577 548 198 912 × 2 = 1 + 0.155 096 397 824;
  • 13) 0.155 096 397 824 × 2 = 0 + 0.310 192 795 648;
  • 14) 0.310 192 795 648 × 2 = 0 + 0.620 385 591 296;
  • 15) 0.620 385 591 296 × 2 = 1 + 0.240 771 182 592;
  • 16) 0.240 771 182 592 × 2 = 0 + 0.481 542 365 184;
  • 17) 0.481 542 365 184 × 2 = 0 + 0.963 084 730 368;
  • 18) 0.963 084 730 368 × 2 = 1 + 0.926 169 460 736;
  • 19) 0.926 169 460 736 × 2 = 1 + 0.852 338 921 472;
  • 20) 0.852 338 921 472 × 2 = 1 + 0.704 677 842 944;
  • 21) 0.704 677 842 944 × 2 = 1 + 0.409 355 685 888;
  • 22) 0.409 355 685 888 × 2 = 0 + 0.818 711 371 776;
  • 23) 0.818 711 371 776 × 2 = 1 + 0.637 422 743 552;
  • 24) 0.637 422 743 552 × 2 = 1 + 0.274 845 487 104;
  • 25) 0.274 845 487 104 × 2 = 0 + 0.549 690 974 208;
  • 26) 0.549 690 974 208 × 2 = 1 + 0.099 381 948 416;
  • 27) 0.099 381 948 416 × 2 = 0 + 0.198 763 896 832;
  • 28) 0.198 763 896 832 × 2 = 0 + 0.397 527 793 664;
  • 29) 0.397 527 793 664 × 2 = 0 + 0.795 055 587 328;
  • 30) 0.795 055 587 328 × 2 = 1 + 0.590 111 174 656;
  • 31) 0.590 111 174 656 × 2 = 1 + 0.180 222 349 312;
  • 32) 0.180 222 349 312 × 2 = 0 + 0.360 444 698 624;
  • 33) 0.360 444 698 624 × 2 = 0 + 0.720 889 397 248;
  • 34) 0.720 889 397 248 × 2 = 1 + 0.441 778 794 496;
  • 35) 0.441 778 794 496 × 2 = 0 + 0.883 557 588 992;
  • 36) 0.883 557 588 992 × 2 = 1 + 0.767 115 177 984;
  • 37) 0.767 115 177 984 × 2 = 1 + 0.534 230 355 968;
  • 38) 0.534 230 355 968 × 2 = 1 + 0.068 460 711 936;
  • 39) 0.068 460 711 936 × 2 = 0 + 0.136 921 423 872;
  • 40) 0.136 921 423 872 × 2 = 0 + 0.273 842 847 744;
  • 41) 0.273 842 847 744 × 2 = 0 + 0.547 685 695 488;
  • 42) 0.547 685 695 488 × 2 = 1 + 0.095 371 390 976;
  • 43) 0.095 371 390 976 × 2 = 0 + 0.190 742 781 952;
  • 44) 0.190 742 781 952 × 2 = 0 + 0.381 485 563 904;
  • 45) 0.381 485 563 904 × 2 = 0 + 0.762 971 127 808;
  • 46) 0.762 971 127 808 × 2 = 1 + 0.525 942 255 616;
  • 47) 0.525 942 255 616 × 2 = 1 + 0.051 884 511 232;
  • 48) 0.051 884 511 232 × 2 = 0 + 0.103 769 022 464;
  • 49) 0.103 769 022 464 × 2 = 0 + 0.207 538 044 928;
  • 50) 0.207 538 044 928 × 2 = 0 + 0.415 076 089 856;
  • 51) 0.415 076 089 856 × 2 = 0 + 0.830 152 179 712;
  • 52) 0.830 152 179 712 × 2 = 1 + 0.660 304 359 424;
  • 53) 0.660 304 359 424 × 2 = 1 + 0.320 608 718 848;
  • 54) 0.320 608 718 848 × 2 = 0 + 0.641 217 437 696;
  • 55) 0.641 217 437 696 × 2 = 1 + 0.282 434 875 392;
  • 56) 0.282 434 875 392 × 2 = 0 + 0.564 869 750 784;
  • 57) 0.564 869 750 784 × 2 = 1 + 0.129 739 501 568;
  • 58) 0.129 739 501 568 × 2 = 0 + 0.259 479 003 136;
  • 59) 0.259 479 003 136 × 2 = 0 + 0.518 958 006 272;
  • 60) 0.518 958 006 272 × 2 = 1 + 0.037 916 012 544;
  • 61) 0.037 916 012 544 × 2 = 0 + 0.075 832 025 088;
  • 62) 0.075 832 025 088 × 2 = 0 + 0.151 664 050 176;
  • 63) 0.151 664 050 176 × 2 = 0 + 0.303 328 100 352;
  • 64) 0.303 328 100 352 × 2 = 0 + 0.606 656 200 704;

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:


0.000 282 005 956 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 0100 0110 0001 1010 1001 0000(2)

6. Positive number before normalization:

0.000 282 005 956 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 0100 0110 0001 1010 1001 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:


0.000 282 005 956 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 0100 0110 0001 1010 1001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 0100 0110 0001 1010 1001 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0101 1100 0100 0110 0001 1010 1001 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 0101 1100 0100 0110 0001 1010 1001 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 0101 1100 0100 0110 0001 1010 1001 0000 =

0010 0111 1011 0100 0110 0101 1100 0100 0110 0001 1010 1001 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 0101 1100 0100 0110 0001 1010 1001 0000


Decimal number -0.000 282 005 956 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0101 1100 0100 0110 0001 1010 1001 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