-0.000 282 005 914 279 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 279(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
-0.000 282 005 914 279(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:

|-0.000 282 005 914 279| = 0.000 282 005 914 279


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: 0.000 282 005 914 279.

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) 0.000 282 005 914 279 × 2 = 0 + 0.000 564 011 828 558;
  • 2) 0.000 564 011 828 558 × 2 = 0 + 0.001 128 023 657 116;
  • 3) 0.001 128 023 657 116 × 2 = 0 + 0.002 256 047 314 232;
  • 4) 0.002 256 047 314 232 × 2 = 0 + 0.004 512 094 628 464;
  • 5) 0.004 512 094 628 464 × 2 = 0 + 0.009 024 189 256 928;
  • 6) 0.009 024 189 256 928 × 2 = 0 + 0.018 048 378 513 856;
  • 7) 0.018 048 378 513 856 × 2 = 0 + 0.036 096 757 027 712;
  • 8) 0.036 096 757 027 712 × 2 = 0 + 0.072 193 514 055 424;
  • 9) 0.072 193 514 055 424 × 2 = 0 + 0.144 387 028 110 848;
  • 10) 0.144 387 028 110 848 × 2 = 0 + 0.288 774 056 221 696;
  • 11) 0.288 774 056 221 696 × 2 = 0 + 0.577 548 112 443 392;
  • 12) 0.577 548 112 443 392 × 2 = 1 + 0.155 096 224 886 784;
  • 13) 0.155 096 224 886 784 × 2 = 0 + 0.310 192 449 773 568;
  • 14) 0.310 192 449 773 568 × 2 = 0 + 0.620 384 899 547 136;
  • 15) 0.620 384 899 547 136 × 2 = 1 + 0.240 769 799 094 272;
  • 16) 0.240 769 799 094 272 × 2 = 0 + 0.481 539 598 188 544;
  • 17) 0.481 539 598 188 544 × 2 = 0 + 0.963 079 196 377 088;
  • 18) 0.963 079 196 377 088 × 2 = 1 + 0.926 158 392 754 176;
  • 19) 0.926 158 392 754 176 × 2 = 1 + 0.852 316 785 508 352;
  • 20) 0.852 316 785 508 352 × 2 = 1 + 0.704 633 571 016 704;
  • 21) 0.704 633 571 016 704 × 2 = 1 + 0.409 267 142 033 408;
  • 22) 0.409 267 142 033 408 × 2 = 0 + 0.818 534 284 066 816;
  • 23) 0.818 534 284 066 816 × 2 = 1 + 0.637 068 568 133 632;
  • 24) 0.637 068 568 133 632 × 2 = 1 + 0.274 137 136 267 264;
  • 25) 0.274 137 136 267 264 × 2 = 0 + 0.548 274 272 534 528;
  • 26) 0.548 274 272 534 528 × 2 = 1 + 0.096 548 545 069 056;
  • 27) 0.096 548 545 069 056 × 2 = 0 + 0.193 097 090 138 112;
  • 28) 0.193 097 090 138 112 × 2 = 0 + 0.386 194 180 276 224;
  • 29) 0.386 194 180 276 224 × 2 = 0 + 0.772 388 360 552 448;
  • 30) 0.772 388 360 552 448 × 2 = 1 + 0.544 776 721 104 896;
  • 31) 0.544 776 721 104 896 × 2 = 1 + 0.089 553 442 209 792;
  • 32) 0.089 553 442 209 792 × 2 = 0 + 0.179 106 884 419 584;
  • 33) 0.179 106 884 419 584 × 2 = 0 + 0.358 213 768 839 168;
  • 34) 0.358 213 768 839 168 × 2 = 0 + 0.716 427 537 678 336;
  • 35) 0.716 427 537 678 336 × 2 = 1 + 0.432 855 075 356 672;
  • 36) 0.432 855 075 356 672 × 2 = 0 + 0.865 710 150 713 344;
  • 37) 0.865 710 150 713 344 × 2 = 1 + 0.731 420 301 426 688;
  • 38) 0.731 420 301 426 688 × 2 = 1 + 0.462 840 602 853 376;
  • 39) 0.462 840 602 853 376 × 2 = 0 + 0.925 681 205 706 752;
  • 40) 0.925 681 205 706 752 × 2 = 1 + 0.851 362 411 413 504;
  • 41) 0.851 362 411 413 504 × 2 = 1 + 0.702 724 822 827 008;
  • 42) 0.702 724 822 827 008 × 2 = 1 + 0.405 449 645 654 016;
  • 43) 0.405 449 645 654 016 × 2 = 0 + 0.810 899 291 308 032;
  • 44) 0.810 899 291 308 032 × 2 = 1 + 0.621 798 582 616 064;
  • 45) 0.621 798 582 616 064 × 2 = 1 + 0.243 597 165 232 128;
  • 46) 0.243 597 165 232 128 × 2 = 0 + 0.487 194 330 464 256;
  • 47) 0.487 194 330 464 256 × 2 = 0 + 0.974 388 660 928 512;
  • 48) 0.974 388 660 928 512 × 2 = 1 + 0.948 777 321 857 024;
  • 49) 0.948 777 321 857 024 × 2 = 1 + 0.897 554 643 714 048;
  • 50) 0.897 554 643 714 048 × 2 = 1 + 0.795 109 287 428 096;
  • 51) 0.795 109 287 428 096 × 2 = 1 + 0.590 218 574 856 192;
  • 52) 0.590 218 574 856 192 × 2 = 1 + 0.180 437 149 712 384;
  • 53) 0.180 437 149 712 384 × 2 = 0 + 0.360 874 299 424 768;
  • 54) 0.360 874 299 424 768 × 2 = 0 + 0.721 748 598 849 536;
  • 55) 0.721 748 598 849 536 × 2 = 1 + 0.443 497 197 699 072;
  • 56) 0.443 497 197 699 072 × 2 = 0 + 0.886 994 395 398 144;
  • 57) 0.886 994 395 398 144 × 2 = 1 + 0.773 988 790 796 288;
  • 58) 0.773 988 790 796 288 × 2 = 1 + 0.547 977 581 592 576;
  • 59) 0.547 977 581 592 576 × 2 = 1 + 0.095 955 163 185 152;
  • 60) 0.095 955 163 185 152 × 2 = 0 + 0.191 910 326 370 304;
  • 61) 0.191 910 326 370 304 × 2 = 0 + 0.383 820 652 740 608;
  • 62) 0.383 820 652 740 608 × 2 = 0 + 0.767 641 305 481 216;
  • 63) 0.767 641 305 481 216 × 2 = 1 + 0.535 282 610 962 432;
  • 64) 0.535 282 610 962 432 × 2 = 1 + 0.070 565 221 924 864;

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:


0.000 282 005 914 279(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 1001 1111 0010 1110 0011(2)

6. Positive number before normalization:

0.000 282 005 914 279(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 1001 1111 0010 1110 0011(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:


0.000 282 005 914 279(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 1001 1111 0010 1110 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 1001 1111 0010 1110 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1101 1001 1111 0010 1110 0011(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 1001 1111 0010 1110 0011


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 1001 1111 0010 1110 0011 =

0010 0111 1011 0100 0110 0010 1101 1101 1001 1111 0010 1110 0011


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 1001 1111 0010 1110 0011


Decimal number -0.000 282 005 914 279 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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