-0.000 282 005 914 446 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 446(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
-0.000 282 005 914 446(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:

|-0.000 282 005 914 446| = 0.000 282 005 914 446


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: 0.000 282 005 914 446.

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) 0.000 282 005 914 446 × 2 = 0 + 0.000 564 011 828 892;
  • 2) 0.000 564 011 828 892 × 2 = 0 + 0.001 128 023 657 784;
  • 3) 0.001 128 023 657 784 × 2 = 0 + 0.002 256 047 315 568;
  • 4) 0.002 256 047 315 568 × 2 = 0 + 0.004 512 094 631 136;
  • 5) 0.004 512 094 631 136 × 2 = 0 + 0.009 024 189 262 272;
  • 6) 0.009 024 189 262 272 × 2 = 0 + 0.018 048 378 524 544;
  • 7) 0.018 048 378 524 544 × 2 = 0 + 0.036 096 757 049 088;
  • 8) 0.036 096 757 049 088 × 2 = 0 + 0.072 193 514 098 176;
  • 9) 0.072 193 514 098 176 × 2 = 0 + 0.144 387 028 196 352;
  • 10) 0.144 387 028 196 352 × 2 = 0 + 0.288 774 056 392 704;
  • 11) 0.288 774 056 392 704 × 2 = 0 + 0.577 548 112 785 408;
  • 12) 0.577 548 112 785 408 × 2 = 1 + 0.155 096 225 570 816;
  • 13) 0.155 096 225 570 816 × 2 = 0 + 0.310 192 451 141 632;
  • 14) 0.310 192 451 141 632 × 2 = 0 + 0.620 384 902 283 264;
  • 15) 0.620 384 902 283 264 × 2 = 1 + 0.240 769 804 566 528;
  • 16) 0.240 769 804 566 528 × 2 = 0 + 0.481 539 609 133 056;
  • 17) 0.481 539 609 133 056 × 2 = 0 + 0.963 079 218 266 112;
  • 18) 0.963 079 218 266 112 × 2 = 1 + 0.926 158 436 532 224;
  • 19) 0.926 158 436 532 224 × 2 = 1 + 0.852 316 873 064 448;
  • 20) 0.852 316 873 064 448 × 2 = 1 + 0.704 633 746 128 896;
  • 21) 0.704 633 746 128 896 × 2 = 1 + 0.409 267 492 257 792;
  • 22) 0.409 267 492 257 792 × 2 = 0 + 0.818 534 984 515 584;
  • 23) 0.818 534 984 515 584 × 2 = 1 + 0.637 069 969 031 168;
  • 24) 0.637 069 969 031 168 × 2 = 1 + 0.274 139 938 062 336;
  • 25) 0.274 139 938 062 336 × 2 = 0 + 0.548 279 876 124 672;
  • 26) 0.548 279 876 124 672 × 2 = 1 + 0.096 559 752 249 344;
  • 27) 0.096 559 752 249 344 × 2 = 0 + 0.193 119 504 498 688;
  • 28) 0.193 119 504 498 688 × 2 = 0 + 0.386 239 008 997 376;
  • 29) 0.386 239 008 997 376 × 2 = 0 + 0.772 478 017 994 752;
  • 30) 0.772 478 017 994 752 × 2 = 1 + 0.544 956 035 989 504;
  • 31) 0.544 956 035 989 504 × 2 = 1 + 0.089 912 071 979 008;
  • 32) 0.089 912 071 979 008 × 2 = 0 + 0.179 824 143 958 016;
  • 33) 0.179 824 143 958 016 × 2 = 0 + 0.359 648 287 916 032;
  • 34) 0.359 648 287 916 032 × 2 = 0 + 0.719 296 575 832 064;
  • 35) 0.719 296 575 832 064 × 2 = 1 + 0.438 593 151 664 128;
  • 36) 0.438 593 151 664 128 × 2 = 0 + 0.877 186 303 328 256;
  • 37) 0.877 186 303 328 256 × 2 = 1 + 0.754 372 606 656 512;
  • 38) 0.754 372 606 656 512 × 2 = 1 + 0.508 745 213 313 024;
  • 39) 0.508 745 213 313 024 × 2 = 1 + 0.017 490 426 626 048;
  • 40) 0.017 490 426 626 048 × 2 = 0 + 0.034 980 853 252 096;
  • 41) 0.034 980 853 252 096 × 2 = 0 + 0.069 961 706 504 192;
  • 42) 0.069 961 706 504 192 × 2 = 0 + 0.139 923 413 008 384;
  • 43) 0.139 923 413 008 384 × 2 = 0 + 0.279 846 826 016 768;
  • 44) 0.279 846 826 016 768 × 2 = 0 + 0.559 693 652 033 536;
  • 45) 0.559 693 652 033 536 × 2 = 1 + 0.119 387 304 067 072;
  • 46) 0.119 387 304 067 072 × 2 = 0 + 0.238 774 608 134 144;
  • 47) 0.238 774 608 134 144 × 2 = 0 + 0.477 549 216 268 288;
  • 48) 0.477 549 216 268 288 × 2 = 0 + 0.955 098 432 536 576;
  • 49) 0.955 098 432 536 576 × 2 = 1 + 0.910 196 865 073 152;
  • 50) 0.910 196 865 073 152 × 2 = 1 + 0.820 393 730 146 304;
  • 51) 0.820 393 730 146 304 × 2 = 1 + 0.640 787 460 292 608;
  • 52) 0.640 787 460 292 608 × 2 = 1 + 0.281 574 920 585 216;
  • 53) 0.281 574 920 585 216 × 2 = 0 + 0.563 149 841 170 432;
  • 54) 0.563 149 841 170 432 × 2 = 1 + 0.126 299 682 340 864;
  • 55) 0.126 299 682 340 864 × 2 = 0 + 0.252 599 364 681 728;
  • 56) 0.252 599 364 681 728 × 2 = 0 + 0.505 198 729 363 456;
  • 57) 0.505 198 729 363 456 × 2 = 1 + 0.010 397 458 726 912;
  • 58) 0.010 397 458 726 912 × 2 = 0 + 0.020 794 917 453 824;
  • 59) 0.020 794 917 453 824 × 2 = 0 + 0.041 589 834 907 648;
  • 60) 0.041 589 834 907 648 × 2 = 0 + 0.083 179 669 815 296;
  • 61) 0.083 179 669 815 296 × 2 = 0 + 0.166 359 339 630 592;
  • 62) 0.166 359 339 630 592 × 2 = 0 + 0.332 718 679 261 184;
  • 63) 0.332 718 679 261 184 × 2 = 0 + 0.665 437 358 522 368;
  • 64) 0.665 437 358 522 368 × 2 = 1 + 0.330 874 717 044 736;

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:


0.000 282 005 914 446(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1000 1111 0100 1000 0001(2)

6. Positive number before normalization:

0.000 282 005 914 446(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1000 1111 0100 1000 0001(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:


0.000 282 005 914 446(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1000 1111 0100 1000 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1000 1111 0100 1000 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0000 1000 1111 0100 1000 0001(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 1000 1111 0100 1000 0001


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 1000 1111 0100 1000 0001 =

0010 0111 1011 0100 0110 0010 1110 0000 1000 1111 0100 1000 0001


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 1000 1111 0100 1000 0001


Decimal number -0.000 282 005 914 446 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 1000 1111 0100 1000 0001

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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