0.000 000 061 023 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 061 023(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
0.000 000 061 023(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: 0.000 000 061 023.

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) 0.000 000 061 023 × 2 = 0 + 0.000 000 122 046;
  • 2) 0.000 000 122 046 × 2 = 0 + 0.000 000 244 092;
  • 3) 0.000 000 244 092 × 2 = 0 + 0.000 000 488 184;
  • 4) 0.000 000 488 184 × 2 = 0 + 0.000 000 976 368;
  • 5) 0.000 000 976 368 × 2 = 0 + 0.000 001 952 736;
  • 6) 0.000 001 952 736 × 2 = 0 + 0.000 003 905 472;
  • 7) 0.000 003 905 472 × 2 = 0 + 0.000 007 810 944;
  • 8) 0.000 007 810 944 × 2 = 0 + 0.000 015 621 888;
  • 9) 0.000 015 621 888 × 2 = 0 + 0.000 031 243 776;
  • 10) 0.000 031 243 776 × 2 = 0 + 0.000 062 487 552;
  • 11) 0.000 062 487 552 × 2 = 0 + 0.000 124 975 104;
  • 12) 0.000 124 975 104 × 2 = 0 + 0.000 249 950 208;
  • 13) 0.000 249 950 208 × 2 = 0 + 0.000 499 900 416;
  • 14) 0.000 499 900 416 × 2 = 0 + 0.000 999 800 832;
  • 15) 0.000 999 800 832 × 2 = 0 + 0.001 999 601 664;
  • 16) 0.001 999 601 664 × 2 = 0 + 0.003 999 203 328;
  • 17) 0.003 999 203 328 × 2 = 0 + 0.007 998 406 656;
  • 18) 0.007 998 406 656 × 2 = 0 + 0.015 996 813 312;
  • 19) 0.015 996 813 312 × 2 = 0 + 0.031 993 626 624;
  • 20) 0.031 993 626 624 × 2 = 0 + 0.063 987 253 248;
  • 21) 0.063 987 253 248 × 2 = 0 + 0.127 974 506 496;
  • 22) 0.127 974 506 496 × 2 = 0 + 0.255 949 012 992;
  • 23) 0.255 949 012 992 × 2 = 0 + 0.511 898 025 984;
  • 24) 0.511 898 025 984 × 2 = 1 + 0.023 796 051 968;
  • 25) 0.023 796 051 968 × 2 = 0 + 0.047 592 103 936;
  • 26) 0.047 592 103 936 × 2 = 0 + 0.095 184 207 872;
  • 27) 0.095 184 207 872 × 2 = 0 + 0.190 368 415 744;
  • 28) 0.190 368 415 744 × 2 = 0 + 0.380 736 831 488;
  • 29) 0.380 736 831 488 × 2 = 0 + 0.761 473 662 976;
  • 30) 0.761 473 662 976 × 2 = 1 + 0.522 947 325 952;
  • 31) 0.522 947 325 952 × 2 = 1 + 0.045 894 651 904;
  • 32) 0.045 894 651 904 × 2 = 0 + 0.091 789 303 808;
  • 33) 0.091 789 303 808 × 2 = 0 + 0.183 578 607 616;
  • 34) 0.183 578 607 616 × 2 = 0 + 0.367 157 215 232;
  • 35) 0.367 157 215 232 × 2 = 0 + 0.734 314 430 464;
  • 36) 0.734 314 430 464 × 2 = 1 + 0.468 628 860 928;
  • 37) 0.468 628 860 928 × 2 = 0 + 0.937 257 721 856;
  • 38) 0.937 257 721 856 × 2 = 1 + 0.874 515 443 712;
  • 39) 0.874 515 443 712 × 2 = 1 + 0.749 030 887 424;
  • 40) 0.749 030 887 424 × 2 = 1 + 0.498 061 774 848;
  • 41) 0.498 061 774 848 × 2 = 0 + 0.996 123 549 696;
  • 42) 0.996 123 549 696 × 2 = 1 + 0.992 247 099 392;
  • 43) 0.992 247 099 392 × 2 = 1 + 0.984 494 198 784;
  • 44) 0.984 494 198 784 × 2 = 1 + 0.968 988 397 568;
  • 45) 0.968 988 397 568 × 2 = 1 + 0.937 976 795 136;
  • 46) 0.937 976 795 136 × 2 = 1 + 0.875 953 590 272;
  • 47) 0.875 953 590 272 × 2 = 1 + 0.751 907 180 544;
  • 48) 0.751 907 180 544 × 2 = 1 + 0.503 814 361 088;
  • 49) 0.503 814 361 088 × 2 = 1 + 0.007 628 722 176;
  • 50) 0.007 628 722 176 × 2 = 0 + 0.015 257 444 352;
  • 51) 0.015 257 444 352 × 2 = 0 + 0.030 514 888 704;
  • 52) 0.030 514 888 704 × 2 = 0 + 0.061 029 777 408;
  • 53) 0.061 029 777 408 × 2 = 0 + 0.122 059 554 816;
  • 54) 0.122 059 554 816 × 2 = 0 + 0.244 119 109 632;
  • 55) 0.244 119 109 632 × 2 = 0 + 0.488 238 219 264;
  • 56) 0.488 238 219 264 × 2 = 0 + 0.976 476 438 528;
  • 57) 0.976 476 438 528 × 2 = 1 + 0.952 952 877 056;
  • 58) 0.952 952 877 056 × 2 = 1 + 0.905 905 754 112;
  • 59) 0.905 905 754 112 × 2 = 1 + 0.811 811 508 224;
  • 60) 0.811 811 508 224 × 2 = 1 + 0.623 623 016 448;
  • 61) 0.623 623 016 448 × 2 = 1 + 0.247 246 032 896;
  • 62) 0.247 246 032 896 × 2 = 0 + 0.494 492 065 792;
  • 63) 0.494 492 065 792 × 2 = 0 + 0.988 984 131 584;
  • 64) 0.988 984 131 584 × 2 = 1 + 0.977 968 263 168;
  • 65) 0.977 968 263 168 × 2 = 1 + 0.955 936 526 336;
  • 66) 0.955 936 526 336 × 2 = 1 + 0.911 873 052 672;
  • 67) 0.911 873 052 672 × 2 = 1 + 0.823 746 105 344;
  • 68) 0.823 746 105 344 × 2 = 1 + 0.647 492 210 688;
  • 69) 0.647 492 210 688 × 2 = 1 + 0.294 984 421 376;
  • 70) 0.294 984 421 376 × 2 = 0 + 0.589 968 842 752;
  • 71) 0.589 968 842 752 × 2 = 1 + 0.179 937 685 504;
  • 72) 0.179 937 685 504 × 2 = 0 + 0.359 875 371 008;
  • 73) 0.359 875 371 008 × 2 = 0 + 0.719 750 742 016;
  • 74) 0.719 750 742 016 × 2 = 1 + 0.439 501 484 032;
  • 75) 0.439 501 484 032 × 2 = 0 + 0.879 002 968 064;
  • 76) 0.879 002 968 064 × 2 = 1 + 0.758 005 936 128;

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:


0.000 000 061 023(10) =

0.0000 0000 0000 0000 0000 0001 0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101(2)

5. Positive number before normalization:

0.000 000 061 023(10) =

0.0000 0000 0000 0000 0000 0001 0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 061 023(10) =

0.0000 0000 0000 0000 0000 0001 0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101(2) =

0.0000 0000 0000 0000 0000 0001 0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101(2) × 20 =

1.0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101(2) × 2-24


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): -24

Mantissa (not normalized):
1.0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(11-1) - 1 =

(-24 + 1 023)(10) =

999(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;
  • 999 ÷ 2 = 499 + 1;
  • 499 ÷ 2 = 249 + 1;
  • 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) =

999(10) =

011 1110 0111(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. 0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101 =

0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101


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 0111

Mantissa (52 bits) =
0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101


Decimal number 0.000 000 061 023 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0111 - 0000 0110 0001 0111 0111 1111 1000 0000 1111 1001 1111 1010 0101

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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