0.000 000 000 000 021 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 021(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
0.000 000 000 000 021(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: 0.000 000 000 000 021.

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) 0.000 000 000 000 021 × 2 = 0 + 0.000 000 000 000 042;
  • 2) 0.000 000 000 000 042 × 2 = 0 + 0.000 000 000 000 084;
  • 3) 0.000 000 000 000 084 × 2 = 0 + 0.000 000 000 000 168;
  • 4) 0.000 000 000 000 168 × 2 = 0 + 0.000 000 000 000 336;
  • 5) 0.000 000 000 000 336 × 2 = 0 + 0.000 000 000 000 672;
  • 6) 0.000 000 000 000 672 × 2 = 0 + 0.000 000 000 001 344;
  • 7) 0.000 000 000 001 344 × 2 = 0 + 0.000 000 000 002 688;
  • 8) 0.000 000 000 002 688 × 2 = 0 + 0.000 000 000 005 376;
  • 9) 0.000 000 000 005 376 × 2 = 0 + 0.000 000 000 010 752;
  • 10) 0.000 000 000 010 752 × 2 = 0 + 0.000 000 000 021 504;
  • 11) 0.000 000 000 021 504 × 2 = 0 + 0.000 000 000 043 008;
  • 12) 0.000 000 000 043 008 × 2 = 0 + 0.000 000 000 086 016;
  • 13) 0.000 000 000 086 016 × 2 = 0 + 0.000 000 000 172 032;
  • 14) 0.000 000 000 172 032 × 2 = 0 + 0.000 000 000 344 064;
  • 15) 0.000 000 000 344 064 × 2 = 0 + 0.000 000 000 688 128;
  • 16) 0.000 000 000 688 128 × 2 = 0 + 0.000 000 001 376 256;
  • 17) 0.000 000 001 376 256 × 2 = 0 + 0.000 000 002 752 512;
  • 18) 0.000 000 002 752 512 × 2 = 0 + 0.000 000 005 505 024;
  • 19) 0.000 000 005 505 024 × 2 = 0 + 0.000 000 011 010 048;
  • 20) 0.000 000 011 010 048 × 2 = 0 + 0.000 000 022 020 096;
  • 21) 0.000 000 022 020 096 × 2 = 0 + 0.000 000 044 040 192;
  • 22) 0.000 000 044 040 192 × 2 = 0 + 0.000 000 088 080 384;
  • 23) 0.000 000 088 080 384 × 2 = 0 + 0.000 000 176 160 768;
  • 24) 0.000 000 176 160 768 × 2 = 0 + 0.000 000 352 321 536;
  • 25) 0.000 000 352 321 536 × 2 = 0 + 0.000 000 704 643 072;
  • 26) 0.000 000 704 643 072 × 2 = 0 + 0.000 001 409 286 144;
  • 27) 0.000 001 409 286 144 × 2 = 0 + 0.000 002 818 572 288;
  • 28) 0.000 002 818 572 288 × 2 = 0 + 0.000 005 637 144 576;
  • 29) 0.000 005 637 144 576 × 2 = 0 + 0.000 011 274 289 152;
  • 30) 0.000 011 274 289 152 × 2 = 0 + 0.000 022 548 578 304;
  • 31) 0.000 022 548 578 304 × 2 = 0 + 0.000 045 097 156 608;
  • 32) 0.000 045 097 156 608 × 2 = 0 + 0.000 090 194 313 216;
  • 33) 0.000 090 194 313 216 × 2 = 0 + 0.000 180 388 626 432;
  • 34) 0.000 180 388 626 432 × 2 = 0 + 0.000 360 777 252 864;
  • 35) 0.000 360 777 252 864 × 2 = 0 + 0.000 721 554 505 728;
  • 36) 0.000 721 554 505 728 × 2 = 0 + 0.001 443 109 011 456;
  • 37) 0.001 443 109 011 456 × 2 = 0 + 0.002 886 218 022 912;
  • 38) 0.002 886 218 022 912 × 2 = 0 + 0.005 772 436 045 824;
  • 39) 0.005 772 436 045 824 × 2 = 0 + 0.011 544 872 091 648;
  • 40) 0.011 544 872 091 648 × 2 = 0 + 0.023 089 744 183 296;
  • 41) 0.023 089 744 183 296 × 2 = 0 + 0.046 179 488 366 592;
  • 42) 0.046 179 488 366 592 × 2 = 0 + 0.092 358 976 733 184;
  • 43) 0.092 358 976 733 184 × 2 = 0 + 0.184 717 953 466 368;
  • 44) 0.184 717 953 466 368 × 2 = 0 + 0.369 435 906 932 736;
  • 45) 0.369 435 906 932 736 × 2 = 0 + 0.738 871 813 865 472;
  • 46) 0.738 871 813 865 472 × 2 = 1 + 0.477 743 627 730 944;
  • 47) 0.477 743 627 730 944 × 2 = 0 + 0.955 487 255 461 888;
  • 48) 0.955 487 255 461 888 × 2 = 1 + 0.910 974 510 923 776;
  • 49) 0.910 974 510 923 776 × 2 = 1 + 0.821 949 021 847 552;
  • 50) 0.821 949 021 847 552 × 2 = 1 + 0.643 898 043 695 104;
  • 51) 0.643 898 043 695 104 × 2 = 1 + 0.287 796 087 390 208;
  • 52) 0.287 796 087 390 208 × 2 = 0 + 0.575 592 174 780 416;
  • 53) 0.575 592 174 780 416 × 2 = 1 + 0.151 184 349 560 832;
  • 54) 0.151 184 349 560 832 × 2 = 0 + 0.302 368 699 121 664;
  • 55) 0.302 368 699 121 664 × 2 = 0 + 0.604 737 398 243 328;
  • 56) 0.604 737 398 243 328 × 2 = 1 + 0.209 474 796 486 656;
  • 57) 0.209 474 796 486 656 × 2 = 0 + 0.418 949 592 973 312;
  • 58) 0.418 949 592 973 312 × 2 = 0 + 0.837 899 185 946 624;
  • 59) 0.837 899 185 946 624 × 2 = 1 + 0.675 798 371 893 248;
  • 60) 0.675 798 371 893 248 × 2 = 1 + 0.351 596 743 786 496;
  • 61) 0.351 596 743 786 496 × 2 = 0 + 0.703 193 487 572 992;
  • 62) 0.703 193 487 572 992 × 2 = 1 + 0.406 386 975 145 984;
  • 63) 0.406 386 975 145 984 × 2 = 0 + 0.812 773 950 291 968;
  • 64) 0.812 773 950 291 968 × 2 = 1 + 0.625 547 900 583 936;
  • 65) 0.625 547 900 583 936 × 2 = 1 + 0.251 095 801 167 872;
  • 66) 0.251 095 801 167 872 × 2 = 0 + 0.502 191 602 335 744;
  • 67) 0.502 191 602 335 744 × 2 = 1 + 0.004 383 204 671 488;
  • 68) 0.004 383 204 671 488 × 2 = 0 + 0.008 766 409 342 976;
  • 69) 0.008 766 409 342 976 × 2 = 0 + 0.017 532 818 685 952;
  • 70) 0.017 532 818 685 952 × 2 = 0 + 0.035 065 637 371 904;
  • 71) 0.035 065 637 371 904 × 2 = 0 + 0.070 131 274 743 808;
  • 72) 0.070 131 274 743 808 × 2 = 0 + 0.140 262 549 487 616;
  • 73) 0.140 262 549 487 616 × 2 = 0 + 0.280 525 098 975 232;
  • 74) 0.280 525 098 975 232 × 2 = 0 + 0.561 050 197 950 464;
  • 75) 0.561 050 197 950 464 × 2 = 1 + 0.122 100 395 900 928;
  • 76) 0.122 100 395 900 928 × 2 = 0 + 0.244 200 791 801 856;
  • 77) 0.244 200 791 801 856 × 2 = 0 + 0.488 401 583 603 712;
  • 78) 0.488 401 583 603 712 × 2 = 0 + 0.976 803 167 207 424;
  • 79) 0.976 803 167 207 424 × 2 = 1 + 0.953 606 334 414 848;
  • 80) 0.953 606 334 414 848 × 2 = 1 + 0.907 212 668 829 696;
  • 81) 0.907 212 668 829 696 × 2 = 1 + 0.814 425 337 659 392;
  • 82) 0.814 425 337 659 392 × 2 = 1 + 0.628 850 675 318 784;
  • 83) 0.628 850 675 318 784 × 2 = 1 + 0.257 701 350 637 568;
  • 84) 0.257 701 350 637 568 × 2 = 0 + 0.515 402 701 275 136;
  • 85) 0.515 402 701 275 136 × 2 = 1 + 0.030 805 402 550 272;
  • 86) 0.030 805 402 550 272 × 2 = 0 + 0.061 610 805 100 544;
  • 87) 0.061 610 805 100 544 × 2 = 0 + 0.123 221 610 201 088;
  • 88) 0.123 221 610 201 088 × 2 = 0 + 0.246 443 220 402 176;
  • 89) 0.246 443 220 402 176 × 2 = 0 + 0.492 886 440 804 352;
  • 90) 0.492 886 440 804 352 × 2 = 0 + 0.985 772 881 608 704;
  • 91) 0.985 772 881 608 704 × 2 = 1 + 0.971 545 763 217 408;
  • 92) 0.971 545 763 217 408 × 2 = 1 + 0.943 091 526 434 816;
  • 93) 0.943 091 526 434 816 × 2 = 1 + 0.886 183 052 869 632;
  • 94) 0.886 183 052 869 632 × 2 = 1 + 0.772 366 105 739 264;
  • 95) 0.772 366 105 739 264 × 2 = 1 + 0.544 732 211 478 528;
  • 96) 0.544 732 211 478 528 × 2 = 1 + 0.089 464 422 957 056;
  • 97) 0.089 464 422 957 056 × 2 = 0 + 0.178 928 845 914 112;
  • 98) 0.178 928 845 914 112 × 2 = 0 + 0.357 857 691 828 224;

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:


0.000 000 000 000 021(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1110 1001 0011 0101 1010 0000 0010 0011 1110 1000 0011 1111 00(2)

5. Positive number before normalization:

0.000 000 000 000 021(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1110 1001 0011 0101 1010 0000 0010 0011 1110 1000 0011 1111 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 021(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1110 1001 0011 0101 1010 0000 0010 0011 1110 1000 0011 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1110 1001 0011 0101 1010 0000 0010 0011 1110 1000 0011 1111 00(2) × 20 =

1.0111 1010 0100 1101 0110 1000 0000 1000 1111 1010 0000 1111 1100(2) × 2-46


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): -46

Mantissa (not normalized):
1.0111 1010 0100 1101 0110 1000 0000 1000 1111 1010 0000 1111 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 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) =

977(10) =

011 1101 0001(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. 0111 1010 0100 1101 0110 1000 0000 1000 1111 1010 0000 1111 1100 =

0111 1010 0100 1101 0110 1000 0000 1000 1111 1010 0000 1111 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 0001

Mantissa (52 bits) =
0111 1010 0100 1101 0110 1000 0000 1000 1111 1010 0000 1111 1100


Decimal number 0.000 000 000 000 021 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0001 - 0111 1010 0100 1101 0110 1000 0000 1000 1111 1010 0000 1111 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