-0.000 035 666 878 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 878(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
-0.000 035 666 878(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:

|-0.000 035 666 878| = 0.000 035 666 878


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: 0.000 035 666 878.

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) 0.000 035 666 878 × 2 = 0 + 0.000 071 333 756;
  • 2) 0.000 071 333 756 × 2 = 0 + 0.000 142 667 512;
  • 3) 0.000 142 667 512 × 2 = 0 + 0.000 285 335 024;
  • 4) 0.000 285 335 024 × 2 = 0 + 0.000 570 670 048;
  • 5) 0.000 570 670 048 × 2 = 0 + 0.001 141 340 096;
  • 6) 0.001 141 340 096 × 2 = 0 + 0.002 282 680 192;
  • 7) 0.002 282 680 192 × 2 = 0 + 0.004 565 360 384;
  • 8) 0.004 565 360 384 × 2 = 0 + 0.009 130 720 768;
  • 9) 0.009 130 720 768 × 2 = 0 + 0.018 261 441 536;
  • 10) 0.018 261 441 536 × 2 = 0 + 0.036 522 883 072;
  • 11) 0.036 522 883 072 × 2 = 0 + 0.073 045 766 144;
  • 12) 0.073 045 766 144 × 2 = 0 + 0.146 091 532 288;
  • 13) 0.146 091 532 288 × 2 = 0 + 0.292 183 064 576;
  • 14) 0.292 183 064 576 × 2 = 0 + 0.584 366 129 152;
  • 15) 0.584 366 129 152 × 2 = 1 + 0.168 732 258 304;
  • 16) 0.168 732 258 304 × 2 = 0 + 0.337 464 516 608;
  • 17) 0.337 464 516 608 × 2 = 0 + 0.674 929 033 216;
  • 18) 0.674 929 033 216 × 2 = 1 + 0.349 858 066 432;
  • 19) 0.349 858 066 432 × 2 = 0 + 0.699 716 132 864;
  • 20) 0.699 716 132 864 × 2 = 1 + 0.399 432 265 728;
  • 21) 0.399 432 265 728 × 2 = 0 + 0.798 864 531 456;
  • 22) 0.798 864 531 456 × 2 = 1 + 0.597 729 062 912;
  • 23) 0.597 729 062 912 × 2 = 1 + 0.195 458 125 824;
  • 24) 0.195 458 125 824 × 2 = 0 + 0.390 916 251 648;
  • 25) 0.390 916 251 648 × 2 = 0 + 0.781 832 503 296;
  • 26) 0.781 832 503 296 × 2 = 1 + 0.563 665 006 592;
  • 27) 0.563 665 006 592 × 2 = 1 + 0.127 330 013 184;
  • 28) 0.127 330 013 184 × 2 = 0 + 0.254 660 026 368;
  • 29) 0.254 660 026 368 × 2 = 0 + 0.509 320 052 736;
  • 30) 0.509 320 052 736 × 2 = 1 + 0.018 640 105 472;
  • 31) 0.018 640 105 472 × 2 = 0 + 0.037 280 210 944;
  • 32) 0.037 280 210 944 × 2 = 0 + 0.074 560 421 888;
  • 33) 0.074 560 421 888 × 2 = 0 + 0.149 120 843 776;
  • 34) 0.149 120 843 776 × 2 = 0 + 0.298 241 687 552;
  • 35) 0.298 241 687 552 × 2 = 0 + 0.596 483 375 104;
  • 36) 0.596 483 375 104 × 2 = 1 + 0.192 966 750 208;
  • 37) 0.192 966 750 208 × 2 = 0 + 0.385 933 500 416;
  • 38) 0.385 933 500 416 × 2 = 0 + 0.771 867 000 832;
  • 39) 0.771 867 000 832 × 2 = 1 + 0.543 734 001 664;
  • 40) 0.543 734 001 664 × 2 = 1 + 0.087 468 003 328;
  • 41) 0.087 468 003 328 × 2 = 0 + 0.174 936 006 656;
  • 42) 0.174 936 006 656 × 2 = 0 + 0.349 872 013 312;
  • 43) 0.349 872 013 312 × 2 = 0 + 0.699 744 026 624;
  • 44) 0.699 744 026 624 × 2 = 1 + 0.399 488 053 248;
  • 45) 0.399 488 053 248 × 2 = 0 + 0.798 976 106 496;
  • 46) 0.798 976 106 496 × 2 = 1 + 0.597 952 212 992;
  • 47) 0.597 952 212 992 × 2 = 1 + 0.195 904 425 984;
  • 48) 0.195 904 425 984 × 2 = 0 + 0.391 808 851 968;
  • 49) 0.391 808 851 968 × 2 = 0 + 0.783 617 703 936;
  • 50) 0.783 617 703 936 × 2 = 1 + 0.567 235 407 872;
  • 51) 0.567 235 407 872 × 2 = 1 + 0.134 470 815 744;
  • 52) 0.134 470 815 744 × 2 = 0 + 0.268 941 631 488;
  • 53) 0.268 941 631 488 × 2 = 0 + 0.537 883 262 976;
  • 54) 0.537 883 262 976 × 2 = 1 + 0.075 766 525 952;
  • 55) 0.075 766 525 952 × 2 = 0 + 0.151 533 051 904;
  • 56) 0.151 533 051 904 × 2 = 0 + 0.303 066 103 808;
  • 57) 0.303 066 103 808 × 2 = 0 + 0.606 132 207 616;
  • 58) 0.606 132 207 616 × 2 = 1 + 0.212 264 415 232;
  • 59) 0.212 264 415 232 × 2 = 0 + 0.424 528 830 464;
  • 60) 0.424 528 830 464 × 2 = 0 + 0.849 057 660 928;
  • 61) 0.849 057 660 928 × 2 = 1 + 0.698 115 321 856;
  • 62) 0.698 115 321 856 × 2 = 1 + 0.396 230 643 712;
  • 63) 0.396 230 643 712 × 2 = 0 + 0.792 461 287 424;
  • 64) 0.792 461 287 424 × 2 = 1 + 0.584 922 574 848;
  • 65) 0.584 922 574 848 × 2 = 1 + 0.169 845 149 696;
  • 66) 0.169 845 149 696 × 2 = 0 + 0.339 690 299 392;
  • 67) 0.339 690 299 392 × 2 = 0 + 0.679 380 598 784;

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:


0.000 035 666 878(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0001 0011 0001 0110 0110 0100 0100 1101 100(2)

6. Positive number before normalization:

0.000 035 666 878(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0001 0011 0001 0110 0110 0100 0100 1101 100(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:


0.000 035 666 878(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0001 0011 0001 0110 0110 0100 0100 1101 100(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0001 0011 0001 0110 0110 0100 0100 1101 100(2) × 20 =

1.0010 1011 0011 0010 0000 1001 1000 1011 0011 0010 0010 0110 1100(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 0010 0000 1001 1000 1011 0011 0010 0010 0110 1100


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 0010 0000 1001 1000 1011 0011 0010 0010 0110 1100 =

0010 1011 0011 0010 0000 1001 1000 1011 0011 0010 0010 0110 1100


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 0010 0000 1001 1000 1011 0011 0010 0010 0110 1100


Decimal number -0.000 035 666 878 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0000 1001 1000 1011 0011 0010 0010 0110 1100

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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