-0.000 282 005 914 401 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 401(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
-0.000 282 005 914 401(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:

|-0.000 282 005 914 401| = 0.000 282 005 914 401


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: 0.000 282 005 914 401.

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) 0.000 282 005 914 401 × 2 = 0 + 0.000 564 011 828 802;
  • 2) 0.000 564 011 828 802 × 2 = 0 + 0.001 128 023 657 604;
  • 3) 0.001 128 023 657 604 × 2 = 0 + 0.002 256 047 315 208;
  • 4) 0.002 256 047 315 208 × 2 = 0 + 0.004 512 094 630 416;
  • 5) 0.004 512 094 630 416 × 2 = 0 + 0.009 024 189 260 832;
  • 6) 0.009 024 189 260 832 × 2 = 0 + 0.018 048 378 521 664;
  • 7) 0.018 048 378 521 664 × 2 = 0 + 0.036 096 757 043 328;
  • 8) 0.036 096 757 043 328 × 2 = 0 + 0.072 193 514 086 656;
  • 9) 0.072 193 514 086 656 × 2 = 0 + 0.144 387 028 173 312;
  • 10) 0.144 387 028 173 312 × 2 = 0 + 0.288 774 056 346 624;
  • 11) 0.288 774 056 346 624 × 2 = 0 + 0.577 548 112 693 248;
  • 12) 0.577 548 112 693 248 × 2 = 1 + 0.155 096 225 386 496;
  • 13) 0.155 096 225 386 496 × 2 = 0 + 0.310 192 450 772 992;
  • 14) 0.310 192 450 772 992 × 2 = 0 + 0.620 384 901 545 984;
  • 15) 0.620 384 901 545 984 × 2 = 1 + 0.240 769 803 091 968;
  • 16) 0.240 769 803 091 968 × 2 = 0 + 0.481 539 606 183 936;
  • 17) 0.481 539 606 183 936 × 2 = 0 + 0.963 079 212 367 872;
  • 18) 0.963 079 212 367 872 × 2 = 1 + 0.926 158 424 735 744;
  • 19) 0.926 158 424 735 744 × 2 = 1 + 0.852 316 849 471 488;
  • 20) 0.852 316 849 471 488 × 2 = 1 + 0.704 633 698 942 976;
  • 21) 0.704 633 698 942 976 × 2 = 1 + 0.409 267 397 885 952;
  • 22) 0.409 267 397 885 952 × 2 = 0 + 0.818 534 795 771 904;
  • 23) 0.818 534 795 771 904 × 2 = 1 + 0.637 069 591 543 808;
  • 24) 0.637 069 591 543 808 × 2 = 1 + 0.274 139 183 087 616;
  • 25) 0.274 139 183 087 616 × 2 = 0 + 0.548 278 366 175 232;
  • 26) 0.548 278 366 175 232 × 2 = 1 + 0.096 556 732 350 464;
  • 27) 0.096 556 732 350 464 × 2 = 0 + 0.193 113 464 700 928;
  • 28) 0.193 113 464 700 928 × 2 = 0 + 0.386 226 929 401 856;
  • 29) 0.386 226 929 401 856 × 2 = 0 + 0.772 453 858 803 712;
  • 30) 0.772 453 858 803 712 × 2 = 1 + 0.544 907 717 607 424;
  • 31) 0.544 907 717 607 424 × 2 = 1 + 0.089 815 435 214 848;
  • 32) 0.089 815 435 214 848 × 2 = 0 + 0.179 630 870 429 696;
  • 33) 0.179 630 870 429 696 × 2 = 0 + 0.359 261 740 859 392;
  • 34) 0.359 261 740 859 392 × 2 = 0 + 0.718 523 481 718 784;
  • 35) 0.718 523 481 718 784 × 2 = 1 + 0.437 046 963 437 568;
  • 36) 0.437 046 963 437 568 × 2 = 0 + 0.874 093 926 875 136;
  • 37) 0.874 093 926 875 136 × 2 = 1 + 0.748 187 853 750 272;
  • 38) 0.748 187 853 750 272 × 2 = 1 + 0.496 375 707 500 544;
  • 39) 0.496 375 707 500 544 × 2 = 0 + 0.992 751 415 001 088;
  • 40) 0.992 751 415 001 088 × 2 = 1 + 0.985 502 830 002 176;
  • 41) 0.985 502 830 002 176 × 2 = 1 + 0.971 005 660 004 352;
  • 42) 0.971 005 660 004 352 × 2 = 1 + 0.942 011 320 008 704;
  • 43) 0.942 011 320 008 704 × 2 = 1 + 0.884 022 640 017 408;
  • 44) 0.884 022 640 017 408 × 2 = 1 + 0.768 045 280 034 816;
  • 45) 0.768 045 280 034 816 × 2 = 1 + 0.536 090 560 069 632;
  • 46) 0.536 090 560 069 632 × 2 = 1 + 0.072 181 120 139 264;
  • 47) 0.072 181 120 139 264 × 2 = 0 + 0.144 362 240 278 528;
  • 48) 0.144 362 240 278 528 × 2 = 0 + 0.288 724 480 557 056;
  • 49) 0.288 724 480 557 056 × 2 = 0 + 0.577 448 961 114 112;
  • 50) 0.577 448 961 114 112 × 2 = 1 + 0.154 897 922 228 224;
  • 51) 0.154 897 922 228 224 × 2 = 0 + 0.309 795 844 456 448;
  • 52) 0.309 795 844 456 448 × 2 = 0 + 0.619 591 688 912 896;
  • 53) 0.619 591 688 912 896 × 2 = 1 + 0.239 183 377 825 792;
  • 54) 0.239 183 377 825 792 × 2 = 0 + 0.478 366 755 651 584;
  • 55) 0.478 366 755 651 584 × 2 = 0 + 0.956 733 511 303 168;
  • 56) 0.956 733 511 303 168 × 2 = 1 + 0.913 467 022 606 336;
  • 57) 0.913 467 022 606 336 × 2 = 1 + 0.826 934 045 212 672;
  • 58) 0.826 934 045 212 672 × 2 = 1 + 0.653 868 090 425 344;
  • 59) 0.653 868 090 425 344 × 2 = 1 + 0.307 736 180 850 688;
  • 60) 0.307 736 180 850 688 × 2 = 0 + 0.615 472 361 701 376;
  • 61) 0.615 472 361 701 376 × 2 = 1 + 0.230 944 723 402 752;
  • 62) 0.230 944 723 402 752 × 2 = 0 + 0.461 889 446 805 504;
  • 63) 0.461 889 446 805 504 × 2 = 0 + 0.923 778 893 611 008;
  • 64) 0.923 778 893 611 008 × 2 = 1 + 0.847 557 787 222 016;

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:


0.000 282 005 914 401(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1100 0100 1001 1110 1001(2)

6. Positive number before normalization:

0.000 282 005 914 401(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1100 0100 1001 1110 1001(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:


0.000 282 005 914 401(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1100 0100 1001 1110 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1100 0100 1001 1110 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1100 0100 1001 1110 1001(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 1101 1111 1100 0100 1001 1110 1001


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 1101 1111 1100 0100 1001 1110 1001 =

0010 0111 1011 0100 0110 0010 1101 1111 1100 0100 1001 1110 1001


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 1101 1111 1100 0100 1001 1110 1001


Decimal number -0.000 282 005 914 401 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 1100 0100 1001 1110 1001

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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