-0.000 282 005 914 224 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 224(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
-0.000 282 005 914 224(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:

|-0.000 282 005 914 224| = 0.000 282 005 914 224


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: 0.000 282 005 914 224.

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) 0.000 282 005 914 224 × 2 = 0 + 0.000 564 011 828 448;
  • 2) 0.000 564 011 828 448 × 2 = 0 + 0.001 128 023 656 896;
  • 3) 0.001 128 023 656 896 × 2 = 0 + 0.002 256 047 313 792;
  • 4) 0.002 256 047 313 792 × 2 = 0 + 0.004 512 094 627 584;
  • 5) 0.004 512 094 627 584 × 2 = 0 + 0.009 024 189 255 168;
  • 6) 0.009 024 189 255 168 × 2 = 0 + 0.018 048 378 510 336;
  • 7) 0.018 048 378 510 336 × 2 = 0 + 0.036 096 757 020 672;
  • 8) 0.036 096 757 020 672 × 2 = 0 + 0.072 193 514 041 344;
  • 9) 0.072 193 514 041 344 × 2 = 0 + 0.144 387 028 082 688;
  • 10) 0.144 387 028 082 688 × 2 = 0 + 0.288 774 056 165 376;
  • 11) 0.288 774 056 165 376 × 2 = 0 + 0.577 548 112 330 752;
  • 12) 0.577 548 112 330 752 × 2 = 1 + 0.155 096 224 661 504;
  • 13) 0.155 096 224 661 504 × 2 = 0 + 0.310 192 449 323 008;
  • 14) 0.310 192 449 323 008 × 2 = 0 + 0.620 384 898 646 016;
  • 15) 0.620 384 898 646 016 × 2 = 1 + 0.240 769 797 292 032;
  • 16) 0.240 769 797 292 032 × 2 = 0 + 0.481 539 594 584 064;
  • 17) 0.481 539 594 584 064 × 2 = 0 + 0.963 079 189 168 128;
  • 18) 0.963 079 189 168 128 × 2 = 1 + 0.926 158 378 336 256;
  • 19) 0.926 158 378 336 256 × 2 = 1 + 0.852 316 756 672 512;
  • 20) 0.852 316 756 672 512 × 2 = 1 + 0.704 633 513 345 024;
  • 21) 0.704 633 513 345 024 × 2 = 1 + 0.409 267 026 690 048;
  • 22) 0.409 267 026 690 048 × 2 = 0 + 0.818 534 053 380 096;
  • 23) 0.818 534 053 380 096 × 2 = 1 + 0.637 068 106 760 192;
  • 24) 0.637 068 106 760 192 × 2 = 1 + 0.274 136 213 520 384;
  • 25) 0.274 136 213 520 384 × 2 = 0 + 0.548 272 427 040 768;
  • 26) 0.548 272 427 040 768 × 2 = 1 + 0.096 544 854 081 536;
  • 27) 0.096 544 854 081 536 × 2 = 0 + 0.193 089 708 163 072;
  • 28) 0.193 089 708 163 072 × 2 = 0 + 0.386 179 416 326 144;
  • 29) 0.386 179 416 326 144 × 2 = 0 + 0.772 358 832 652 288;
  • 30) 0.772 358 832 652 288 × 2 = 1 + 0.544 717 665 304 576;
  • 31) 0.544 717 665 304 576 × 2 = 1 + 0.089 435 330 609 152;
  • 32) 0.089 435 330 609 152 × 2 = 0 + 0.178 870 661 218 304;
  • 33) 0.178 870 661 218 304 × 2 = 0 + 0.357 741 322 436 608;
  • 34) 0.357 741 322 436 608 × 2 = 0 + 0.715 482 644 873 216;
  • 35) 0.715 482 644 873 216 × 2 = 1 + 0.430 965 289 746 432;
  • 36) 0.430 965 289 746 432 × 2 = 0 + 0.861 930 579 492 864;
  • 37) 0.861 930 579 492 864 × 2 = 1 + 0.723 861 158 985 728;
  • 38) 0.723 861 158 985 728 × 2 = 1 + 0.447 722 317 971 456;
  • 39) 0.447 722 317 971 456 × 2 = 0 + 0.895 444 635 942 912;
  • 40) 0.895 444 635 942 912 × 2 = 1 + 0.790 889 271 885 824;
  • 41) 0.790 889 271 885 824 × 2 = 1 + 0.581 778 543 771 648;
  • 42) 0.581 778 543 771 648 × 2 = 1 + 0.163 557 087 543 296;
  • 43) 0.163 557 087 543 296 × 2 = 0 + 0.327 114 175 086 592;
  • 44) 0.327 114 175 086 592 × 2 = 0 + 0.654 228 350 173 184;
  • 45) 0.654 228 350 173 184 × 2 = 1 + 0.308 456 700 346 368;
  • 46) 0.308 456 700 346 368 × 2 = 0 + 0.616 913 400 692 736;
  • 47) 0.616 913 400 692 736 × 2 = 1 + 0.233 826 801 385 472;
  • 48) 0.233 826 801 385 472 × 2 = 0 + 0.467 653 602 770 944;
  • 49) 0.467 653 602 770 944 × 2 = 0 + 0.935 307 205 541 888;
  • 50) 0.935 307 205 541 888 × 2 = 1 + 0.870 614 411 083 776;
  • 51) 0.870 614 411 083 776 × 2 = 1 + 0.741 228 822 167 552;
  • 52) 0.741 228 822 167 552 × 2 = 1 + 0.482 457 644 335 104;
  • 53) 0.482 457 644 335 104 × 2 = 0 + 0.964 915 288 670 208;
  • 54) 0.964 915 288 670 208 × 2 = 1 + 0.929 830 577 340 416;
  • 55) 0.929 830 577 340 416 × 2 = 1 + 0.859 661 154 680 832;
  • 56) 0.859 661 154 680 832 × 2 = 1 + 0.719 322 309 361 664;
  • 57) 0.719 322 309 361 664 × 2 = 1 + 0.438 644 618 723 328;
  • 58) 0.438 644 618 723 328 × 2 = 0 + 0.877 289 237 446 656;
  • 59) 0.877 289 237 446 656 × 2 = 1 + 0.754 578 474 893 312;
  • 60) 0.754 578 474 893 312 × 2 = 1 + 0.509 156 949 786 624;
  • 61) 0.509 156 949 786 624 × 2 = 1 + 0.018 313 899 573 248;
  • 62) 0.018 313 899 573 248 × 2 = 0 + 0.036 627 799 146 496;
  • 63) 0.036 627 799 146 496 × 2 = 0 + 0.073 255 598 292 992;
  • 64) 0.073 255 598 292 992 × 2 = 0 + 0.146 511 196 585 984;

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:


0.000 282 005 914 224(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1010 0111 0111 1011 1000(2)

6. Positive number before normalization:

0.000 282 005 914 224(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1010 0111 0111 1011 1000(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:


0.000 282 005 914 224(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1010 0111 0111 1011 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1010 0111 0111 1011 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1100 1010 0111 0111 1011 1000(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 1100 1010 0111 0111 1011 1000


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 1100 1010 0111 0111 1011 1000 =

0010 0111 1011 0100 0110 0010 1101 1100 1010 0111 0111 1011 1000


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 1100 1010 0111 0111 1011 1000


Decimal number -0.000 282 005 914 224 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1100 1010 0111 0111 1011 1000

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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