-0.000 282 005 914 528 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 528(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
-0.000 282 005 914 528(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:

|-0.000 282 005 914 528| = 0.000 282 005 914 528


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: 0.000 282 005 914 528.

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) 0.000 282 005 914 528 × 2 = 0 + 0.000 564 011 829 056;
  • 2) 0.000 564 011 829 056 × 2 = 0 + 0.001 128 023 658 112;
  • 3) 0.001 128 023 658 112 × 2 = 0 + 0.002 256 047 316 224;
  • 4) 0.002 256 047 316 224 × 2 = 0 + 0.004 512 094 632 448;
  • 5) 0.004 512 094 632 448 × 2 = 0 + 0.009 024 189 264 896;
  • 6) 0.009 024 189 264 896 × 2 = 0 + 0.018 048 378 529 792;
  • 7) 0.018 048 378 529 792 × 2 = 0 + 0.036 096 757 059 584;
  • 8) 0.036 096 757 059 584 × 2 = 0 + 0.072 193 514 119 168;
  • 9) 0.072 193 514 119 168 × 2 = 0 + 0.144 387 028 238 336;
  • 10) 0.144 387 028 238 336 × 2 = 0 + 0.288 774 056 476 672;
  • 11) 0.288 774 056 476 672 × 2 = 0 + 0.577 548 112 953 344;
  • 12) 0.577 548 112 953 344 × 2 = 1 + 0.155 096 225 906 688;
  • 13) 0.155 096 225 906 688 × 2 = 0 + 0.310 192 451 813 376;
  • 14) 0.310 192 451 813 376 × 2 = 0 + 0.620 384 903 626 752;
  • 15) 0.620 384 903 626 752 × 2 = 1 + 0.240 769 807 253 504;
  • 16) 0.240 769 807 253 504 × 2 = 0 + 0.481 539 614 507 008;
  • 17) 0.481 539 614 507 008 × 2 = 0 + 0.963 079 229 014 016;
  • 18) 0.963 079 229 014 016 × 2 = 1 + 0.926 158 458 028 032;
  • 19) 0.926 158 458 028 032 × 2 = 1 + 0.852 316 916 056 064;
  • 20) 0.852 316 916 056 064 × 2 = 1 + 0.704 633 832 112 128;
  • 21) 0.704 633 832 112 128 × 2 = 1 + 0.409 267 664 224 256;
  • 22) 0.409 267 664 224 256 × 2 = 0 + 0.818 535 328 448 512;
  • 23) 0.818 535 328 448 512 × 2 = 1 + 0.637 070 656 897 024;
  • 24) 0.637 070 656 897 024 × 2 = 1 + 0.274 141 313 794 048;
  • 25) 0.274 141 313 794 048 × 2 = 0 + 0.548 282 627 588 096;
  • 26) 0.548 282 627 588 096 × 2 = 1 + 0.096 565 255 176 192;
  • 27) 0.096 565 255 176 192 × 2 = 0 + 0.193 130 510 352 384;
  • 28) 0.193 130 510 352 384 × 2 = 0 + 0.386 261 020 704 768;
  • 29) 0.386 261 020 704 768 × 2 = 0 + 0.772 522 041 409 536;
  • 30) 0.772 522 041 409 536 × 2 = 1 + 0.545 044 082 819 072;
  • 31) 0.545 044 082 819 072 × 2 = 1 + 0.090 088 165 638 144;
  • 32) 0.090 088 165 638 144 × 2 = 0 + 0.180 176 331 276 288;
  • 33) 0.180 176 331 276 288 × 2 = 0 + 0.360 352 662 552 576;
  • 34) 0.360 352 662 552 576 × 2 = 0 + 0.720 705 325 105 152;
  • 35) 0.720 705 325 105 152 × 2 = 1 + 0.441 410 650 210 304;
  • 36) 0.441 410 650 210 304 × 2 = 0 + 0.882 821 300 420 608;
  • 37) 0.882 821 300 420 608 × 2 = 1 + 0.765 642 600 841 216;
  • 38) 0.765 642 600 841 216 × 2 = 1 + 0.531 285 201 682 432;
  • 39) 0.531 285 201 682 432 × 2 = 1 + 0.062 570 403 364 864;
  • 40) 0.062 570 403 364 864 × 2 = 0 + 0.125 140 806 729 728;
  • 41) 0.125 140 806 729 728 × 2 = 0 + 0.250 281 613 459 456;
  • 42) 0.250 281 613 459 456 × 2 = 0 + 0.500 563 226 918 912;
  • 43) 0.500 563 226 918 912 × 2 = 1 + 0.001 126 453 837 824;
  • 44) 0.001 126 453 837 824 × 2 = 0 + 0.002 252 907 675 648;
  • 45) 0.002 252 907 675 648 × 2 = 0 + 0.004 505 815 351 296;
  • 46) 0.004 505 815 351 296 × 2 = 0 + 0.009 011 630 702 592;
  • 47) 0.009 011 630 702 592 × 2 = 0 + 0.018 023 261 405 184;
  • 48) 0.018 023 261 405 184 × 2 = 0 + 0.036 046 522 810 368;
  • 49) 0.036 046 522 810 368 × 2 = 0 + 0.072 093 045 620 736;
  • 50) 0.072 093 045 620 736 × 2 = 0 + 0.144 186 091 241 472;
  • 51) 0.144 186 091 241 472 × 2 = 0 + 0.288 372 182 482 944;
  • 52) 0.288 372 182 482 944 × 2 = 0 + 0.576 744 364 965 888;
  • 53) 0.576 744 364 965 888 × 2 = 1 + 0.153 488 729 931 776;
  • 54) 0.153 488 729 931 776 × 2 = 0 + 0.306 977 459 863 552;
  • 55) 0.306 977 459 863 552 × 2 = 0 + 0.613 954 919 727 104;
  • 56) 0.613 954 919 727 104 × 2 = 1 + 0.227 909 839 454 208;
  • 57) 0.227 909 839 454 208 × 2 = 0 + 0.455 819 678 908 416;
  • 58) 0.455 819 678 908 416 × 2 = 0 + 0.911 639 357 816 832;
  • 59) 0.911 639 357 816 832 × 2 = 1 + 0.823 278 715 633 664;
  • 60) 0.823 278 715 633 664 × 2 = 1 + 0.646 557 431 267 328;
  • 61) 0.646 557 431 267 328 × 2 = 1 + 0.293 114 862 534 656;
  • 62) 0.293 114 862 534 656 × 2 = 0 + 0.586 229 725 069 312;
  • 63) 0.586 229 725 069 312 × 2 = 1 + 0.172 459 450 138 624;
  • 64) 0.172 459 450 138 624 × 2 = 0 + 0.344 918 900 277 248;

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:


0.000 282 005 914 528(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0000 0000 1001 0011 1010(2)

6. Positive number before normalization:

0.000 282 005 914 528(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0000 0000 1001 0011 1010(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:


0.000 282 005 914 528(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0000 0000 1001 0011 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0000 0000 1001 0011 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0010 0000 0000 1001 0011 1010(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 0010 0000 0000 1001 0011 1010


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 0010 0000 0000 1001 0011 1010 =

0010 0111 1011 0100 0110 0010 1110 0010 0000 0000 1001 0011 1010


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 0010 0000 0000 1001 0011 1010


Decimal number -0.000 282 005 914 528 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0010 0000 0000 1001 0011 1010

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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