0.000 000 000 000 227 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 227(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
0.000 000 000 000 227(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: 0.000 000 000 000 227.

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) 0.000 000 000 000 227 × 2 = 0 + 0.000 000 000 000 454;
  • 2) 0.000 000 000 000 454 × 2 = 0 + 0.000 000 000 000 908;
  • 3) 0.000 000 000 000 908 × 2 = 0 + 0.000 000 000 001 816;
  • 4) 0.000 000 000 001 816 × 2 = 0 + 0.000 000 000 003 632;
  • 5) 0.000 000 000 003 632 × 2 = 0 + 0.000 000 000 007 264;
  • 6) 0.000 000 000 007 264 × 2 = 0 + 0.000 000 000 014 528;
  • 7) 0.000 000 000 014 528 × 2 = 0 + 0.000 000 000 029 056;
  • 8) 0.000 000 000 029 056 × 2 = 0 + 0.000 000 000 058 112;
  • 9) 0.000 000 000 058 112 × 2 = 0 + 0.000 000 000 116 224;
  • 10) 0.000 000 000 116 224 × 2 = 0 + 0.000 000 000 232 448;
  • 11) 0.000 000 000 232 448 × 2 = 0 + 0.000 000 000 464 896;
  • 12) 0.000 000 000 464 896 × 2 = 0 + 0.000 000 000 929 792;
  • 13) 0.000 000 000 929 792 × 2 = 0 + 0.000 000 001 859 584;
  • 14) 0.000 000 001 859 584 × 2 = 0 + 0.000 000 003 719 168;
  • 15) 0.000 000 003 719 168 × 2 = 0 + 0.000 000 007 438 336;
  • 16) 0.000 000 007 438 336 × 2 = 0 + 0.000 000 014 876 672;
  • 17) 0.000 000 014 876 672 × 2 = 0 + 0.000 000 029 753 344;
  • 18) 0.000 000 029 753 344 × 2 = 0 + 0.000 000 059 506 688;
  • 19) 0.000 000 059 506 688 × 2 = 0 + 0.000 000 119 013 376;
  • 20) 0.000 000 119 013 376 × 2 = 0 + 0.000 000 238 026 752;
  • 21) 0.000 000 238 026 752 × 2 = 0 + 0.000 000 476 053 504;
  • 22) 0.000 000 476 053 504 × 2 = 0 + 0.000 000 952 107 008;
  • 23) 0.000 000 952 107 008 × 2 = 0 + 0.000 001 904 214 016;
  • 24) 0.000 001 904 214 016 × 2 = 0 + 0.000 003 808 428 032;
  • 25) 0.000 003 808 428 032 × 2 = 0 + 0.000 007 616 856 064;
  • 26) 0.000 007 616 856 064 × 2 = 0 + 0.000 015 233 712 128;
  • 27) 0.000 015 233 712 128 × 2 = 0 + 0.000 030 467 424 256;
  • 28) 0.000 030 467 424 256 × 2 = 0 + 0.000 060 934 848 512;
  • 29) 0.000 060 934 848 512 × 2 = 0 + 0.000 121 869 697 024;
  • 30) 0.000 121 869 697 024 × 2 = 0 + 0.000 243 739 394 048;
  • 31) 0.000 243 739 394 048 × 2 = 0 + 0.000 487 478 788 096;
  • 32) 0.000 487 478 788 096 × 2 = 0 + 0.000 974 957 576 192;
  • 33) 0.000 974 957 576 192 × 2 = 0 + 0.001 949 915 152 384;
  • 34) 0.001 949 915 152 384 × 2 = 0 + 0.003 899 830 304 768;
  • 35) 0.003 899 830 304 768 × 2 = 0 + 0.007 799 660 609 536;
  • 36) 0.007 799 660 609 536 × 2 = 0 + 0.015 599 321 219 072;
  • 37) 0.015 599 321 219 072 × 2 = 0 + 0.031 198 642 438 144;
  • 38) 0.031 198 642 438 144 × 2 = 0 + 0.062 397 284 876 288;
  • 39) 0.062 397 284 876 288 × 2 = 0 + 0.124 794 569 752 576;
  • 40) 0.124 794 569 752 576 × 2 = 0 + 0.249 589 139 505 152;
  • 41) 0.249 589 139 505 152 × 2 = 0 + 0.499 178 279 010 304;
  • 42) 0.499 178 279 010 304 × 2 = 0 + 0.998 356 558 020 608;
  • 43) 0.998 356 558 020 608 × 2 = 1 + 0.996 713 116 041 216;
  • 44) 0.996 713 116 041 216 × 2 = 1 + 0.993 426 232 082 432;
  • 45) 0.993 426 232 082 432 × 2 = 1 + 0.986 852 464 164 864;
  • 46) 0.986 852 464 164 864 × 2 = 1 + 0.973 704 928 329 728;
  • 47) 0.973 704 928 329 728 × 2 = 1 + 0.947 409 856 659 456;
  • 48) 0.947 409 856 659 456 × 2 = 1 + 0.894 819 713 318 912;
  • 49) 0.894 819 713 318 912 × 2 = 1 + 0.789 639 426 637 824;
  • 50) 0.789 639 426 637 824 × 2 = 1 + 0.579 278 853 275 648;
  • 51) 0.579 278 853 275 648 × 2 = 1 + 0.158 557 706 551 296;
  • 52) 0.158 557 706 551 296 × 2 = 0 + 0.317 115 413 102 592;
  • 53) 0.317 115 413 102 592 × 2 = 0 + 0.634 230 826 205 184;
  • 54) 0.634 230 826 205 184 × 2 = 1 + 0.268 461 652 410 368;
  • 55) 0.268 461 652 410 368 × 2 = 0 + 0.536 923 304 820 736;
  • 56) 0.536 923 304 820 736 × 2 = 1 + 0.073 846 609 641 472;
  • 57) 0.073 846 609 641 472 × 2 = 0 + 0.147 693 219 282 944;
  • 58) 0.147 693 219 282 944 × 2 = 0 + 0.295 386 438 565 888;
  • 59) 0.295 386 438 565 888 × 2 = 0 + 0.590 772 877 131 776;
  • 60) 0.590 772 877 131 776 × 2 = 1 + 0.181 545 754 263 552;
  • 61) 0.181 545 754 263 552 × 2 = 0 + 0.363 091 508 527 104;
  • 62) 0.363 091 508 527 104 × 2 = 0 + 0.726 183 017 054 208;
  • 63) 0.726 183 017 054 208 × 2 = 1 + 0.452 366 034 108 416;
  • 64) 0.452 366 034 108 416 × 2 = 0 + 0.904 732 068 216 832;
  • 65) 0.904 732 068 216 832 × 2 = 1 + 0.809 464 136 433 664;
  • 66) 0.809 464 136 433 664 × 2 = 1 + 0.618 928 272 867 328;
  • 67) 0.618 928 272 867 328 × 2 = 1 + 0.237 856 545 734 656;
  • 68) 0.237 856 545 734 656 × 2 = 0 + 0.475 713 091 469 312;
  • 69) 0.475 713 091 469 312 × 2 = 0 + 0.951 426 182 938 624;
  • 70) 0.951 426 182 938 624 × 2 = 1 + 0.902 852 365 877 248;
  • 71) 0.902 852 365 877 248 × 2 = 1 + 0.805 704 731 754 496;
  • 72) 0.805 704 731 754 496 × 2 = 1 + 0.611 409 463 508 992;
  • 73) 0.611 409 463 508 992 × 2 = 1 + 0.222 818 927 017 984;
  • 74) 0.222 818 927 017 984 × 2 = 0 + 0.445 637 854 035 968;
  • 75) 0.445 637 854 035 968 × 2 = 0 + 0.891 275 708 071 936;
  • 76) 0.891 275 708 071 936 × 2 = 1 + 0.782 551 416 143 872;
  • 77) 0.782 551 416 143 872 × 2 = 1 + 0.565 102 832 287 744;
  • 78) 0.565 102 832 287 744 × 2 = 1 + 0.130 205 664 575 488;
  • 79) 0.130 205 664 575 488 × 2 = 0 + 0.260 411 329 150 976;
  • 80) 0.260 411 329 150 976 × 2 = 0 + 0.520 822 658 301 952;
  • 81) 0.520 822 658 301 952 × 2 = 1 + 0.041 645 316 603 904;
  • 82) 0.041 645 316 603 904 × 2 = 0 + 0.083 290 633 207 808;
  • 83) 0.083 290 633 207 808 × 2 = 0 + 0.166 581 266 415 616;
  • 84) 0.166 581 266 415 616 × 2 = 0 + 0.333 162 532 831 232;
  • 85) 0.333 162 532 831 232 × 2 = 0 + 0.666 325 065 662 464;
  • 86) 0.666 325 065 662 464 × 2 = 1 + 0.332 650 131 324 928;
  • 87) 0.332 650 131 324 928 × 2 = 0 + 0.665 300 262 649 856;
  • 88) 0.665 300 262 649 856 × 2 = 1 + 0.330 600 525 299 712;
  • 89) 0.330 600 525 299 712 × 2 = 0 + 0.661 201 050 599 424;
  • 90) 0.661 201 050 599 424 × 2 = 1 + 0.322 402 101 198 848;
  • 91) 0.322 402 101 198 848 × 2 = 0 + 0.644 804 202 397 696;
  • 92) 0.644 804 202 397 696 × 2 = 1 + 0.289 608 404 795 392;
  • 93) 0.289 608 404 795 392 × 2 = 0 + 0.579 216 809 590 784;
  • 94) 0.579 216 809 590 784 × 2 = 1 + 0.158 433 619 181 568;
  • 95) 0.158 433 619 181 568 × 2 = 0 + 0.316 867 238 363 136;

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:


0.000 000 000 000 227(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1110 0101 0001 0010 1110 0111 1001 1100 1000 0101 0101 010(2)

5. Positive number before normalization:

0.000 000 000 000 227(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1110 0101 0001 0010 1110 0111 1001 1100 1000 0101 0101 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 227(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1110 0101 0001 0010 1110 0111 1001 1100 1000 0101 0101 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1110 0101 0001 0010 1110 0111 1001 1100 1000 0101 0101 010(2) × 20 =

1.1111 1111 0010 1000 1001 0111 0011 1100 1110 0100 0010 1010 1010(2) × 2-43


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): -43

Mantissa (not normalized):
1.1111 1111 0010 1000 1001 0111 0011 1100 1110 0100 0010 1010 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 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) =

980(10) =

011 1101 0100(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 1111 0010 1000 1001 0111 0011 1100 1110 0100 0010 1010 1010 =

1111 1111 0010 1000 1001 0111 0011 1100 1110 0100 0010 1010 1010


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 0100

Mantissa (52 bits) =
1111 1111 0010 1000 1001 0111 0011 1100 1110 0100 0010 1010 1010


Decimal number 0.000 000 000 000 227 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 1111 1111 0010 1000 1001 0111 0011 1100 1110 0100 0010 1010 1010

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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