0.000 000 000 000 261 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 261(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
0.000 000 000 000 261(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: 0.000 000 000 000 261.

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) 0.000 000 000 000 261 × 2 = 0 + 0.000 000 000 000 522;
  • 2) 0.000 000 000 000 522 × 2 = 0 + 0.000 000 000 001 044;
  • 3) 0.000 000 000 001 044 × 2 = 0 + 0.000 000 000 002 088;
  • 4) 0.000 000 000 002 088 × 2 = 0 + 0.000 000 000 004 176;
  • 5) 0.000 000 000 004 176 × 2 = 0 + 0.000 000 000 008 352;
  • 6) 0.000 000 000 008 352 × 2 = 0 + 0.000 000 000 016 704;
  • 7) 0.000 000 000 016 704 × 2 = 0 + 0.000 000 000 033 408;
  • 8) 0.000 000 000 033 408 × 2 = 0 + 0.000 000 000 066 816;
  • 9) 0.000 000 000 066 816 × 2 = 0 + 0.000 000 000 133 632;
  • 10) 0.000 000 000 133 632 × 2 = 0 + 0.000 000 000 267 264;
  • 11) 0.000 000 000 267 264 × 2 = 0 + 0.000 000 000 534 528;
  • 12) 0.000 000 000 534 528 × 2 = 0 + 0.000 000 001 069 056;
  • 13) 0.000 000 001 069 056 × 2 = 0 + 0.000 000 002 138 112;
  • 14) 0.000 000 002 138 112 × 2 = 0 + 0.000 000 004 276 224;
  • 15) 0.000 000 004 276 224 × 2 = 0 + 0.000 000 008 552 448;
  • 16) 0.000 000 008 552 448 × 2 = 0 + 0.000 000 017 104 896;
  • 17) 0.000 000 017 104 896 × 2 = 0 + 0.000 000 034 209 792;
  • 18) 0.000 000 034 209 792 × 2 = 0 + 0.000 000 068 419 584;
  • 19) 0.000 000 068 419 584 × 2 = 0 + 0.000 000 136 839 168;
  • 20) 0.000 000 136 839 168 × 2 = 0 + 0.000 000 273 678 336;
  • 21) 0.000 000 273 678 336 × 2 = 0 + 0.000 000 547 356 672;
  • 22) 0.000 000 547 356 672 × 2 = 0 + 0.000 001 094 713 344;
  • 23) 0.000 001 094 713 344 × 2 = 0 + 0.000 002 189 426 688;
  • 24) 0.000 002 189 426 688 × 2 = 0 + 0.000 004 378 853 376;
  • 25) 0.000 004 378 853 376 × 2 = 0 + 0.000 008 757 706 752;
  • 26) 0.000 008 757 706 752 × 2 = 0 + 0.000 017 515 413 504;
  • 27) 0.000 017 515 413 504 × 2 = 0 + 0.000 035 030 827 008;
  • 28) 0.000 035 030 827 008 × 2 = 0 + 0.000 070 061 654 016;
  • 29) 0.000 070 061 654 016 × 2 = 0 + 0.000 140 123 308 032;
  • 30) 0.000 140 123 308 032 × 2 = 0 + 0.000 280 246 616 064;
  • 31) 0.000 280 246 616 064 × 2 = 0 + 0.000 560 493 232 128;
  • 32) 0.000 560 493 232 128 × 2 = 0 + 0.001 120 986 464 256;
  • 33) 0.001 120 986 464 256 × 2 = 0 + 0.002 241 972 928 512;
  • 34) 0.002 241 972 928 512 × 2 = 0 + 0.004 483 945 857 024;
  • 35) 0.004 483 945 857 024 × 2 = 0 + 0.008 967 891 714 048;
  • 36) 0.008 967 891 714 048 × 2 = 0 + 0.017 935 783 428 096;
  • 37) 0.017 935 783 428 096 × 2 = 0 + 0.035 871 566 856 192;
  • 38) 0.035 871 566 856 192 × 2 = 0 + 0.071 743 133 712 384;
  • 39) 0.071 743 133 712 384 × 2 = 0 + 0.143 486 267 424 768;
  • 40) 0.143 486 267 424 768 × 2 = 0 + 0.286 972 534 849 536;
  • 41) 0.286 972 534 849 536 × 2 = 0 + 0.573 945 069 699 072;
  • 42) 0.573 945 069 699 072 × 2 = 1 + 0.147 890 139 398 144;
  • 43) 0.147 890 139 398 144 × 2 = 0 + 0.295 780 278 796 288;
  • 44) 0.295 780 278 796 288 × 2 = 0 + 0.591 560 557 592 576;
  • 45) 0.591 560 557 592 576 × 2 = 1 + 0.183 121 115 185 152;
  • 46) 0.183 121 115 185 152 × 2 = 0 + 0.366 242 230 370 304;
  • 47) 0.366 242 230 370 304 × 2 = 0 + 0.732 484 460 740 608;
  • 48) 0.732 484 460 740 608 × 2 = 1 + 0.464 968 921 481 216;
  • 49) 0.464 968 921 481 216 × 2 = 0 + 0.929 937 842 962 432;
  • 50) 0.929 937 842 962 432 × 2 = 1 + 0.859 875 685 924 864;
  • 51) 0.859 875 685 924 864 × 2 = 1 + 0.719 751 371 849 728;
  • 52) 0.719 751 371 849 728 × 2 = 1 + 0.439 502 743 699 456;
  • 53) 0.439 502 743 699 456 × 2 = 0 + 0.879 005 487 398 912;
  • 54) 0.879 005 487 398 912 × 2 = 1 + 0.758 010 974 797 824;
  • 55) 0.758 010 974 797 824 × 2 = 1 + 0.516 021 949 595 648;
  • 56) 0.516 021 949 595 648 × 2 = 1 + 0.032 043 899 191 296;
  • 57) 0.032 043 899 191 296 × 2 = 0 + 0.064 087 798 382 592;
  • 58) 0.064 087 798 382 592 × 2 = 0 + 0.128 175 596 765 184;
  • 59) 0.128 175 596 765 184 × 2 = 0 + 0.256 351 193 530 368;
  • 60) 0.256 351 193 530 368 × 2 = 0 + 0.512 702 387 060 736;
  • 61) 0.512 702 387 060 736 × 2 = 1 + 0.025 404 774 121 472;
  • 62) 0.025 404 774 121 472 × 2 = 0 + 0.050 809 548 242 944;
  • 63) 0.050 809 548 242 944 × 2 = 0 + 0.101 619 096 485 888;
  • 64) 0.101 619 096 485 888 × 2 = 0 + 0.203 238 192 971 776;
  • 65) 0.203 238 192 971 776 × 2 = 0 + 0.406 476 385 943 552;
  • 66) 0.406 476 385 943 552 × 2 = 0 + 0.812 952 771 887 104;
  • 67) 0.812 952 771 887 104 × 2 = 1 + 0.625 905 543 774 208;
  • 68) 0.625 905 543 774 208 × 2 = 1 + 0.251 811 087 548 416;
  • 69) 0.251 811 087 548 416 × 2 = 0 + 0.503 622 175 096 832;
  • 70) 0.503 622 175 096 832 × 2 = 1 + 0.007 244 350 193 664;
  • 71) 0.007 244 350 193 664 × 2 = 0 + 0.014 488 700 387 328;
  • 72) 0.014 488 700 387 328 × 2 = 0 + 0.028 977 400 774 656;
  • 73) 0.028 977 400 774 656 × 2 = 0 + 0.057 954 801 549 312;
  • 74) 0.057 954 801 549 312 × 2 = 0 + 0.115 909 603 098 624;
  • 75) 0.115 909 603 098 624 × 2 = 0 + 0.231 819 206 197 248;
  • 76) 0.231 819 206 197 248 × 2 = 0 + 0.463 638 412 394 496;
  • 77) 0.463 638 412 394 496 × 2 = 0 + 0.927 276 824 788 992;
  • 78) 0.927 276 824 788 992 × 2 = 1 + 0.854 553 649 577 984;
  • 79) 0.854 553 649 577 984 × 2 = 1 + 0.709 107 299 155 968;
  • 80) 0.709 107 299 155 968 × 2 = 1 + 0.418 214 598 311 936;
  • 81) 0.418 214 598 311 936 × 2 = 0 + 0.836 429 196 623 872;
  • 82) 0.836 429 196 623 872 × 2 = 1 + 0.672 858 393 247 744;
  • 83) 0.672 858 393 247 744 × 2 = 1 + 0.345 716 786 495 488;
  • 84) 0.345 716 786 495 488 × 2 = 0 + 0.691 433 572 990 976;
  • 85) 0.691 433 572 990 976 × 2 = 1 + 0.382 867 145 981 952;
  • 86) 0.382 867 145 981 952 × 2 = 0 + 0.765 734 291 963 904;
  • 87) 0.765 734 291 963 904 × 2 = 1 + 0.531 468 583 927 808;
  • 88) 0.531 468 583 927 808 × 2 = 1 + 0.062 937 167 855 616;
  • 89) 0.062 937 167 855 616 × 2 = 0 + 0.125 874 335 711 232;
  • 90) 0.125 874 335 711 232 × 2 = 0 + 0.251 748 671 422 464;
  • 91) 0.251 748 671 422 464 × 2 = 0 + 0.503 497 342 844 928;
  • 92) 0.503 497 342 844 928 × 2 = 1 + 0.006 994 685 689 856;
  • 93) 0.006 994 685 689 856 × 2 = 0 + 0.013 989 371 379 712;
  • 94) 0.013 989 371 379 712 × 2 = 0 + 0.027 978 742 759 424;

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:


0.000 000 000 000 261(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1001 0111 0111 0000 1000 0011 0100 0000 0111 0110 1011 0001 00(2)

5. Positive number before normalization:

0.000 000 000 000 261(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1001 0111 0111 0000 1000 0011 0100 0000 0111 0110 1011 0001 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 261(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1001 0111 0111 0000 1000 0011 0100 0000 0111 0110 1011 0001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1001 0111 0111 0000 1000 0011 0100 0000 0111 0110 1011 0001 00(2) × 20 =

1.0010 0101 1101 1100 0010 0000 1101 0000 0001 1101 1010 1100 0100(2) × 2-42


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): -42

Mantissa (not normalized):
1.0010 0101 1101 1100 0010 0000 1101 0000 0001 1101 1010 1100 0100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(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;
  • 981 ÷ 2 = 490 + 1;
  • 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) =

981(10) =

011 1101 0101(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. 0010 0101 1101 1100 0010 0000 1101 0000 0001 1101 1010 1100 0100 =

0010 0101 1101 1100 0010 0000 1101 0000 0001 1101 1010 1100 0100


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 0101

Mantissa (52 bits) =
0010 0101 1101 1100 0010 0000 1101 0000 0001 1101 1010 1100 0100


Decimal number 0.000 000 000 000 261 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0101 - 0010 0101 1101 1100 0010 0000 1101 0000 0001 1101 1010 1100 0100

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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