-0.066 298 275 497 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.066 298 275 497(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
-0.066 298 275 497(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:

|-0.066 298 275 497| = 0.066 298 275 497


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: 0.066 298 275 497.

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) 0.066 298 275 497 × 2 = 0 + 0.132 596 550 994;
  • 2) 0.132 596 550 994 × 2 = 0 + 0.265 193 101 988;
  • 3) 0.265 193 101 988 × 2 = 0 + 0.530 386 203 976;
  • 4) 0.530 386 203 976 × 2 = 1 + 0.060 772 407 952;
  • 5) 0.060 772 407 952 × 2 = 0 + 0.121 544 815 904;
  • 6) 0.121 544 815 904 × 2 = 0 + 0.243 089 631 808;
  • 7) 0.243 089 631 808 × 2 = 0 + 0.486 179 263 616;
  • 8) 0.486 179 263 616 × 2 = 0 + 0.972 358 527 232;
  • 9) 0.972 358 527 232 × 2 = 1 + 0.944 717 054 464;
  • 10) 0.944 717 054 464 × 2 = 1 + 0.889 434 108 928;
  • 11) 0.889 434 108 928 × 2 = 1 + 0.778 868 217 856;
  • 12) 0.778 868 217 856 × 2 = 1 + 0.557 736 435 712;
  • 13) 0.557 736 435 712 × 2 = 1 + 0.115 472 871 424;
  • 14) 0.115 472 871 424 × 2 = 0 + 0.230 945 742 848;
  • 15) 0.230 945 742 848 × 2 = 0 + 0.461 891 485 696;
  • 16) 0.461 891 485 696 × 2 = 0 + 0.923 782 971 392;
  • 17) 0.923 782 971 392 × 2 = 1 + 0.847 565 942 784;
  • 18) 0.847 565 942 784 × 2 = 1 + 0.695 131 885 568;
  • 19) 0.695 131 885 568 × 2 = 1 + 0.390 263 771 136;
  • 20) 0.390 263 771 136 × 2 = 0 + 0.780 527 542 272;
  • 21) 0.780 527 542 272 × 2 = 1 + 0.561 055 084 544;
  • 22) 0.561 055 084 544 × 2 = 1 + 0.122 110 169 088;
  • 23) 0.122 110 169 088 × 2 = 0 + 0.244 220 338 176;
  • 24) 0.244 220 338 176 × 2 = 0 + 0.488 440 676 352;
  • 25) 0.488 440 676 352 × 2 = 0 + 0.976 881 352 704;
  • 26) 0.976 881 352 704 × 2 = 1 + 0.953 762 705 408;
  • 27) 0.953 762 705 408 × 2 = 1 + 0.907 525 410 816;
  • 28) 0.907 525 410 816 × 2 = 1 + 0.815 050 821 632;
  • 29) 0.815 050 821 632 × 2 = 1 + 0.630 101 643 264;
  • 30) 0.630 101 643 264 × 2 = 1 + 0.260 203 286 528;
  • 31) 0.260 203 286 528 × 2 = 0 + 0.520 406 573 056;
  • 32) 0.520 406 573 056 × 2 = 1 + 0.040 813 146 112;
  • 33) 0.040 813 146 112 × 2 = 0 + 0.081 626 292 224;
  • 34) 0.081 626 292 224 × 2 = 0 + 0.163 252 584 448;
  • 35) 0.163 252 584 448 × 2 = 0 + 0.326 505 168 896;
  • 36) 0.326 505 168 896 × 2 = 0 + 0.653 010 337 792;
  • 37) 0.653 010 337 792 × 2 = 1 + 0.306 020 675 584;
  • 38) 0.306 020 675 584 × 2 = 0 + 0.612 041 351 168;
  • 39) 0.612 041 351 168 × 2 = 1 + 0.224 082 702 336;
  • 40) 0.224 082 702 336 × 2 = 0 + 0.448 165 404 672;
  • 41) 0.448 165 404 672 × 2 = 0 + 0.896 330 809 344;
  • 42) 0.896 330 809 344 × 2 = 1 + 0.792 661 618 688;
  • 43) 0.792 661 618 688 × 2 = 1 + 0.585 323 237 376;
  • 44) 0.585 323 237 376 × 2 = 1 + 0.170 646 474 752;
  • 45) 0.170 646 474 752 × 2 = 0 + 0.341 292 949 504;
  • 46) 0.341 292 949 504 × 2 = 0 + 0.682 585 899 008;
  • 47) 0.682 585 899 008 × 2 = 1 + 0.365 171 798 016;
  • 48) 0.365 171 798 016 × 2 = 0 + 0.730 343 596 032;
  • 49) 0.730 343 596 032 × 2 = 1 + 0.460 687 192 064;
  • 50) 0.460 687 192 064 × 2 = 0 + 0.921 374 384 128;
  • 51) 0.921 374 384 128 × 2 = 1 + 0.842 748 768 256;
  • 52) 0.842 748 768 256 × 2 = 1 + 0.685 497 536 512;
  • 53) 0.685 497 536 512 × 2 = 1 + 0.370 995 073 024;
  • 54) 0.370 995 073 024 × 2 = 0 + 0.741 990 146 048;
  • 55) 0.741 990 146 048 × 2 = 1 + 0.483 980 292 096;
  • 56) 0.483 980 292 096 × 2 = 0 + 0.967 960 584 192;

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:


0.066 298 275 497(10) =

0.0001 0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010(2)

6. Positive number before normalization:

0.066 298 275 497(10) =

0.0001 0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.066 298 275 497(10) =

0.0001 0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010(2) =

0.0001 0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010(2) × 20 =

1.0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010(2) × 2-4


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):
1.0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010 =

0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010


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 1011

Mantissa (52 bits) =
0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010


Decimal number -0.066 298 275 497 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0000 1111 1000 1110 1100 0111 1101 0000 1010 0111 0010 1011 1010

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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