-0.000 035 666 844 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 844 3(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
-0.000 035 666 844 3(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:

|-0.000 035 666 844 3| = 0.000 035 666 844 3


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: 0.000 035 666 844 3.

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) 0.000 035 666 844 3 × 2 = 0 + 0.000 071 333 688 6;
  • 2) 0.000 071 333 688 6 × 2 = 0 + 0.000 142 667 377 2;
  • 3) 0.000 142 667 377 2 × 2 = 0 + 0.000 285 334 754 4;
  • 4) 0.000 285 334 754 4 × 2 = 0 + 0.000 570 669 508 8;
  • 5) 0.000 570 669 508 8 × 2 = 0 + 0.001 141 339 017 6;
  • 6) 0.001 141 339 017 6 × 2 = 0 + 0.002 282 678 035 2;
  • 7) 0.002 282 678 035 2 × 2 = 0 + 0.004 565 356 070 4;
  • 8) 0.004 565 356 070 4 × 2 = 0 + 0.009 130 712 140 8;
  • 9) 0.009 130 712 140 8 × 2 = 0 + 0.018 261 424 281 6;
  • 10) 0.018 261 424 281 6 × 2 = 0 + 0.036 522 848 563 2;
  • 11) 0.036 522 848 563 2 × 2 = 0 + 0.073 045 697 126 4;
  • 12) 0.073 045 697 126 4 × 2 = 0 + 0.146 091 394 252 8;
  • 13) 0.146 091 394 252 8 × 2 = 0 + 0.292 182 788 505 6;
  • 14) 0.292 182 788 505 6 × 2 = 0 + 0.584 365 577 011 2;
  • 15) 0.584 365 577 011 2 × 2 = 1 + 0.168 731 154 022 4;
  • 16) 0.168 731 154 022 4 × 2 = 0 + 0.337 462 308 044 8;
  • 17) 0.337 462 308 044 8 × 2 = 0 + 0.674 924 616 089 6;
  • 18) 0.674 924 616 089 6 × 2 = 1 + 0.349 849 232 179 2;
  • 19) 0.349 849 232 179 2 × 2 = 0 + 0.699 698 464 358 4;
  • 20) 0.699 698 464 358 4 × 2 = 1 + 0.399 396 928 716 8;
  • 21) 0.399 396 928 716 8 × 2 = 0 + 0.798 793 857 433 6;
  • 22) 0.798 793 857 433 6 × 2 = 1 + 0.597 587 714 867 2;
  • 23) 0.597 587 714 867 2 × 2 = 1 + 0.195 175 429 734 4;
  • 24) 0.195 175 429 734 4 × 2 = 0 + 0.390 350 859 468 8;
  • 25) 0.390 350 859 468 8 × 2 = 0 + 0.780 701 718 937 6;
  • 26) 0.780 701 718 937 6 × 2 = 1 + 0.561 403 437 875 2;
  • 27) 0.561 403 437 875 2 × 2 = 1 + 0.122 806 875 750 4;
  • 28) 0.122 806 875 750 4 × 2 = 0 + 0.245 613 751 500 8;
  • 29) 0.245 613 751 500 8 × 2 = 0 + 0.491 227 503 001 6;
  • 30) 0.491 227 503 001 6 × 2 = 0 + 0.982 455 006 003 2;
  • 31) 0.982 455 006 003 2 × 2 = 1 + 0.964 910 012 006 4;
  • 32) 0.964 910 012 006 4 × 2 = 1 + 0.929 820 024 012 8;
  • 33) 0.929 820 024 012 8 × 2 = 1 + 0.859 640 048 025 6;
  • 34) 0.859 640 048 025 6 × 2 = 1 + 0.719 280 096 051 2;
  • 35) 0.719 280 096 051 2 × 2 = 1 + 0.438 560 192 102 4;
  • 36) 0.438 560 192 102 4 × 2 = 0 + 0.877 120 384 204 8;
  • 37) 0.877 120 384 204 8 × 2 = 1 + 0.754 240 768 409 6;
  • 38) 0.754 240 768 409 6 × 2 = 1 + 0.508 481 536 819 2;
  • 39) 0.508 481 536 819 2 × 2 = 1 + 0.016 963 073 638 4;
  • 40) 0.016 963 073 638 4 × 2 = 0 + 0.033 926 147 276 8;
  • 41) 0.033 926 147 276 8 × 2 = 0 + 0.067 852 294 553 6;
  • 42) 0.067 852 294 553 6 × 2 = 0 + 0.135 704 589 107 2;
  • 43) 0.135 704 589 107 2 × 2 = 0 + 0.271 409 178 214 4;
  • 44) 0.271 409 178 214 4 × 2 = 0 + 0.542 818 356 428 8;
  • 45) 0.542 818 356 428 8 × 2 = 1 + 0.085 636 712 857 6;
  • 46) 0.085 636 712 857 6 × 2 = 0 + 0.171 273 425 715 2;
  • 47) 0.171 273 425 715 2 × 2 = 0 + 0.342 546 851 430 4;
  • 48) 0.342 546 851 430 4 × 2 = 0 + 0.685 093 702 860 8;
  • 49) 0.685 093 702 860 8 × 2 = 1 + 0.370 187 405 721 6;
  • 50) 0.370 187 405 721 6 × 2 = 0 + 0.740 374 811 443 2;
  • 51) 0.740 374 811 443 2 × 2 = 1 + 0.480 749 622 886 4;
  • 52) 0.480 749 622 886 4 × 2 = 0 + 0.961 499 245 772 8;
  • 53) 0.961 499 245 772 8 × 2 = 1 + 0.922 998 491 545 6;
  • 54) 0.922 998 491 545 6 × 2 = 1 + 0.845 996 983 091 2;
  • 55) 0.845 996 983 091 2 × 2 = 1 + 0.691 993 966 182 4;
  • 56) 0.691 993 966 182 4 × 2 = 1 + 0.383 987 932 364 8;
  • 57) 0.383 987 932 364 8 × 2 = 0 + 0.767 975 864 729 6;
  • 58) 0.767 975 864 729 6 × 2 = 1 + 0.535 951 729 459 2;
  • 59) 0.535 951 729 459 2 × 2 = 1 + 0.071 903 458 918 4;
  • 60) 0.071 903 458 918 4 × 2 = 0 + 0.143 806 917 836 8;
  • 61) 0.143 806 917 836 8 × 2 = 0 + 0.287 613 835 673 6;
  • 62) 0.287 613 835 673 6 × 2 = 0 + 0.575 227 671 347 2;
  • 63) 0.575 227 671 347 2 × 2 = 1 + 0.150 455 342 694 4;
  • 64) 0.150 455 342 694 4 × 2 = 0 + 0.300 910 685 388 8;
  • 65) 0.300 910 685 388 8 × 2 = 0 + 0.601 821 370 777 6;
  • 66) 0.601 821 370 777 6 × 2 = 1 + 0.203 642 741 555 2;
  • 67) 0.203 642 741 555 2 × 2 = 0 + 0.407 285 483 110 4;

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:


0.000 035 666 844 3(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1110 0000 1000 1010 1111 0110 0010 010(2)

6. Positive number before normalization:

0.000 035 666 844 3(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1110 0000 1000 1010 1111 0110 0010 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 035 666 844 3(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1110 0000 1000 1010 1111 0110 0010 010(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1110 0000 1000 1010 1111 0110 0010 010(2) × 20 =

1.0010 1011 0011 0001 1111 0111 0000 0100 0101 0111 1011 0001 0010(2) × 2-15


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): -15

Mantissa (not normalized):
1.0010 1011 0011 0001 1111 0111 0000 0100 0101 0111 1011 0001 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 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) =

1008(10) =

011 1111 0000(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 1011 0011 0001 1111 0111 0000 0100 0101 0111 1011 0001 0010 =

0010 1011 0011 0001 1111 0111 0000 0100 0101 0111 1011 0001 0010


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 0000

Mantissa (52 bits) =
0010 1011 0011 0001 1111 0111 0000 0100 0101 0111 1011 0001 0010


Decimal number -0.000 035 666 844 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0111 0000 0100 0101 0111 1011 0001 0010

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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