-0.000 282 006 071 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 071(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
-0.000 282 006 071(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:

|-0.000 282 006 071| = 0.000 282 006 071


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: 0.000 282 006 071.

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) 0.000 282 006 071 × 2 = 0 + 0.000 564 012 142;
  • 2) 0.000 564 012 142 × 2 = 0 + 0.001 128 024 284;
  • 3) 0.001 128 024 284 × 2 = 0 + 0.002 256 048 568;
  • 4) 0.002 256 048 568 × 2 = 0 + 0.004 512 097 136;
  • 5) 0.004 512 097 136 × 2 = 0 + 0.009 024 194 272;
  • 6) 0.009 024 194 272 × 2 = 0 + 0.018 048 388 544;
  • 7) 0.018 048 388 544 × 2 = 0 + 0.036 096 777 088;
  • 8) 0.036 096 777 088 × 2 = 0 + 0.072 193 554 176;
  • 9) 0.072 193 554 176 × 2 = 0 + 0.144 387 108 352;
  • 10) 0.144 387 108 352 × 2 = 0 + 0.288 774 216 704;
  • 11) 0.288 774 216 704 × 2 = 0 + 0.577 548 433 408;
  • 12) 0.577 548 433 408 × 2 = 1 + 0.155 096 866 816;
  • 13) 0.155 096 866 816 × 2 = 0 + 0.310 193 733 632;
  • 14) 0.310 193 733 632 × 2 = 0 + 0.620 387 467 264;
  • 15) 0.620 387 467 264 × 2 = 1 + 0.240 774 934 528;
  • 16) 0.240 774 934 528 × 2 = 0 + 0.481 549 869 056;
  • 17) 0.481 549 869 056 × 2 = 0 + 0.963 099 738 112;
  • 18) 0.963 099 738 112 × 2 = 1 + 0.926 199 476 224;
  • 19) 0.926 199 476 224 × 2 = 1 + 0.852 398 952 448;
  • 20) 0.852 398 952 448 × 2 = 1 + 0.704 797 904 896;
  • 21) 0.704 797 904 896 × 2 = 1 + 0.409 595 809 792;
  • 22) 0.409 595 809 792 × 2 = 0 + 0.819 191 619 584;
  • 23) 0.819 191 619 584 × 2 = 1 + 0.638 383 239 168;
  • 24) 0.638 383 239 168 × 2 = 1 + 0.276 766 478 336;
  • 25) 0.276 766 478 336 × 2 = 0 + 0.553 532 956 672;
  • 26) 0.553 532 956 672 × 2 = 1 + 0.107 065 913 344;
  • 27) 0.107 065 913 344 × 2 = 0 + 0.214 131 826 688;
  • 28) 0.214 131 826 688 × 2 = 0 + 0.428 263 653 376;
  • 29) 0.428 263 653 376 × 2 = 0 + 0.856 527 306 752;
  • 30) 0.856 527 306 752 × 2 = 1 + 0.713 054 613 504;
  • 31) 0.713 054 613 504 × 2 = 1 + 0.426 109 227 008;
  • 32) 0.426 109 227 008 × 2 = 0 + 0.852 218 454 016;
  • 33) 0.852 218 454 016 × 2 = 1 + 0.704 436 908 032;
  • 34) 0.704 436 908 032 × 2 = 1 + 0.408 873 816 064;
  • 35) 0.408 873 816 064 × 2 = 0 + 0.817 747 632 128;
  • 36) 0.817 747 632 128 × 2 = 1 + 0.635 495 264 256;
  • 37) 0.635 495 264 256 × 2 = 1 + 0.270 990 528 512;
  • 38) 0.270 990 528 512 × 2 = 0 + 0.541 981 057 024;
  • 39) 0.541 981 057 024 × 2 = 1 + 0.083 962 114 048;
  • 40) 0.083 962 114 048 × 2 = 0 + 0.167 924 228 096;
  • 41) 0.167 924 228 096 × 2 = 0 + 0.335 848 456 192;
  • 42) 0.335 848 456 192 × 2 = 0 + 0.671 696 912 384;
  • 43) 0.671 696 912 384 × 2 = 1 + 0.343 393 824 768;
  • 44) 0.343 393 824 768 × 2 = 0 + 0.686 787 649 536;
  • 45) 0.686 787 649 536 × 2 = 1 + 0.373 575 299 072;
  • 46) 0.373 575 299 072 × 2 = 0 + 0.747 150 598 144;
  • 47) 0.747 150 598 144 × 2 = 1 + 0.494 301 196 288;
  • 48) 0.494 301 196 288 × 2 = 0 + 0.988 602 392 576;
  • 49) 0.988 602 392 576 × 2 = 1 + 0.977 204 785 152;
  • 50) 0.977 204 785 152 × 2 = 1 + 0.954 409 570 304;
  • 51) 0.954 409 570 304 × 2 = 1 + 0.908 819 140 608;
  • 52) 0.908 819 140 608 × 2 = 1 + 0.817 638 281 216;
  • 53) 0.817 638 281 216 × 2 = 1 + 0.635 276 562 432;
  • 54) 0.635 276 562 432 × 2 = 1 + 0.270 553 124 864;
  • 55) 0.270 553 124 864 × 2 = 0 + 0.541 106 249 728;
  • 56) 0.541 106 249 728 × 2 = 1 + 0.082 212 499 456;
  • 57) 0.082 212 499 456 × 2 = 0 + 0.164 424 998 912;
  • 58) 0.164 424 998 912 × 2 = 0 + 0.328 849 997 824;
  • 59) 0.328 849 997 824 × 2 = 0 + 0.657 699 995 648;
  • 60) 0.657 699 995 648 × 2 = 1 + 0.315 399 991 296;
  • 61) 0.315 399 991 296 × 2 = 0 + 0.630 799 982 592;
  • 62) 0.630 799 982 592 × 2 = 1 + 0.261 599 965 184;
  • 63) 0.261 599 965 184 × 2 = 0 + 0.523 199 930 368;
  • 64) 0.523 199 930 368 × 2 = 1 + 0.046 399 860 736;

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:


0.000 282 006 071(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1101 1010 0010 1010 1111 1101 0001 0101(2)

6. Positive number before normalization:

0.000 282 006 071(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1101 1010 0010 1010 1111 1101 0001 0101(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:


0.000 282 006 071(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1101 1010 0010 1010 1111 1101 0001 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1101 1010 0010 1010 1111 1101 0001 0101(2) × 20 =

1.0010 0111 1011 0100 0110 1101 1010 0010 1010 1111 1101 0001 0101(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 1101 1010 0010 1010 1111 1101 0001 0101


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 1101 1010 0010 1010 1111 1101 0001 0101 =

0010 0111 1011 0100 0110 1101 1010 0010 1010 1111 1101 0001 0101


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 1101 1010 0010 1010 1111 1101 0001 0101


Decimal number -0.000 282 006 071 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1101 1010 0010 1010 1111 1101 0001 0101

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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