-0.000 282 005 914 417 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 417(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
-0.000 282 005 914 417(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:

|-0.000 282 005 914 417| = 0.000 282 005 914 417


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: 0.000 282 005 914 417.

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) 0.000 282 005 914 417 × 2 = 0 + 0.000 564 011 828 834;
  • 2) 0.000 564 011 828 834 × 2 = 0 + 0.001 128 023 657 668;
  • 3) 0.001 128 023 657 668 × 2 = 0 + 0.002 256 047 315 336;
  • 4) 0.002 256 047 315 336 × 2 = 0 + 0.004 512 094 630 672;
  • 5) 0.004 512 094 630 672 × 2 = 0 + 0.009 024 189 261 344;
  • 6) 0.009 024 189 261 344 × 2 = 0 + 0.018 048 378 522 688;
  • 7) 0.018 048 378 522 688 × 2 = 0 + 0.036 096 757 045 376;
  • 8) 0.036 096 757 045 376 × 2 = 0 + 0.072 193 514 090 752;
  • 9) 0.072 193 514 090 752 × 2 = 0 + 0.144 387 028 181 504;
  • 10) 0.144 387 028 181 504 × 2 = 0 + 0.288 774 056 363 008;
  • 11) 0.288 774 056 363 008 × 2 = 0 + 0.577 548 112 726 016;
  • 12) 0.577 548 112 726 016 × 2 = 1 + 0.155 096 225 452 032;
  • 13) 0.155 096 225 452 032 × 2 = 0 + 0.310 192 450 904 064;
  • 14) 0.310 192 450 904 064 × 2 = 0 + 0.620 384 901 808 128;
  • 15) 0.620 384 901 808 128 × 2 = 1 + 0.240 769 803 616 256;
  • 16) 0.240 769 803 616 256 × 2 = 0 + 0.481 539 607 232 512;
  • 17) 0.481 539 607 232 512 × 2 = 0 + 0.963 079 214 465 024;
  • 18) 0.963 079 214 465 024 × 2 = 1 + 0.926 158 428 930 048;
  • 19) 0.926 158 428 930 048 × 2 = 1 + 0.852 316 857 860 096;
  • 20) 0.852 316 857 860 096 × 2 = 1 + 0.704 633 715 720 192;
  • 21) 0.704 633 715 720 192 × 2 = 1 + 0.409 267 431 440 384;
  • 22) 0.409 267 431 440 384 × 2 = 0 + 0.818 534 862 880 768;
  • 23) 0.818 534 862 880 768 × 2 = 1 + 0.637 069 725 761 536;
  • 24) 0.637 069 725 761 536 × 2 = 1 + 0.274 139 451 523 072;
  • 25) 0.274 139 451 523 072 × 2 = 0 + 0.548 278 903 046 144;
  • 26) 0.548 278 903 046 144 × 2 = 1 + 0.096 557 806 092 288;
  • 27) 0.096 557 806 092 288 × 2 = 0 + 0.193 115 612 184 576;
  • 28) 0.193 115 612 184 576 × 2 = 0 + 0.386 231 224 369 152;
  • 29) 0.386 231 224 369 152 × 2 = 0 + 0.772 462 448 738 304;
  • 30) 0.772 462 448 738 304 × 2 = 1 + 0.544 924 897 476 608;
  • 31) 0.544 924 897 476 608 × 2 = 1 + 0.089 849 794 953 216;
  • 32) 0.089 849 794 953 216 × 2 = 0 + 0.179 699 589 906 432;
  • 33) 0.179 699 589 906 432 × 2 = 0 + 0.359 399 179 812 864;
  • 34) 0.359 399 179 812 864 × 2 = 0 + 0.718 798 359 625 728;
  • 35) 0.718 798 359 625 728 × 2 = 1 + 0.437 596 719 251 456;
  • 36) 0.437 596 719 251 456 × 2 = 0 + 0.875 193 438 502 912;
  • 37) 0.875 193 438 502 912 × 2 = 1 + 0.750 386 877 005 824;
  • 38) 0.750 386 877 005 824 × 2 = 1 + 0.500 773 754 011 648;
  • 39) 0.500 773 754 011 648 × 2 = 1 + 0.001 547 508 023 296;
  • 40) 0.001 547 508 023 296 × 2 = 0 + 0.003 095 016 046 592;
  • 41) 0.003 095 016 046 592 × 2 = 0 + 0.006 190 032 093 184;
  • 42) 0.006 190 032 093 184 × 2 = 0 + 0.012 380 064 186 368;
  • 43) 0.012 380 064 186 368 × 2 = 0 + 0.024 760 128 372 736;
  • 44) 0.024 760 128 372 736 × 2 = 0 + 0.049 520 256 745 472;
  • 45) 0.049 520 256 745 472 × 2 = 0 + 0.099 040 513 490 944;
  • 46) 0.099 040 513 490 944 × 2 = 0 + 0.198 081 026 981 888;
  • 47) 0.198 081 026 981 888 × 2 = 0 + 0.396 162 053 963 776;
  • 48) 0.396 162 053 963 776 × 2 = 0 + 0.792 324 107 927 552;
  • 49) 0.792 324 107 927 552 × 2 = 1 + 0.584 648 215 855 104;
  • 50) 0.584 648 215 855 104 × 2 = 1 + 0.169 296 431 710 208;
  • 51) 0.169 296 431 710 208 × 2 = 0 + 0.338 592 863 420 416;
  • 52) 0.338 592 863 420 416 × 2 = 0 + 0.677 185 726 840 832;
  • 53) 0.677 185 726 840 832 × 2 = 1 + 0.354 371 453 681 664;
  • 54) 0.354 371 453 681 664 × 2 = 0 + 0.708 742 907 363 328;
  • 55) 0.708 742 907 363 328 × 2 = 1 + 0.417 485 814 726 656;
  • 56) 0.417 485 814 726 656 × 2 = 0 + 0.834 971 629 453 312;
  • 57) 0.834 971 629 453 312 × 2 = 1 + 0.669 943 258 906 624;
  • 58) 0.669 943 258 906 624 × 2 = 1 + 0.339 886 517 813 248;
  • 59) 0.339 886 517 813 248 × 2 = 0 + 0.679 773 035 626 496;
  • 60) 0.679 773 035 626 496 × 2 = 1 + 0.359 546 071 252 992;
  • 61) 0.359 546 071 252 992 × 2 = 0 + 0.719 092 142 505 984;
  • 62) 0.719 092 142 505 984 × 2 = 1 + 0.438 184 285 011 968;
  • 63) 0.438 184 285 011 968 × 2 = 0 + 0.876 368 570 023 936;
  • 64) 0.876 368 570 023 936 × 2 = 1 + 0.752 737 140 047 872;

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:


0.000 282 005 914 417(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0000 1100 1010 1101 0101(2)

6. Positive number before normalization:

0.000 282 005 914 417(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0000 1100 1010 1101 0101(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:


0.000 282 005 914 417(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0000 1100 1010 1101 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0000 1100 1010 1101 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0000 0000 1100 1010 1101 0101(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 0010 1110 0000 0000 1100 1010 1101 0101


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 0010 1110 0000 0000 1100 1010 1101 0101 =

0010 0111 1011 0100 0110 0010 1110 0000 0000 1100 1010 1101 0101


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 0010 1110 0000 0000 1100 1010 1101 0101


Decimal number -0.000 282 005 914 417 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 0000 1100 1010 1101 0101

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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