0.000 000 021 980 16 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 980 16(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
0.000 000 021 980 16(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: 0.000 000 021 980 16.

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) 0.000 000 021 980 16 × 2 = 0 + 0.000 000 043 960 32;
  • 2) 0.000 000 043 960 32 × 2 = 0 + 0.000 000 087 920 64;
  • 3) 0.000 000 087 920 64 × 2 = 0 + 0.000 000 175 841 28;
  • 4) 0.000 000 175 841 28 × 2 = 0 + 0.000 000 351 682 56;
  • 5) 0.000 000 351 682 56 × 2 = 0 + 0.000 000 703 365 12;
  • 6) 0.000 000 703 365 12 × 2 = 0 + 0.000 001 406 730 24;
  • 7) 0.000 001 406 730 24 × 2 = 0 + 0.000 002 813 460 48;
  • 8) 0.000 002 813 460 48 × 2 = 0 + 0.000 005 626 920 96;
  • 9) 0.000 005 626 920 96 × 2 = 0 + 0.000 011 253 841 92;
  • 10) 0.000 011 253 841 92 × 2 = 0 + 0.000 022 507 683 84;
  • 11) 0.000 022 507 683 84 × 2 = 0 + 0.000 045 015 367 68;
  • 12) 0.000 045 015 367 68 × 2 = 0 + 0.000 090 030 735 36;
  • 13) 0.000 090 030 735 36 × 2 = 0 + 0.000 180 061 470 72;
  • 14) 0.000 180 061 470 72 × 2 = 0 + 0.000 360 122 941 44;
  • 15) 0.000 360 122 941 44 × 2 = 0 + 0.000 720 245 882 88;
  • 16) 0.000 720 245 882 88 × 2 = 0 + 0.001 440 491 765 76;
  • 17) 0.001 440 491 765 76 × 2 = 0 + 0.002 880 983 531 52;
  • 18) 0.002 880 983 531 52 × 2 = 0 + 0.005 761 967 063 04;
  • 19) 0.005 761 967 063 04 × 2 = 0 + 0.011 523 934 126 08;
  • 20) 0.011 523 934 126 08 × 2 = 0 + 0.023 047 868 252 16;
  • 21) 0.023 047 868 252 16 × 2 = 0 + 0.046 095 736 504 32;
  • 22) 0.046 095 736 504 32 × 2 = 0 + 0.092 191 473 008 64;
  • 23) 0.092 191 473 008 64 × 2 = 0 + 0.184 382 946 017 28;
  • 24) 0.184 382 946 017 28 × 2 = 0 + 0.368 765 892 034 56;
  • 25) 0.368 765 892 034 56 × 2 = 0 + 0.737 531 784 069 12;
  • 26) 0.737 531 784 069 12 × 2 = 1 + 0.475 063 568 138 24;
  • 27) 0.475 063 568 138 24 × 2 = 0 + 0.950 127 136 276 48;
  • 28) 0.950 127 136 276 48 × 2 = 1 + 0.900 254 272 552 96;
  • 29) 0.900 254 272 552 96 × 2 = 1 + 0.800 508 545 105 92;
  • 30) 0.800 508 545 105 92 × 2 = 1 + 0.601 017 090 211 84;
  • 31) 0.601 017 090 211 84 × 2 = 1 + 0.202 034 180 423 68;
  • 32) 0.202 034 180 423 68 × 2 = 0 + 0.404 068 360 847 36;
  • 33) 0.404 068 360 847 36 × 2 = 0 + 0.808 136 721 694 72;
  • 34) 0.808 136 721 694 72 × 2 = 1 + 0.616 273 443 389 44;
  • 35) 0.616 273 443 389 44 × 2 = 1 + 0.232 546 886 778 88;
  • 36) 0.232 546 886 778 88 × 2 = 0 + 0.465 093 773 557 76;
  • 37) 0.465 093 773 557 76 × 2 = 0 + 0.930 187 547 115 52;
  • 38) 0.930 187 547 115 52 × 2 = 1 + 0.860 375 094 231 04;
  • 39) 0.860 375 094 231 04 × 2 = 1 + 0.720 750 188 462 08;
  • 40) 0.720 750 188 462 08 × 2 = 1 + 0.441 500 376 924 16;
  • 41) 0.441 500 376 924 16 × 2 = 0 + 0.883 000 753 848 32;
  • 42) 0.883 000 753 848 32 × 2 = 1 + 0.766 001 507 696 64;
  • 43) 0.766 001 507 696 64 × 2 = 1 + 0.532 003 015 393 28;
  • 44) 0.532 003 015 393 28 × 2 = 1 + 0.064 006 030 786 56;
  • 45) 0.064 006 030 786 56 × 2 = 0 + 0.128 012 061 573 12;
  • 46) 0.128 012 061 573 12 × 2 = 0 + 0.256 024 123 146 24;
  • 47) 0.256 024 123 146 24 × 2 = 0 + 0.512 048 246 292 48;
  • 48) 0.512 048 246 292 48 × 2 = 1 + 0.024 096 492 584 96;
  • 49) 0.024 096 492 584 96 × 2 = 0 + 0.048 192 985 169 92;
  • 50) 0.048 192 985 169 92 × 2 = 0 + 0.096 385 970 339 84;
  • 51) 0.096 385 970 339 84 × 2 = 0 + 0.192 771 940 679 68;
  • 52) 0.192 771 940 679 68 × 2 = 0 + 0.385 543 881 359 36;
  • 53) 0.385 543 881 359 36 × 2 = 0 + 0.771 087 762 718 72;
  • 54) 0.771 087 762 718 72 × 2 = 1 + 0.542 175 525 437 44;
  • 55) 0.542 175 525 437 44 × 2 = 1 + 0.084 351 050 874 88;
  • 56) 0.084 351 050 874 88 × 2 = 0 + 0.168 702 101 749 76;
  • 57) 0.168 702 101 749 76 × 2 = 0 + 0.337 404 203 499 52;
  • 58) 0.337 404 203 499 52 × 2 = 0 + 0.674 808 406 999 04;
  • 59) 0.674 808 406 999 04 × 2 = 1 + 0.349 616 813 998 08;
  • 60) 0.349 616 813 998 08 × 2 = 0 + 0.699 233 627 996 16;
  • 61) 0.699 233 627 996 16 × 2 = 1 + 0.398 467 255 992 32;
  • 62) 0.398 467 255 992 32 × 2 = 0 + 0.796 934 511 984 64;
  • 63) 0.796 934 511 984 64 × 2 = 1 + 0.593 869 023 969 28;
  • 64) 0.593 869 023 969 28 × 2 = 1 + 0.187 738 047 938 56;
  • 65) 0.187 738 047 938 56 × 2 = 0 + 0.375 476 095 877 12;
  • 66) 0.375 476 095 877 12 × 2 = 0 + 0.750 952 191 754 24;
  • 67) 0.750 952 191 754 24 × 2 = 1 + 0.501 904 383 508 48;
  • 68) 0.501 904 383 508 48 × 2 = 1 + 0.003 808 767 016 96;
  • 69) 0.003 808 767 016 96 × 2 = 0 + 0.007 617 534 033 92;
  • 70) 0.007 617 534 033 92 × 2 = 0 + 0.015 235 068 067 84;
  • 71) 0.015 235 068 067 84 × 2 = 0 + 0.030 470 136 135 68;
  • 72) 0.030 470 136 135 68 × 2 = 0 + 0.060 940 272 271 36;
  • 73) 0.060 940 272 271 36 × 2 = 0 + 0.121 880 544 542 72;
  • 74) 0.121 880 544 542 72 × 2 = 0 + 0.243 761 089 085 44;
  • 75) 0.243 761 089 085 44 × 2 = 0 + 0.487 522 178 170 88;
  • 76) 0.487 522 178 170 88 × 2 = 0 + 0.975 044 356 341 76;
  • 77) 0.975 044 356 341 76 × 2 = 1 + 0.950 088 712 683 52;
  • 78) 0.950 088 712 683 52 × 2 = 1 + 0.900 177 425 367 04;

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:


0.000 000 021 980 16(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0111 0111 0001 0000 0110 0010 1011 0011 0000 0000 11(2)

5. Positive number before normalization:

0.000 000 021 980 16(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0111 0111 0001 0000 0110 0010 1011 0011 0000 0000 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 021 980 16(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0111 0111 0001 0000 0110 0010 1011 0011 0000 0000 11(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0111 0111 0001 0000 0110 0010 1011 0011 0000 0000 11(2) × 20 =

1.0111 1001 1001 1101 1100 0100 0001 1000 1010 1100 1100 0000 0011(2) × 2-26


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): -26

Mantissa (not normalized):
1.0111 1001 1001 1101 1100 0100 0001 1000 1010 1100 1100 0000 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

997(10) =

011 1110 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. 0111 1001 1001 1101 1100 0100 0001 1000 1010 1100 1100 0000 0011 =

0111 1001 1001 1101 1100 0100 0001 1000 1010 1100 1100 0000 0011


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 1110 0101

Mantissa (52 bits) =
0111 1001 1001 1101 1100 0100 0001 1000 1010 1100 1100 0000 0011


Decimal number 0.000 000 021 980 16 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 0111 1001 1001 1101 1100 0100 0001 1000 1010 1100 1100 0000 0011

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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