-0.000 282 005 571 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 571(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
-0.000 282 005 571(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:

|-0.000 282 005 571| = 0.000 282 005 571


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: 0.000 282 005 571.

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) 0.000 282 005 571 × 2 = 0 + 0.000 564 011 142;
  • 2) 0.000 564 011 142 × 2 = 0 + 0.001 128 022 284;
  • 3) 0.001 128 022 284 × 2 = 0 + 0.002 256 044 568;
  • 4) 0.002 256 044 568 × 2 = 0 + 0.004 512 089 136;
  • 5) 0.004 512 089 136 × 2 = 0 + 0.009 024 178 272;
  • 6) 0.009 024 178 272 × 2 = 0 + 0.018 048 356 544;
  • 7) 0.018 048 356 544 × 2 = 0 + 0.036 096 713 088;
  • 8) 0.036 096 713 088 × 2 = 0 + 0.072 193 426 176;
  • 9) 0.072 193 426 176 × 2 = 0 + 0.144 386 852 352;
  • 10) 0.144 386 852 352 × 2 = 0 + 0.288 773 704 704;
  • 11) 0.288 773 704 704 × 2 = 0 + 0.577 547 409 408;
  • 12) 0.577 547 409 408 × 2 = 1 + 0.155 094 818 816;
  • 13) 0.155 094 818 816 × 2 = 0 + 0.310 189 637 632;
  • 14) 0.310 189 637 632 × 2 = 0 + 0.620 379 275 264;
  • 15) 0.620 379 275 264 × 2 = 1 + 0.240 758 550 528;
  • 16) 0.240 758 550 528 × 2 = 0 + 0.481 517 101 056;
  • 17) 0.481 517 101 056 × 2 = 0 + 0.963 034 202 112;
  • 18) 0.963 034 202 112 × 2 = 1 + 0.926 068 404 224;
  • 19) 0.926 068 404 224 × 2 = 1 + 0.852 136 808 448;
  • 20) 0.852 136 808 448 × 2 = 1 + 0.704 273 616 896;
  • 21) 0.704 273 616 896 × 2 = 1 + 0.408 547 233 792;
  • 22) 0.408 547 233 792 × 2 = 0 + 0.817 094 467 584;
  • 23) 0.817 094 467 584 × 2 = 1 + 0.634 188 935 168;
  • 24) 0.634 188 935 168 × 2 = 1 + 0.268 377 870 336;
  • 25) 0.268 377 870 336 × 2 = 0 + 0.536 755 740 672;
  • 26) 0.536 755 740 672 × 2 = 1 + 0.073 511 481 344;
  • 27) 0.073 511 481 344 × 2 = 0 + 0.147 022 962 688;
  • 28) 0.147 022 962 688 × 2 = 0 + 0.294 045 925 376;
  • 29) 0.294 045 925 376 × 2 = 0 + 0.588 091 850 752;
  • 30) 0.588 091 850 752 × 2 = 1 + 0.176 183 701 504;
  • 31) 0.176 183 701 504 × 2 = 0 + 0.352 367 403 008;
  • 32) 0.352 367 403 008 × 2 = 0 + 0.704 734 806 016;
  • 33) 0.704 734 806 016 × 2 = 1 + 0.409 469 612 032;
  • 34) 0.409 469 612 032 × 2 = 0 + 0.818 939 224 064;
  • 35) 0.818 939 224 064 × 2 = 1 + 0.637 878 448 128;
  • 36) 0.637 878 448 128 × 2 = 1 + 0.275 756 896 256;
  • 37) 0.275 756 896 256 × 2 = 0 + 0.551 513 792 512;
  • 38) 0.551 513 792 512 × 2 = 1 + 0.103 027 585 024;
  • 39) 0.103 027 585 024 × 2 = 0 + 0.206 055 170 048;
  • 40) 0.206 055 170 048 × 2 = 0 + 0.412 110 340 096;
  • 41) 0.412 110 340 096 × 2 = 0 + 0.824 220 680 192;
  • 42) 0.824 220 680 192 × 2 = 1 + 0.648 441 360 384;
  • 43) 0.648 441 360 384 × 2 = 1 + 0.296 882 720 768;
  • 44) 0.296 882 720 768 × 2 = 0 + 0.593 765 441 536;
  • 45) 0.593 765 441 536 × 2 = 1 + 0.187 530 883 072;
  • 46) 0.187 530 883 072 × 2 = 0 + 0.375 061 766 144;
  • 47) 0.375 061 766 144 × 2 = 0 + 0.750 123 532 288;
  • 48) 0.750 123 532 288 × 2 = 1 + 0.500 247 064 576;
  • 49) 0.500 247 064 576 × 2 = 1 + 0.000 494 129 152;
  • 50) 0.000 494 129 152 × 2 = 0 + 0.000 988 258 304;
  • 51) 0.000 988 258 304 × 2 = 0 + 0.001 976 516 608;
  • 52) 0.001 976 516 608 × 2 = 0 + 0.003 953 033 216;
  • 53) 0.003 953 033 216 × 2 = 0 + 0.007 906 066 432;
  • 54) 0.007 906 066 432 × 2 = 0 + 0.015 812 132 864;
  • 55) 0.015 812 132 864 × 2 = 0 + 0.031 624 265 728;
  • 56) 0.031 624 265 728 × 2 = 0 + 0.063 248 531 456;
  • 57) 0.063 248 531 456 × 2 = 0 + 0.126 497 062 912;
  • 58) 0.126 497 062 912 × 2 = 0 + 0.252 994 125 824;
  • 59) 0.252 994 125 824 × 2 = 0 + 0.505 988 251 648;
  • 60) 0.505 988 251 648 × 2 = 1 + 0.011 976 503 296;
  • 61) 0.011 976 503 296 × 2 = 0 + 0.023 953 006 592;
  • 62) 0.023 953 006 592 × 2 = 0 + 0.047 906 013 184;
  • 63) 0.047 906 013 184 × 2 = 0 + 0.095 812 026 368;
  • 64) 0.095 812 026 368 × 2 = 0 + 0.191 624 052 736;

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:


0.000 282 005 571(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1011 0100 0110 1001 1000 0000 0001 0000(2)

6. Positive number before normalization:

0.000 282 005 571(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1011 0100 0110 1001 1000 0000 0001 0000(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:


0.000 282 005 571(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1011 0100 0110 1001 1000 0000 0001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1011 0100 0110 1001 1000 0000 0001 0000(2) × 20 =

1.0010 0111 1011 0100 0100 1011 0100 0110 1001 1000 0000 0001 0000(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 0100 1011 0100 0110 1001 1000 0000 0001 0000


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 0100 1011 0100 0110 1001 1000 0000 0001 0000 =

0010 0111 1011 0100 0100 1011 0100 0110 1001 1000 0000 0001 0000


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 0100 1011 0100 0110 1001 1000 0000 0001 0000


Decimal number -0.000 282 005 571 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 1011 0100 0110 1001 1000 0000 0001 0000

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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