-0.000 035 666 849 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 849 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
-0.000 035 666 849 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:

|-0.000 035 666 849 3| = 0.000 035 666 849 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: 0.000 035 666 849 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) 0.000 035 666 849 3 × 2 = 0 + 0.000 071 333 698 6;
  • 2) 0.000 071 333 698 6 × 2 = 0 + 0.000 142 667 397 2;
  • 3) 0.000 142 667 397 2 × 2 = 0 + 0.000 285 334 794 4;
  • 4) 0.000 285 334 794 4 × 2 = 0 + 0.000 570 669 588 8;
  • 5) 0.000 570 669 588 8 × 2 = 0 + 0.001 141 339 177 6;
  • 6) 0.001 141 339 177 6 × 2 = 0 + 0.002 282 678 355 2;
  • 7) 0.002 282 678 355 2 × 2 = 0 + 0.004 565 356 710 4;
  • 8) 0.004 565 356 710 4 × 2 = 0 + 0.009 130 713 420 8;
  • 9) 0.009 130 713 420 8 × 2 = 0 + 0.018 261 426 841 6;
  • 10) 0.018 261 426 841 6 × 2 = 0 + 0.036 522 853 683 2;
  • 11) 0.036 522 853 683 2 × 2 = 0 + 0.073 045 707 366 4;
  • 12) 0.073 045 707 366 4 × 2 = 0 + 0.146 091 414 732 8;
  • 13) 0.146 091 414 732 8 × 2 = 0 + 0.292 182 829 465 6;
  • 14) 0.292 182 829 465 6 × 2 = 0 + 0.584 365 658 931 2;
  • 15) 0.584 365 658 931 2 × 2 = 1 + 0.168 731 317 862 4;
  • 16) 0.168 731 317 862 4 × 2 = 0 + 0.337 462 635 724 8;
  • 17) 0.337 462 635 724 8 × 2 = 0 + 0.674 925 271 449 6;
  • 18) 0.674 925 271 449 6 × 2 = 1 + 0.349 850 542 899 2;
  • 19) 0.349 850 542 899 2 × 2 = 0 + 0.699 701 085 798 4;
  • 20) 0.699 701 085 798 4 × 2 = 1 + 0.399 402 171 596 8;
  • 21) 0.399 402 171 596 8 × 2 = 0 + 0.798 804 343 193 6;
  • 22) 0.798 804 343 193 6 × 2 = 1 + 0.597 608 686 387 2;
  • 23) 0.597 608 686 387 2 × 2 = 1 + 0.195 217 372 774 4;
  • 24) 0.195 217 372 774 4 × 2 = 0 + 0.390 434 745 548 8;
  • 25) 0.390 434 745 548 8 × 2 = 0 + 0.780 869 491 097 6;
  • 26) 0.780 869 491 097 6 × 2 = 1 + 0.561 738 982 195 2;
  • 27) 0.561 738 982 195 2 × 2 = 1 + 0.123 477 964 390 4;
  • 28) 0.123 477 964 390 4 × 2 = 0 + 0.246 955 928 780 8;
  • 29) 0.246 955 928 780 8 × 2 = 0 + 0.493 911 857 561 6;
  • 30) 0.493 911 857 561 6 × 2 = 0 + 0.987 823 715 123 2;
  • 31) 0.987 823 715 123 2 × 2 = 1 + 0.975 647 430 246 4;
  • 32) 0.975 647 430 246 4 × 2 = 1 + 0.951 294 860 492 8;
  • 33) 0.951 294 860 492 8 × 2 = 1 + 0.902 589 720 985 6;
  • 34) 0.902 589 720 985 6 × 2 = 1 + 0.805 179 441 971 2;
  • 35) 0.805 179 441 971 2 × 2 = 1 + 0.610 358 883 942 4;
  • 36) 0.610 358 883 942 4 × 2 = 1 + 0.220 717 767 884 8;
  • 37) 0.220 717 767 884 8 × 2 = 0 + 0.441 435 535 769 6;
  • 38) 0.441 435 535 769 6 × 2 = 0 + 0.882 871 071 539 2;
  • 39) 0.882 871 071 539 2 × 2 = 1 + 0.765 742 143 078 4;
  • 40) 0.765 742 143 078 4 × 2 = 1 + 0.531 484 286 156 8;
  • 41) 0.531 484 286 156 8 × 2 = 1 + 0.062 968 572 313 6;
  • 42) 0.062 968 572 313 6 × 2 = 0 + 0.125 937 144 627 2;
  • 43) 0.125 937 144 627 2 × 2 = 0 + 0.251 874 289 254 4;
  • 44) 0.251 874 289 254 4 × 2 = 0 + 0.503 748 578 508 8;
  • 45) 0.503 748 578 508 8 × 2 = 1 + 0.007 497 157 017 6;
  • 46) 0.007 497 157 017 6 × 2 = 0 + 0.014 994 314 035 2;
  • 47) 0.014 994 314 035 2 × 2 = 0 + 0.029 988 628 070 4;
  • 48) 0.029 988 628 070 4 × 2 = 0 + 0.059 977 256 140 8;
  • 49) 0.059 977 256 140 8 × 2 = 0 + 0.119 954 512 281 6;
  • 50) 0.119 954 512 281 6 × 2 = 0 + 0.239 909 024 563 2;
  • 51) 0.239 909 024 563 2 × 2 = 0 + 0.479 818 049 126 4;
  • 52) 0.479 818 049 126 4 × 2 = 0 + 0.959 636 098 252 8;
  • 53) 0.959 636 098 252 8 × 2 = 1 + 0.919 272 196 505 6;
  • 54) 0.919 272 196 505 6 × 2 = 1 + 0.838 544 393 011 2;
  • 55) 0.838 544 393 011 2 × 2 = 1 + 0.677 088 786 022 4;
  • 56) 0.677 088 786 022 4 × 2 = 1 + 0.354 177 572 044 8;
  • 57) 0.354 177 572 044 8 × 2 = 0 + 0.708 355 144 089 6;
  • 58) 0.708 355 144 089 6 × 2 = 1 + 0.416 710 288 179 2;
  • 59) 0.416 710 288 179 2 × 2 = 0 + 0.833 420 576 358 4;
  • 60) 0.833 420 576 358 4 × 2 = 1 + 0.666 841 152 716 8;
  • 61) 0.666 841 152 716 8 × 2 = 1 + 0.333 682 305 433 6;
  • 62) 0.333 682 305 433 6 × 2 = 0 + 0.667 364 610 867 2;
  • 63) 0.667 364 610 867 2 × 2 = 1 + 0.334 729 221 734 4;
  • 64) 0.334 729 221 734 4 × 2 = 0 + 0.669 458 443 468 8;
  • 65) 0.669 458 443 468 8 × 2 = 1 + 0.338 916 886 937 6;
  • 66) 0.338 916 886 937 6 × 2 = 0 + 0.677 833 773 875 2;
  • 67) 0.677 833 773 875 2 × 2 = 1 + 0.355 667 547 750 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:


0.000 035 666 849 3(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1111 0011 1000 1000 0000 1111 0101 1010 101(2)

6. Positive number before normalization:

0.000 035 666 849 3(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1111 0011 1000 1000 0000 1111 0101 1010 101(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:


0.000 035 666 849 3(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1111 0011 1000 1000 0000 1111 0101 1010 101(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1111 0011 1000 1000 0000 1111 0101 1010 101(2) × 20 =

1.0010 1011 0011 0001 1111 1001 1100 0100 0000 0111 1010 1101 0101(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 1001 1100 0100 0000 0111 1010 1101 0101


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 1001 1100 0100 0000 0111 1010 1101 0101 =

0010 1011 0011 0001 1111 1001 1100 0100 0000 0111 1010 1101 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 0000

Mantissa (52 bits) =
0010 1011 0011 0001 1111 1001 1100 0100 0000 0111 1010 1101 0101


Decimal number -0.000 035 666 849 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 1001 1100 0100 0000 0111 1010 1101 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