-0.000 282 005 914 399 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 399(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
-0.000 282 005 914 399(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:

|-0.000 282 005 914 399| = 0.000 282 005 914 399


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: 0.000 282 005 914 399.

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) 0.000 282 005 914 399 × 2 = 0 + 0.000 564 011 828 798;
  • 2) 0.000 564 011 828 798 × 2 = 0 + 0.001 128 023 657 596;
  • 3) 0.001 128 023 657 596 × 2 = 0 + 0.002 256 047 315 192;
  • 4) 0.002 256 047 315 192 × 2 = 0 + 0.004 512 094 630 384;
  • 5) 0.004 512 094 630 384 × 2 = 0 + 0.009 024 189 260 768;
  • 6) 0.009 024 189 260 768 × 2 = 0 + 0.018 048 378 521 536;
  • 7) 0.018 048 378 521 536 × 2 = 0 + 0.036 096 757 043 072;
  • 8) 0.036 096 757 043 072 × 2 = 0 + 0.072 193 514 086 144;
  • 9) 0.072 193 514 086 144 × 2 = 0 + 0.144 387 028 172 288;
  • 10) 0.144 387 028 172 288 × 2 = 0 + 0.288 774 056 344 576;
  • 11) 0.288 774 056 344 576 × 2 = 0 + 0.577 548 112 689 152;
  • 12) 0.577 548 112 689 152 × 2 = 1 + 0.155 096 225 378 304;
  • 13) 0.155 096 225 378 304 × 2 = 0 + 0.310 192 450 756 608;
  • 14) 0.310 192 450 756 608 × 2 = 0 + 0.620 384 901 513 216;
  • 15) 0.620 384 901 513 216 × 2 = 1 + 0.240 769 803 026 432;
  • 16) 0.240 769 803 026 432 × 2 = 0 + 0.481 539 606 052 864;
  • 17) 0.481 539 606 052 864 × 2 = 0 + 0.963 079 212 105 728;
  • 18) 0.963 079 212 105 728 × 2 = 1 + 0.926 158 424 211 456;
  • 19) 0.926 158 424 211 456 × 2 = 1 + 0.852 316 848 422 912;
  • 20) 0.852 316 848 422 912 × 2 = 1 + 0.704 633 696 845 824;
  • 21) 0.704 633 696 845 824 × 2 = 1 + 0.409 267 393 691 648;
  • 22) 0.409 267 393 691 648 × 2 = 0 + 0.818 534 787 383 296;
  • 23) 0.818 534 787 383 296 × 2 = 1 + 0.637 069 574 766 592;
  • 24) 0.637 069 574 766 592 × 2 = 1 + 0.274 139 149 533 184;
  • 25) 0.274 139 149 533 184 × 2 = 0 + 0.548 278 299 066 368;
  • 26) 0.548 278 299 066 368 × 2 = 1 + 0.096 556 598 132 736;
  • 27) 0.096 556 598 132 736 × 2 = 0 + 0.193 113 196 265 472;
  • 28) 0.193 113 196 265 472 × 2 = 0 + 0.386 226 392 530 944;
  • 29) 0.386 226 392 530 944 × 2 = 0 + 0.772 452 785 061 888;
  • 30) 0.772 452 785 061 888 × 2 = 1 + 0.544 905 570 123 776;
  • 31) 0.544 905 570 123 776 × 2 = 1 + 0.089 811 140 247 552;
  • 32) 0.089 811 140 247 552 × 2 = 0 + 0.179 622 280 495 104;
  • 33) 0.179 622 280 495 104 × 2 = 0 + 0.359 244 560 990 208;
  • 34) 0.359 244 560 990 208 × 2 = 0 + 0.718 489 121 980 416;
  • 35) 0.718 489 121 980 416 × 2 = 1 + 0.436 978 243 960 832;
  • 36) 0.436 978 243 960 832 × 2 = 0 + 0.873 956 487 921 664;
  • 37) 0.873 956 487 921 664 × 2 = 1 + 0.747 912 975 843 328;
  • 38) 0.747 912 975 843 328 × 2 = 1 + 0.495 825 951 686 656;
  • 39) 0.495 825 951 686 656 × 2 = 0 + 0.991 651 903 373 312;
  • 40) 0.991 651 903 373 312 × 2 = 1 + 0.983 303 806 746 624;
  • 41) 0.983 303 806 746 624 × 2 = 1 + 0.966 607 613 493 248;
  • 42) 0.966 607 613 493 248 × 2 = 1 + 0.933 215 226 986 496;
  • 43) 0.933 215 226 986 496 × 2 = 1 + 0.866 430 453 972 992;
  • 44) 0.866 430 453 972 992 × 2 = 1 + 0.732 860 907 945 984;
  • 45) 0.732 860 907 945 984 × 2 = 1 + 0.465 721 815 891 968;
  • 46) 0.465 721 815 891 968 × 2 = 0 + 0.931 443 631 783 936;
  • 47) 0.931 443 631 783 936 × 2 = 1 + 0.862 887 263 567 872;
  • 48) 0.862 887 263 567 872 × 2 = 1 + 0.725 774 527 135 744;
  • 49) 0.725 774 527 135 744 × 2 = 1 + 0.451 549 054 271 488;
  • 50) 0.451 549 054 271 488 × 2 = 0 + 0.903 098 108 542 976;
  • 51) 0.903 098 108 542 976 × 2 = 1 + 0.806 196 217 085 952;
  • 52) 0.806 196 217 085 952 × 2 = 1 + 0.612 392 434 171 904;
  • 53) 0.612 392 434 171 904 × 2 = 1 + 0.224 784 868 343 808;
  • 54) 0.224 784 868 343 808 × 2 = 0 + 0.449 569 736 687 616;
  • 55) 0.449 569 736 687 616 × 2 = 0 + 0.899 139 473 375 232;
  • 56) 0.899 139 473 375 232 × 2 = 1 + 0.798 278 946 750 464;
  • 57) 0.798 278 946 750 464 × 2 = 1 + 0.596 557 893 500 928;
  • 58) 0.596 557 893 500 928 × 2 = 1 + 0.193 115 787 001 856;
  • 59) 0.193 115 787 001 856 × 2 = 0 + 0.386 231 574 003 712;
  • 60) 0.386 231 574 003 712 × 2 = 0 + 0.772 463 148 007 424;
  • 61) 0.772 463 148 007 424 × 2 = 1 + 0.544 926 296 014 848;
  • 62) 0.544 926 296 014 848 × 2 = 1 + 0.089 852 592 029 696;
  • 63) 0.089 852 592 029 696 × 2 = 0 + 0.179 705 184 059 392;
  • 64) 0.179 705 184 059 392 × 2 = 0 + 0.359 410 368 118 784;

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:


0.000 282 005 914 399(10) =

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

6. Positive number before normalization:

0.000 282 005 914 399(10) =

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


0.000 282 005 914 399(10) =

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

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

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


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 1011 1011 1001 1100 1100 =

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


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 1011 1011 1001 1100 1100


Decimal number -0.000 282 005 914 399 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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