-0.000 035 666 839 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 839(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
-0.000 035 666 839(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:

|-0.000 035 666 839| = 0.000 035 666 839


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: 0.000 035 666 839.

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) 0.000 035 666 839 × 2 = 0 + 0.000 071 333 678;
  • 2) 0.000 071 333 678 × 2 = 0 + 0.000 142 667 356;
  • 3) 0.000 142 667 356 × 2 = 0 + 0.000 285 334 712;
  • 4) 0.000 285 334 712 × 2 = 0 + 0.000 570 669 424;
  • 5) 0.000 570 669 424 × 2 = 0 + 0.001 141 338 848;
  • 6) 0.001 141 338 848 × 2 = 0 + 0.002 282 677 696;
  • 7) 0.002 282 677 696 × 2 = 0 + 0.004 565 355 392;
  • 8) 0.004 565 355 392 × 2 = 0 + 0.009 130 710 784;
  • 9) 0.009 130 710 784 × 2 = 0 + 0.018 261 421 568;
  • 10) 0.018 261 421 568 × 2 = 0 + 0.036 522 843 136;
  • 11) 0.036 522 843 136 × 2 = 0 + 0.073 045 686 272;
  • 12) 0.073 045 686 272 × 2 = 0 + 0.146 091 372 544;
  • 13) 0.146 091 372 544 × 2 = 0 + 0.292 182 745 088;
  • 14) 0.292 182 745 088 × 2 = 0 + 0.584 365 490 176;
  • 15) 0.584 365 490 176 × 2 = 1 + 0.168 730 980 352;
  • 16) 0.168 730 980 352 × 2 = 0 + 0.337 461 960 704;
  • 17) 0.337 461 960 704 × 2 = 0 + 0.674 923 921 408;
  • 18) 0.674 923 921 408 × 2 = 1 + 0.349 847 842 816;
  • 19) 0.349 847 842 816 × 2 = 0 + 0.699 695 685 632;
  • 20) 0.699 695 685 632 × 2 = 1 + 0.399 391 371 264;
  • 21) 0.399 391 371 264 × 2 = 0 + 0.798 782 742 528;
  • 22) 0.798 782 742 528 × 2 = 1 + 0.597 565 485 056;
  • 23) 0.597 565 485 056 × 2 = 1 + 0.195 130 970 112;
  • 24) 0.195 130 970 112 × 2 = 0 + 0.390 261 940 224;
  • 25) 0.390 261 940 224 × 2 = 0 + 0.780 523 880 448;
  • 26) 0.780 523 880 448 × 2 = 1 + 0.561 047 760 896;
  • 27) 0.561 047 760 896 × 2 = 1 + 0.122 095 521 792;
  • 28) 0.122 095 521 792 × 2 = 0 + 0.244 191 043 584;
  • 29) 0.244 191 043 584 × 2 = 0 + 0.488 382 087 168;
  • 30) 0.488 382 087 168 × 2 = 0 + 0.976 764 174 336;
  • 31) 0.976 764 174 336 × 2 = 1 + 0.953 528 348 672;
  • 32) 0.953 528 348 672 × 2 = 1 + 0.907 056 697 344;
  • 33) 0.907 056 697 344 × 2 = 1 + 0.814 113 394 688;
  • 34) 0.814 113 394 688 × 2 = 1 + 0.628 226 789 376;
  • 35) 0.628 226 789 376 × 2 = 1 + 0.256 453 578 752;
  • 36) 0.256 453 578 752 × 2 = 0 + 0.512 907 157 504;
  • 37) 0.512 907 157 504 × 2 = 1 + 0.025 814 315 008;
  • 38) 0.025 814 315 008 × 2 = 0 + 0.051 628 630 016;
  • 39) 0.051 628 630 016 × 2 = 0 + 0.103 257 260 032;
  • 40) 0.103 257 260 032 × 2 = 0 + 0.206 514 520 064;
  • 41) 0.206 514 520 064 × 2 = 0 + 0.413 029 040 128;
  • 42) 0.413 029 040 128 × 2 = 0 + 0.826 058 080 256;
  • 43) 0.826 058 080 256 × 2 = 1 + 0.652 116 160 512;
  • 44) 0.652 116 160 512 × 2 = 1 + 0.304 232 321 024;
  • 45) 0.304 232 321 024 × 2 = 0 + 0.608 464 642 048;
  • 46) 0.608 464 642 048 × 2 = 1 + 0.216 929 284 096;
  • 47) 0.216 929 284 096 × 2 = 0 + 0.433 858 568 192;
  • 48) 0.433 858 568 192 × 2 = 0 + 0.867 717 136 384;
  • 49) 0.867 717 136 384 × 2 = 1 + 0.735 434 272 768;
  • 50) 0.735 434 272 768 × 2 = 1 + 0.470 868 545 536;
  • 51) 0.470 868 545 536 × 2 = 0 + 0.941 737 091 072;
  • 52) 0.941 737 091 072 × 2 = 1 + 0.883 474 182 144;
  • 53) 0.883 474 182 144 × 2 = 1 + 0.766 948 364 288;
  • 54) 0.766 948 364 288 × 2 = 1 + 0.533 896 728 576;
  • 55) 0.533 896 728 576 × 2 = 1 + 0.067 793 457 152;
  • 56) 0.067 793 457 152 × 2 = 0 + 0.135 586 914 304;
  • 57) 0.135 586 914 304 × 2 = 0 + 0.271 173 828 608;
  • 58) 0.271 173 828 608 × 2 = 0 + 0.542 347 657 216;
  • 59) 0.542 347 657 216 × 2 = 1 + 0.084 695 314 432;
  • 60) 0.084 695 314 432 × 2 = 0 + 0.169 390 628 864;
  • 61) 0.169 390 628 864 × 2 = 0 + 0.338 781 257 728;
  • 62) 0.338 781 257 728 × 2 = 0 + 0.677 562 515 456;
  • 63) 0.677 562 515 456 × 2 = 1 + 0.355 125 030 912;
  • 64) 0.355 125 030 912 × 2 = 0 + 0.710 250 061 824;
  • 65) 0.710 250 061 824 × 2 = 1 + 0.420 500 123 648;
  • 66) 0.420 500 123 648 × 2 = 0 + 0.841 000 247 296;
  • 67) 0.841 000 247 296 × 2 = 1 + 0.682 000 494 592;

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:


0.000 035 666 839(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1000 0011 0100 1101 1110 0010 0010 101(2)

6. Positive number before normalization:

0.000 035 666 839(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1000 0011 0100 1101 1110 0010 0010 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:


0.000 035 666 839(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1000 0011 0100 1101 1110 0010 0010 101(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1000 0011 0100 1101 1110 0010 0010 101(2) × 20 =

1.0010 1011 0011 0001 1111 0100 0001 1010 0110 1111 0001 0001 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 0100 0001 1010 0110 1111 0001 0001 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 0100 0001 1010 0110 1111 0001 0001 0101 =

0010 1011 0011 0001 1111 0100 0001 1010 0110 1111 0001 0001 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 0100 0001 1010 0110 1111 0001 0001 0101


Decimal number -0.000 035 666 839 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0100 0001 1010 0110 1111 0001 0001 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