-0.000 035 666 809 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 809(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
-0.000 035 666 809(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:

|-0.000 035 666 809| = 0.000 035 666 809


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: 0.000 035 666 809.

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) 0.000 035 666 809 × 2 = 0 + 0.000 071 333 618;
  • 2) 0.000 071 333 618 × 2 = 0 + 0.000 142 667 236;
  • 3) 0.000 142 667 236 × 2 = 0 + 0.000 285 334 472;
  • 4) 0.000 285 334 472 × 2 = 0 + 0.000 570 668 944;
  • 5) 0.000 570 668 944 × 2 = 0 + 0.001 141 337 888;
  • 6) 0.001 141 337 888 × 2 = 0 + 0.002 282 675 776;
  • 7) 0.002 282 675 776 × 2 = 0 + 0.004 565 351 552;
  • 8) 0.004 565 351 552 × 2 = 0 + 0.009 130 703 104;
  • 9) 0.009 130 703 104 × 2 = 0 + 0.018 261 406 208;
  • 10) 0.018 261 406 208 × 2 = 0 + 0.036 522 812 416;
  • 11) 0.036 522 812 416 × 2 = 0 + 0.073 045 624 832;
  • 12) 0.073 045 624 832 × 2 = 0 + 0.146 091 249 664;
  • 13) 0.146 091 249 664 × 2 = 0 + 0.292 182 499 328;
  • 14) 0.292 182 499 328 × 2 = 0 + 0.584 364 998 656;
  • 15) 0.584 364 998 656 × 2 = 1 + 0.168 729 997 312;
  • 16) 0.168 729 997 312 × 2 = 0 + 0.337 459 994 624;
  • 17) 0.337 459 994 624 × 2 = 0 + 0.674 919 989 248;
  • 18) 0.674 919 989 248 × 2 = 1 + 0.349 839 978 496;
  • 19) 0.349 839 978 496 × 2 = 0 + 0.699 679 956 992;
  • 20) 0.699 679 956 992 × 2 = 1 + 0.399 359 913 984;
  • 21) 0.399 359 913 984 × 2 = 0 + 0.798 719 827 968;
  • 22) 0.798 719 827 968 × 2 = 1 + 0.597 439 655 936;
  • 23) 0.597 439 655 936 × 2 = 1 + 0.194 879 311 872;
  • 24) 0.194 879 311 872 × 2 = 0 + 0.389 758 623 744;
  • 25) 0.389 758 623 744 × 2 = 0 + 0.779 517 247 488;
  • 26) 0.779 517 247 488 × 2 = 1 + 0.559 034 494 976;
  • 27) 0.559 034 494 976 × 2 = 1 + 0.118 068 989 952;
  • 28) 0.118 068 989 952 × 2 = 0 + 0.236 137 979 904;
  • 29) 0.236 137 979 904 × 2 = 0 + 0.472 275 959 808;
  • 30) 0.472 275 959 808 × 2 = 0 + 0.944 551 919 616;
  • 31) 0.944 551 919 616 × 2 = 1 + 0.889 103 839 232;
  • 32) 0.889 103 839 232 × 2 = 1 + 0.778 207 678 464;
  • 33) 0.778 207 678 464 × 2 = 1 + 0.556 415 356 928;
  • 34) 0.556 415 356 928 × 2 = 1 + 0.112 830 713 856;
  • 35) 0.112 830 713 856 × 2 = 0 + 0.225 661 427 712;
  • 36) 0.225 661 427 712 × 2 = 0 + 0.451 322 855 424;
  • 37) 0.451 322 855 424 × 2 = 0 + 0.902 645 710 848;
  • 38) 0.902 645 710 848 × 2 = 1 + 0.805 291 421 696;
  • 39) 0.805 291 421 696 × 2 = 1 + 0.610 582 843 392;
  • 40) 0.610 582 843 392 × 2 = 1 + 0.221 165 686 784;
  • 41) 0.221 165 686 784 × 2 = 0 + 0.442 331 373 568;
  • 42) 0.442 331 373 568 × 2 = 0 + 0.884 662 747 136;
  • 43) 0.884 662 747 136 × 2 = 1 + 0.769 325 494 272;
  • 44) 0.769 325 494 272 × 2 = 1 + 0.538 650 988 544;
  • 45) 0.538 650 988 544 × 2 = 1 + 0.077 301 977 088;
  • 46) 0.077 301 977 088 × 2 = 0 + 0.154 603 954 176;
  • 47) 0.154 603 954 176 × 2 = 0 + 0.309 207 908 352;
  • 48) 0.309 207 908 352 × 2 = 0 + 0.618 415 816 704;
  • 49) 0.618 415 816 704 × 2 = 1 + 0.236 831 633 408;
  • 50) 0.236 831 633 408 × 2 = 0 + 0.473 663 266 816;
  • 51) 0.473 663 266 816 × 2 = 0 + 0.947 326 533 632;
  • 52) 0.947 326 533 632 × 2 = 1 + 0.894 653 067 264;
  • 53) 0.894 653 067 264 × 2 = 1 + 0.789 306 134 528;
  • 54) 0.789 306 134 528 × 2 = 1 + 0.578 612 269 056;
  • 55) 0.578 612 269 056 × 2 = 1 + 0.157 224 538 112;
  • 56) 0.157 224 538 112 × 2 = 0 + 0.314 449 076 224;
  • 57) 0.314 449 076 224 × 2 = 0 + 0.628 898 152 448;
  • 58) 0.628 898 152 448 × 2 = 1 + 0.257 796 304 896;
  • 59) 0.257 796 304 896 × 2 = 0 + 0.515 592 609 792;
  • 60) 0.515 592 609 792 × 2 = 1 + 0.031 185 219 584;
  • 61) 0.031 185 219 584 × 2 = 0 + 0.062 370 439 168;
  • 62) 0.062 370 439 168 × 2 = 0 + 0.124 740 878 336;
  • 63) 0.124 740 878 336 × 2 = 0 + 0.249 481 756 672;
  • 64) 0.249 481 756 672 × 2 = 0 + 0.498 963 513 344;
  • 65) 0.498 963 513 344 × 2 = 0 + 0.997 927 026 688;
  • 66) 0.997 927 026 688 × 2 = 1 + 0.995 854 053 376;
  • 67) 0.995 854 053 376 × 2 = 1 + 0.991 708 106 752;

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:


0.000 035 666 809(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1100 0111 0011 1000 1001 1110 0101 0000 011(2)

6. Positive number before normalization:

0.000 035 666 809(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1100 0111 0011 1000 1001 1110 0101 0000 011(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:


0.000 035 666 809(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1100 0111 0011 1000 1001 1110 0101 0000 011(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1100 0111 0011 1000 1001 1110 0101 0000 011(2) × 20 =

1.0010 1011 0011 0001 1110 0011 1001 1100 0100 1111 0010 1000 0011(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 1110 0011 1001 1100 0100 1111 0010 1000 0011


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 1110 0011 1001 1100 0100 1111 0010 1000 0011 =

0010 1011 0011 0001 1110 0011 1001 1100 0100 1111 0010 1000 0011


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 1110 0011 1001 1100 0100 1111 0010 1000 0011


Decimal number -0.000 035 666 809 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1110 0011 1001 1100 0100 1111 0010 1000 0011

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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