-0.000 282 005 914 294 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 294(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
-0.000 282 005 914 294(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:

|-0.000 282 005 914 294| = 0.000 282 005 914 294


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: 0.000 282 005 914 294.

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) 0.000 282 005 914 294 × 2 = 0 + 0.000 564 011 828 588;
  • 2) 0.000 564 011 828 588 × 2 = 0 + 0.001 128 023 657 176;
  • 3) 0.001 128 023 657 176 × 2 = 0 + 0.002 256 047 314 352;
  • 4) 0.002 256 047 314 352 × 2 = 0 + 0.004 512 094 628 704;
  • 5) 0.004 512 094 628 704 × 2 = 0 + 0.009 024 189 257 408;
  • 6) 0.009 024 189 257 408 × 2 = 0 + 0.018 048 378 514 816;
  • 7) 0.018 048 378 514 816 × 2 = 0 + 0.036 096 757 029 632;
  • 8) 0.036 096 757 029 632 × 2 = 0 + 0.072 193 514 059 264;
  • 9) 0.072 193 514 059 264 × 2 = 0 + 0.144 387 028 118 528;
  • 10) 0.144 387 028 118 528 × 2 = 0 + 0.288 774 056 237 056;
  • 11) 0.288 774 056 237 056 × 2 = 0 + 0.577 548 112 474 112;
  • 12) 0.577 548 112 474 112 × 2 = 1 + 0.155 096 224 948 224;
  • 13) 0.155 096 224 948 224 × 2 = 0 + 0.310 192 449 896 448;
  • 14) 0.310 192 449 896 448 × 2 = 0 + 0.620 384 899 792 896;
  • 15) 0.620 384 899 792 896 × 2 = 1 + 0.240 769 799 585 792;
  • 16) 0.240 769 799 585 792 × 2 = 0 + 0.481 539 599 171 584;
  • 17) 0.481 539 599 171 584 × 2 = 0 + 0.963 079 198 343 168;
  • 18) 0.963 079 198 343 168 × 2 = 1 + 0.926 158 396 686 336;
  • 19) 0.926 158 396 686 336 × 2 = 1 + 0.852 316 793 372 672;
  • 20) 0.852 316 793 372 672 × 2 = 1 + 0.704 633 586 745 344;
  • 21) 0.704 633 586 745 344 × 2 = 1 + 0.409 267 173 490 688;
  • 22) 0.409 267 173 490 688 × 2 = 0 + 0.818 534 346 981 376;
  • 23) 0.818 534 346 981 376 × 2 = 1 + 0.637 068 693 962 752;
  • 24) 0.637 068 693 962 752 × 2 = 1 + 0.274 137 387 925 504;
  • 25) 0.274 137 387 925 504 × 2 = 0 + 0.548 274 775 851 008;
  • 26) 0.548 274 775 851 008 × 2 = 1 + 0.096 549 551 702 016;
  • 27) 0.096 549 551 702 016 × 2 = 0 + 0.193 099 103 404 032;
  • 28) 0.193 099 103 404 032 × 2 = 0 + 0.386 198 206 808 064;
  • 29) 0.386 198 206 808 064 × 2 = 0 + 0.772 396 413 616 128;
  • 30) 0.772 396 413 616 128 × 2 = 1 + 0.544 792 827 232 256;
  • 31) 0.544 792 827 232 256 × 2 = 1 + 0.089 585 654 464 512;
  • 32) 0.089 585 654 464 512 × 2 = 0 + 0.179 171 308 929 024;
  • 33) 0.179 171 308 929 024 × 2 = 0 + 0.358 342 617 858 048;
  • 34) 0.358 342 617 858 048 × 2 = 0 + 0.716 685 235 716 096;
  • 35) 0.716 685 235 716 096 × 2 = 1 + 0.433 370 471 432 192;
  • 36) 0.433 370 471 432 192 × 2 = 0 + 0.866 740 942 864 384;
  • 37) 0.866 740 942 864 384 × 2 = 1 + 0.733 481 885 728 768;
  • 38) 0.733 481 885 728 768 × 2 = 1 + 0.466 963 771 457 536;
  • 39) 0.466 963 771 457 536 × 2 = 0 + 0.933 927 542 915 072;
  • 40) 0.933 927 542 915 072 × 2 = 1 + 0.867 855 085 830 144;
  • 41) 0.867 855 085 830 144 × 2 = 1 + 0.735 710 171 660 288;
  • 42) 0.735 710 171 660 288 × 2 = 1 + 0.471 420 343 320 576;
  • 43) 0.471 420 343 320 576 × 2 = 0 + 0.942 840 686 641 152;
  • 44) 0.942 840 686 641 152 × 2 = 1 + 0.885 681 373 282 304;
  • 45) 0.885 681 373 282 304 × 2 = 1 + 0.771 362 746 564 608;
  • 46) 0.771 362 746 564 608 × 2 = 1 + 0.542 725 493 129 216;
  • 47) 0.542 725 493 129 216 × 2 = 1 + 0.085 450 986 258 432;
  • 48) 0.085 450 986 258 432 × 2 = 0 + 0.170 901 972 516 864;
  • 49) 0.170 901 972 516 864 × 2 = 0 + 0.341 803 945 033 728;
  • 50) 0.341 803 945 033 728 × 2 = 0 + 0.683 607 890 067 456;
  • 51) 0.683 607 890 067 456 × 2 = 1 + 0.367 215 780 134 912;
  • 52) 0.367 215 780 134 912 × 2 = 0 + 0.734 431 560 269 824;
  • 53) 0.734 431 560 269 824 × 2 = 1 + 0.468 863 120 539 648;
  • 54) 0.468 863 120 539 648 × 2 = 0 + 0.937 726 241 079 296;
  • 55) 0.937 726 241 079 296 × 2 = 1 + 0.875 452 482 158 592;
  • 56) 0.875 452 482 158 592 × 2 = 1 + 0.750 904 964 317 184;
  • 57) 0.750 904 964 317 184 × 2 = 1 + 0.501 809 928 634 368;
  • 58) 0.501 809 928 634 368 × 2 = 1 + 0.003 619 857 268 736;
  • 59) 0.003 619 857 268 736 × 2 = 0 + 0.007 239 714 537 472;
  • 60) 0.007 239 714 537 472 × 2 = 0 + 0.014 479 429 074 944;
  • 61) 0.014 479 429 074 944 × 2 = 0 + 0.028 958 858 149 888;
  • 62) 0.028 958 858 149 888 × 2 = 0 + 0.057 917 716 299 776;
  • 63) 0.057 917 716 299 776 × 2 = 0 + 0.115 835 432 599 552;
  • 64) 0.115 835 432 599 552 × 2 = 0 + 0.231 670 865 199 104;

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:


0.000 282 005 914 294(10) =

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

6. Positive number before normalization:

0.000 282 005 914 294(10) =

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


0.000 282 005 914 294(10) =

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

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

1.0010 0111 1011 0100 0110 0010 1101 1101 1110 0010 1011 1100 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 0010 1101 1101 1110 0010 1011 1100 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 0010 1101 1101 1110 0010 1011 1100 0000 =

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


Decimal number -0.000 282 005 914 294 converted to 64 bit double precision IEEE 754 binary floating point representation:

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