0.000 000 000 000 113 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 113(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
0.000 000 000 000 113(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: 0.000 000 000 000 113.

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) 0.000 000 000 000 113 × 2 = 0 + 0.000 000 000 000 226;
  • 2) 0.000 000 000 000 226 × 2 = 0 + 0.000 000 000 000 452;
  • 3) 0.000 000 000 000 452 × 2 = 0 + 0.000 000 000 000 904;
  • 4) 0.000 000 000 000 904 × 2 = 0 + 0.000 000 000 001 808;
  • 5) 0.000 000 000 001 808 × 2 = 0 + 0.000 000 000 003 616;
  • 6) 0.000 000 000 003 616 × 2 = 0 + 0.000 000 000 007 232;
  • 7) 0.000 000 000 007 232 × 2 = 0 + 0.000 000 000 014 464;
  • 8) 0.000 000 000 014 464 × 2 = 0 + 0.000 000 000 028 928;
  • 9) 0.000 000 000 028 928 × 2 = 0 + 0.000 000 000 057 856;
  • 10) 0.000 000 000 057 856 × 2 = 0 + 0.000 000 000 115 712;
  • 11) 0.000 000 000 115 712 × 2 = 0 + 0.000 000 000 231 424;
  • 12) 0.000 000 000 231 424 × 2 = 0 + 0.000 000 000 462 848;
  • 13) 0.000 000 000 462 848 × 2 = 0 + 0.000 000 000 925 696;
  • 14) 0.000 000 000 925 696 × 2 = 0 + 0.000 000 001 851 392;
  • 15) 0.000 000 001 851 392 × 2 = 0 + 0.000 000 003 702 784;
  • 16) 0.000 000 003 702 784 × 2 = 0 + 0.000 000 007 405 568;
  • 17) 0.000 000 007 405 568 × 2 = 0 + 0.000 000 014 811 136;
  • 18) 0.000 000 014 811 136 × 2 = 0 + 0.000 000 029 622 272;
  • 19) 0.000 000 029 622 272 × 2 = 0 + 0.000 000 059 244 544;
  • 20) 0.000 000 059 244 544 × 2 = 0 + 0.000 000 118 489 088;
  • 21) 0.000 000 118 489 088 × 2 = 0 + 0.000 000 236 978 176;
  • 22) 0.000 000 236 978 176 × 2 = 0 + 0.000 000 473 956 352;
  • 23) 0.000 000 473 956 352 × 2 = 0 + 0.000 000 947 912 704;
  • 24) 0.000 000 947 912 704 × 2 = 0 + 0.000 001 895 825 408;
  • 25) 0.000 001 895 825 408 × 2 = 0 + 0.000 003 791 650 816;
  • 26) 0.000 003 791 650 816 × 2 = 0 + 0.000 007 583 301 632;
  • 27) 0.000 007 583 301 632 × 2 = 0 + 0.000 015 166 603 264;
  • 28) 0.000 015 166 603 264 × 2 = 0 + 0.000 030 333 206 528;
  • 29) 0.000 030 333 206 528 × 2 = 0 + 0.000 060 666 413 056;
  • 30) 0.000 060 666 413 056 × 2 = 0 + 0.000 121 332 826 112;
  • 31) 0.000 121 332 826 112 × 2 = 0 + 0.000 242 665 652 224;
  • 32) 0.000 242 665 652 224 × 2 = 0 + 0.000 485 331 304 448;
  • 33) 0.000 485 331 304 448 × 2 = 0 + 0.000 970 662 608 896;
  • 34) 0.000 970 662 608 896 × 2 = 0 + 0.001 941 325 217 792;
  • 35) 0.001 941 325 217 792 × 2 = 0 + 0.003 882 650 435 584;
  • 36) 0.003 882 650 435 584 × 2 = 0 + 0.007 765 300 871 168;
  • 37) 0.007 765 300 871 168 × 2 = 0 + 0.015 530 601 742 336;
  • 38) 0.015 530 601 742 336 × 2 = 0 + 0.031 061 203 484 672;
  • 39) 0.031 061 203 484 672 × 2 = 0 + 0.062 122 406 969 344;
  • 40) 0.062 122 406 969 344 × 2 = 0 + 0.124 244 813 938 688;
  • 41) 0.124 244 813 938 688 × 2 = 0 + 0.248 489 627 877 376;
  • 42) 0.248 489 627 877 376 × 2 = 0 + 0.496 979 255 754 752;
  • 43) 0.496 979 255 754 752 × 2 = 0 + 0.993 958 511 509 504;
  • 44) 0.993 958 511 509 504 × 2 = 1 + 0.987 917 023 019 008;
  • 45) 0.987 917 023 019 008 × 2 = 1 + 0.975 834 046 038 016;
  • 46) 0.975 834 046 038 016 × 2 = 1 + 0.951 668 092 076 032;
  • 47) 0.951 668 092 076 032 × 2 = 1 + 0.903 336 184 152 064;
  • 48) 0.903 336 184 152 064 × 2 = 1 + 0.806 672 368 304 128;
  • 49) 0.806 672 368 304 128 × 2 = 1 + 0.613 344 736 608 256;
  • 50) 0.613 344 736 608 256 × 2 = 1 + 0.226 689 473 216 512;
  • 51) 0.226 689 473 216 512 × 2 = 0 + 0.453 378 946 433 024;
  • 52) 0.453 378 946 433 024 × 2 = 0 + 0.906 757 892 866 048;
  • 53) 0.906 757 892 866 048 × 2 = 1 + 0.813 515 785 732 096;
  • 54) 0.813 515 785 732 096 × 2 = 1 + 0.627 031 571 464 192;
  • 55) 0.627 031 571 464 192 × 2 = 1 + 0.254 063 142 928 384;
  • 56) 0.254 063 142 928 384 × 2 = 0 + 0.508 126 285 856 768;
  • 57) 0.508 126 285 856 768 × 2 = 1 + 0.016 252 571 713 536;
  • 58) 0.016 252 571 713 536 × 2 = 0 + 0.032 505 143 427 072;
  • 59) 0.032 505 143 427 072 × 2 = 0 + 0.065 010 286 854 144;
  • 60) 0.065 010 286 854 144 × 2 = 0 + 0.130 020 573 708 288;
  • 61) 0.130 020 573 708 288 × 2 = 0 + 0.260 041 147 416 576;
  • 62) 0.260 041 147 416 576 × 2 = 0 + 0.520 082 294 833 152;
  • 63) 0.520 082 294 833 152 × 2 = 1 + 0.040 164 589 666 304;
  • 64) 0.040 164 589 666 304 × 2 = 0 + 0.080 329 179 332 608;
  • 65) 0.080 329 179 332 608 × 2 = 0 + 0.160 658 358 665 216;
  • 66) 0.160 658 358 665 216 × 2 = 0 + 0.321 316 717 330 432;
  • 67) 0.321 316 717 330 432 × 2 = 0 + 0.642 633 434 660 864;
  • 68) 0.642 633 434 660 864 × 2 = 1 + 0.285 266 869 321 728;
  • 69) 0.285 266 869 321 728 × 2 = 0 + 0.570 533 738 643 456;
  • 70) 0.570 533 738 643 456 × 2 = 1 + 0.141 067 477 286 912;
  • 71) 0.141 067 477 286 912 × 2 = 0 + 0.282 134 954 573 824;
  • 72) 0.282 134 954 573 824 × 2 = 0 + 0.564 269 909 147 648;
  • 73) 0.564 269 909 147 648 × 2 = 1 + 0.128 539 818 295 296;
  • 74) 0.128 539 818 295 296 × 2 = 0 + 0.257 079 636 590 592;
  • 75) 0.257 079 636 590 592 × 2 = 0 + 0.514 159 273 181 184;
  • 76) 0.514 159 273 181 184 × 2 = 1 + 0.028 318 546 362 368;
  • 77) 0.028 318 546 362 368 × 2 = 0 + 0.056 637 092 724 736;
  • 78) 0.056 637 092 724 736 × 2 = 0 + 0.113 274 185 449 472;
  • 79) 0.113 274 185 449 472 × 2 = 0 + 0.226 548 370 898 944;
  • 80) 0.226 548 370 898 944 × 2 = 0 + 0.453 096 741 797 888;
  • 81) 0.453 096 741 797 888 × 2 = 0 + 0.906 193 483 595 776;
  • 82) 0.906 193 483 595 776 × 2 = 1 + 0.812 386 967 191 552;
  • 83) 0.812 386 967 191 552 × 2 = 1 + 0.624 773 934 383 104;
  • 84) 0.624 773 934 383 104 × 2 = 1 + 0.249 547 868 766 208;
  • 85) 0.249 547 868 766 208 × 2 = 0 + 0.499 095 737 532 416;
  • 86) 0.499 095 737 532 416 × 2 = 0 + 0.998 191 475 064 832;
  • 87) 0.998 191 475 064 832 × 2 = 1 + 0.996 382 950 129 664;
  • 88) 0.996 382 950 129 664 × 2 = 1 + 0.992 765 900 259 328;
  • 89) 0.992 765 900 259 328 × 2 = 1 + 0.985 531 800 518 656;
  • 90) 0.985 531 800 518 656 × 2 = 1 + 0.971 063 601 037 312;
  • 91) 0.971 063 601 037 312 × 2 = 1 + 0.942 127 202 074 624;
  • 92) 0.942 127 202 074 624 × 2 = 1 + 0.884 254 404 149 248;
  • 93) 0.884 254 404 149 248 × 2 = 1 + 0.768 508 808 298 496;
  • 94) 0.768 508 808 298 496 × 2 = 1 + 0.537 017 616 596 992;
  • 95) 0.537 017 616 596 992 × 2 = 1 + 0.074 035 233 193 984;
  • 96) 0.074 035 233 193 984 × 2 = 0 + 0.148 070 466 387 968;

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:


0.000 000 000 000 113(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110(2)

5. Positive number before normalization:

0.000 000 000 000 113(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110(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:


0.000 000 000 000 113(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110(2) × 20 =

1.1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110(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):
1.1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110


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. 1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110 =

1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110


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) =
1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110


Decimal number 0.000 000 000 000 113 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0011 - 1111 1100 1110 1000 0010 0001 0100 1001 0000 0111 0011 1111 1110

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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