0.000 000 000 000 107 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 107(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
0.000 000 000 000 107(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 107.

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) 0.000 000 000 000 107 × 2 = 0 + 0.000 000 000 000 214;
  • 2) 0.000 000 000 000 214 × 2 = 0 + 0.000 000 000 000 428;
  • 3) 0.000 000 000 000 428 × 2 = 0 + 0.000 000 000 000 856;
  • 4) 0.000 000 000 000 856 × 2 = 0 + 0.000 000 000 001 712;
  • 5) 0.000 000 000 001 712 × 2 = 0 + 0.000 000 000 003 424;
  • 6) 0.000 000 000 003 424 × 2 = 0 + 0.000 000 000 006 848;
  • 7) 0.000 000 000 006 848 × 2 = 0 + 0.000 000 000 013 696;
  • 8) 0.000 000 000 013 696 × 2 = 0 + 0.000 000 000 027 392;
  • 9) 0.000 000 000 027 392 × 2 = 0 + 0.000 000 000 054 784;
  • 10) 0.000 000 000 054 784 × 2 = 0 + 0.000 000 000 109 568;
  • 11) 0.000 000 000 109 568 × 2 = 0 + 0.000 000 000 219 136;
  • 12) 0.000 000 000 219 136 × 2 = 0 + 0.000 000 000 438 272;
  • 13) 0.000 000 000 438 272 × 2 = 0 + 0.000 000 000 876 544;
  • 14) 0.000 000 000 876 544 × 2 = 0 + 0.000 000 001 753 088;
  • 15) 0.000 000 001 753 088 × 2 = 0 + 0.000 000 003 506 176;
  • 16) 0.000 000 003 506 176 × 2 = 0 + 0.000 000 007 012 352;
  • 17) 0.000 000 007 012 352 × 2 = 0 + 0.000 000 014 024 704;
  • 18) 0.000 000 014 024 704 × 2 = 0 + 0.000 000 028 049 408;
  • 19) 0.000 000 028 049 408 × 2 = 0 + 0.000 000 056 098 816;
  • 20) 0.000 000 056 098 816 × 2 = 0 + 0.000 000 112 197 632;
  • 21) 0.000 000 112 197 632 × 2 = 0 + 0.000 000 224 395 264;
  • 22) 0.000 000 224 395 264 × 2 = 0 + 0.000 000 448 790 528;
  • 23) 0.000 000 448 790 528 × 2 = 0 + 0.000 000 897 581 056;
  • 24) 0.000 000 897 581 056 × 2 = 0 + 0.000 001 795 162 112;
  • 25) 0.000 001 795 162 112 × 2 = 0 + 0.000 003 590 324 224;
  • 26) 0.000 003 590 324 224 × 2 = 0 + 0.000 007 180 648 448;
  • 27) 0.000 007 180 648 448 × 2 = 0 + 0.000 014 361 296 896;
  • 28) 0.000 014 361 296 896 × 2 = 0 + 0.000 028 722 593 792;
  • 29) 0.000 028 722 593 792 × 2 = 0 + 0.000 057 445 187 584;
  • 30) 0.000 057 445 187 584 × 2 = 0 + 0.000 114 890 375 168;
  • 31) 0.000 114 890 375 168 × 2 = 0 + 0.000 229 780 750 336;
  • 32) 0.000 229 780 750 336 × 2 = 0 + 0.000 459 561 500 672;
  • 33) 0.000 459 561 500 672 × 2 = 0 + 0.000 919 123 001 344;
  • 34) 0.000 919 123 001 344 × 2 = 0 + 0.001 838 246 002 688;
  • 35) 0.001 838 246 002 688 × 2 = 0 + 0.003 676 492 005 376;
  • 36) 0.003 676 492 005 376 × 2 = 0 + 0.007 352 984 010 752;
  • 37) 0.007 352 984 010 752 × 2 = 0 + 0.014 705 968 021 504;
  • 38) 0.014 705 968 021 504 × 2 = 0 + 0.029 411 936 043 008;
  • 39) 0.029 411 936 043 008 × 2 = 0 + 0.058 823 872 086 016;
  • 40) 0.058 823 872 086 016 × 2 = 0 + 0.117 647 744 172 032;
  • 41) 0.117 647 744 172 032 × 2 = 0 + 0.235 295 488 344 064;
  • 42) 0.235 295 488 344 064 × 2 = 0 + 0.470 590 976 688 128;
  • 43) 0.470 590 976 688 128 × 2 = 0 + 0.941 181 953 376 256;
  • 44) 0.941 181 953 376 256 × 2 = 1 + 0.882 363 906 752 512;
  • 45) 0.882 363 906 752 512 × 2 = 1 + 0.764 727 813 505 024;
  • 46) 0.764 727 813 505 024 × 2 = 1 + 0.529 455 627 010 048;
  • 47) 0.529 455 627 010 048 × 2 = 1 + 0.058 911 254 020 096;
  • 48) 0.058 911 254 020 096 × 2 = 0 + 0.117 822 508 040 192;
  • 49) 0.117 822 508 040 192 × 2 = 0 + 0.235 645 016 080 384;
  • 50) 0.235 645 016 080 384 × 2 = 0 + 0.471 290 032 160 768;
  • 51) 0.471 290 032 160 768 × 2 = 0 + 0.942 580 064 321 536;
  • 52) 0.942 580 064 321 536 × 2 = 1 + 0.885 160 128 643 072;
  • 53) 0.885 160 128 643 072 × 2 = 1 + 0.770 320 257 286 144;
  • 54) 0.770 320 257 286 144 × 2 = 1 + 0.540 640 514 572 288;
  • 55) 0.540 640 514 572 288 × 2 = 1 + 0.081 281 029 144 576;
  • 56) 0.081 281 029 144 576 × 2 = 0 + 0.162 562 058 289 152;
  • 57) 0.162 562 058 289 152 × 2 = 0 + 0.325 124 116 578 304;
  • 58) 0.325 124 116 578 304 × 2 = 0 + 0.650 248 233 156 608;
  • 59) 0.650 248 233 156 608 × 2 = 1 + 0.300 496 466 313 216;
  • 60) 0.300 496 466 313 216 × 2 = 0 + 0.600 992 932 626 432;
  • 61) 0.600 992 932 626 432 × 2 = 1 + 0.201 985 865 252 864;
  • 62) 0.201 985 865 252 864 × 2 = 0 + 0.403 971 730 505 728;
  • 63) 0.403 971 730 505 728 × 2 = 0 + 0.807 943 461 011 456;
  • 64) 0.807 943 461 011 456 × 2 = 1 + 0.615 886 922 022 912;
  • 65) 0.615 886 922 022 912 × 2 = 1 + 0.231 773 844 045 824;
  • 66) 0.231 773 844 045 824 × 2 = 0 + 0.463 547 688 091 648;
  • 67) 0.463 547 688 091 648 × 2 = 0 + 0.927 095 376 183 296;
  • 68) 0.927 095 376 183 296 × 2 = 1 + 0.854 190 752 366 592;
  • 69) 0.854 190 752 366 592 × 2 = 1 + 0.708 381 504 733 184;
  • 70) 0.708 381 504 733 184 × 2 = 1 + 0.416 763 009 466 368;
  • 71) 0.416 763 009 466 368 × 2 = 0 + 0.833 526 018 932 736;
  • 72) 0.833 526 018 932 736 × 2 = 1 + 0.667 052 037 865 472;
  • 73) 0.667 052 037 865 472 × 2 = 1 + 0.334 104 075 730 944;
  • 74) 0.334 104 075 730 944 × 2 = 0 + 0.668 208 151 461 888;
  • 75) 0.668 208 151 461 888 × 2 = 1 + 0.336 416 302 923 776;
  • 76) 0.336 416 302 923 776 × 2 = 0 + 0.672 832 605 847 552;
  • 77) 0.672 832 605 847 552 × 2 = 1 + 0.345 665 211 695 104;
  • 78) 0.345 665 211 695 104 × 2 = 0 + 0.691 330 423 390 208;
  • 79) 0.691 330 423 390 208 × 2 = 1 + 0.382 660 846 780 416;
  • 80) 0.382 660 846 780 416 × 2 = 0 + 0.765 321 693 560 832;
  • 81) 0.765 321 693 560 832 × 2 = 1 + 0.530 643 387 121 664;
  • 82) 0.530 643 387 121 664 × 2 = 1 + 0.061 286 774 243 328;
  • 83) 0.061 286 774 243 328 × 2 = 0 + 0.122 573 548 486 656;
  • 84) 0.122 573 548 486 656 × 2 = 0 + 0.245 147 096 973 312;
  • 85) 0.245 147 096 973 312 × 2 = 0 + 0.490 294 193 946 624;
  • 86) 0.490 294 193 946 624 × 2 = 0 + 0.980 588 387 893 248;
  • 87) 0.980 588 387 893 248 × 2 = 1 + 0.961 176 775 786 496;
  • 88) 0.961 176 775 786 496 × 2 = 1 + 0.922 353 551 572 992;
  • 89) 0.922 353 551 572 992 × 2 = 1 + 0.844 707 103 145 984;
  • 90) 0.844 707 103 145 984 × 2 = 1 + 0.689 414 206 291 968;
  • 91) 0.689 414 206 291 968 × 2 = 1 + 0.378 828 412 583 936;
  • 92) 0.378 828 412 583 936 × 2 = 0 + 0.757 656 825 167 872;
  • 93) 0.757 656 825 167 872 × 2 = 1 + 0.515 313 650 335 744;
  • 94) 0.515 313 650 335 744 × 2 = 1 + 0.030 627 300 671 488;
  • 95) 0.030 627 300 671 488 × 2 = 0 + 0.061 254 601 342 976;
  • 96) 0.061 254 601 342 976 × 2 = 0 + 0.122 509 202 685 952;

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).


4. 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:


0.000 000 000 000 107(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100(2)

5. Positive number before normalization:

0.000 000 000 000 107(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 44 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 107(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100(2) × 20 =

1.1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100(2) × 2-44


7. 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 0 (a positive number)

Exponent (unadjusted): -44

Mantissa (not normalized):
1.1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-44 + 2(11-1) - 1 =

(-44 + 1 023)(10) =

979(10)


9. 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;
  • 979 ÷ 2 = 489 + 1;
  • 489 ÷ 2 = 244 + 1;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

979(10) =

011 1101 0011(2)


11. 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. 1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100 =

1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1101 0011

Mantissa (52 bits) =
1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 1100


Decimal number 0.000 000 000 000 107 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0011 - 1110 0001 1110 0010 1001 1001 1101 1010 1010 1100 0011 1110 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