0.000 000 000 015 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 015 3(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
0.000 000 000 015 3(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: 0.000 000 000 015 3.

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) 0.000 000 000 015 3 × 2 = 0 + 0.000 000 000 030 6;
  • 2) 0.000 000 000 030 6 × 2 = 0 + 0.000 000 000 061 2;
  • 3) 0.000 000 000 061 2 × 2 = 0 + 0.000 000 000 122 4;
  • 4) 0.000 000 000 122 4 × 2 = 0 + 0.000 000 000 244 8;
  • 5) 0.000 000 000 244 8 × 2 = 0 + 0.000 000 000 489 6;
  • 6) 0.000 000 000 489 6 × 2 = 0 + 0.000 000 000 979 2;
  • 7) 0.000 000 000 979 2 × 2 = 0 + 0.000 000 001 958 4;
  • 8) 0.000 000 001 958 4 × 2 = 0 + 0.000 000 003 916 8;
  • 9) 0.000 000 003 916 8 × 2 = 0 + 0.000 000 007 833 6;
  • 10) 0.000 000 007 833 6 × 2 = 0 + 0.000 000 015 667 2;
  • 11) 0.000 000 015 667 2 × 2 = 0 + 0.000 000 031 334 4;
  • 12) 0.000 000 031 334 4 × 2 = 0 + 0.000 000 062 668 8;
  • 13) 0.000 000 062 668 8 × 2 = 0 + 0.000 000 125 337 6;
  • 14) 0.000 000 125 337 6 × 2 = 0 + 0.000 000 250 675 2;
  • 15) 0.000 000 250 675 2 × 2 = 0 + 0.000 000 501 350 4;
  • 16) 0.000 000 501 350 4 × 2 = 0 + 0.000 001 002 700 8;
  • 17) 0.000 001 002 700 8 × 2 = 0 + 0.000 002 005 401 6;
  • 18) 0.000 002 005 401 6 × 2 = 0 + 0.000 004 010 803 2;
  • 19) 0.000 004 010 803 2 × 2 = 0 + 0.000 008 021 606 4;
  • 20) 0.000 008 021 606 4 × 2 = 0 + 0.000 016 043 212 8;
  • 21) 0.000 016 043 212 8 × 2 = 0 + 0.000 032 086 425 6;
  • 22) 0.000 032 086 425 6 × 2 = 0 + 0.000 064 172 851 2;
  • 23) 0.000 064 172 851 2 × 2 = 0 + 0.000 128 345 702 4;
  • 24) 0.000 128 345 702 4 × 2 = 0 + 0.000 256 691 404 8;
  • 25) 0.000 256 691 404 8 × 2 = 0 + 0.000 513 382 809 6;
  • 26) 0.000 513 382 809 6 × 2 = 0 + 0.001 026 765 619 2;
  • 27) 0.001 026 765 619 2 × 2 = 0 + 0.002 053 531 238 4;
  • 28) 0.002 053 531 238 4 × 2 = 0 + 0.004 107 062 476 8;
  • 29) 0.004 107 062 476 8 × 2 = 0 + 0.008 214 124 953 6;
  • 30) 0.008 214 124 953 6 × 2 = 0 + 0.016 428 249 907 2;
  • 31) 0.016 428 249 907 2 × 2 = 0 + 0.032 856 499 814 4;
  • 32) 0.032 856 499 814 4 × 2 = 0 + 0.065 712 999 628 8;
  • 33) 0.065 712 999 628 8 × 2 = 0 + 0.131 425 999 257 6;
  • 34) 0.131 425 999 257 6 × 2 = 0 + 0.262 851 998 515 2;
  • 35) 0.262 851 998 515 2 × 2 = 0 + 0.525 703 997 030 4;
  • 36) 0.525 703 997 030 4 × 2 = 1 + 0.051 407 994 060 8;
  • 37) 0.051 407 994 060 8 × 2 = 0 + 0.102 815 988 121 6;
  • 38) 0.102 815 988 121 6 × 2 = 0 + 0.205 631 976 243 2;
  • 39) 0.205 631 976 243 2 × 2 = 0 + 0.411 263 952 486 4;
  • 40) 0.411 263 952 486 4 × 2 = 0 + 0.822 527 904 972 8;
  • 41) 0.822 527 904 972 8 × 2 = 1 + 0.645 055 809 945 6;
  • 42) 0.645 055 809 945 6 × 2 = 1 + 0.290 111 619 891 2;
  • 43) 0.290 111 619 891 2 × 2 = 0 + 0.580 223 239 782 4;
  • 44) 0.580 223 239 782 4 × 2 = 1 + 0.160 446 479 564 8;
  • 45) 0.160 446 479 564 8 × 2 = 0 + 0.320 892 959 129 6;
  • 46) 0.320 892 959 129 6 × 2 = 0 + 0.641 785 918 259 2;
  • 47) 0.641 785 918 259 2 × 2 = 1 + 0.283 571 836 518 4;
  • 48) 0.283 571 836 518 4 × 2 = 0 + 0.567 143 673 036 8;
  • 49) 0.567 143 673 036 8 × 2 = 1 + 0.134 287 346 073 6;
  • 50) 0.134 287 346 073 6 × 2 = 0 + 0.268 574 692 147 2;
  • 51) 0.268 574 692 147 2 × 2 = 0 + 0.537 149 384 294 4;
  • 52) 0.537 149 384 294 4 × 2 = 1 + 0.074 298 768 588 8;
  • 53) 0.074 298 768 588 8 × 2 = 0 + 0.148 597 537 177 6;
  • 54) 0.148 597 537 177 6 × 2 = 0 + 0.297 195 074 355 2;
  • 55) 0.297 195 074 355 2 × 2 = 0 + 0.594 390 148 710 4;
  • 56) 0.594 390 148 710 4 × 2 = 1 + 0.188 780 297 420 8;
  • 57) 0.188 780 297 420 8 × 2 = 0 + 0.377 560 594 841 6;
  • 58) 0.377 560 594 841 6 × 2 = 0 + 0.755 121 189 683 2;
  • 59) 0.755 121 189 683 2 × 2 = 1 + 0.510 242 379 366 4;
  • 60) 0.510 242 379 366 4 × 2 = 1 + 0.020 484 758 732 8;
  • 61) 0.020 484 758 732 8 × 2 = 0 + 0.040 969 517 465 6;
  • 62) 0.040 969 517 465 6 × 2 = 0 + 0.081 939 034 931 2;
  • 63) 0.081 939 034 931 2 × 2 = 0 + 0.163 878 069 862 4;
  • 64) 0.163 878 069 862 4 × 2 = 0 + 0.327 756 139 724 8;
  • 65) 0.327 756 139 724 8 × 2 = 0 + 0.655 512 279 449 6;
  • 66) 0.655 512 279 449 6 × 2 = 1 + 0.311 024 558 899 2;
  • 67) 0.311 024 558 899 2 × 2 = 0 + 0.622 049 117 798 4;
  • 68) 0.622 049 117 798 4 × 2 = 1 + 0.244 098 235 596 8;
  • 69) 0.244 098 235 596 8 × 2 = 0 + 0.488 196 471 193 6;
  • 70) 0.488 196 471 193 6 × 2 = 0 + 0.976 392 942 387 2;
  • 71) 0.976 392 942 387 2 × 2 = 1 + 0.952 785 884 774 4;
  • 72) 0.952 785 884 774 4 × 2 = 1 + 0.905 571 769 548 8;
  • 73) 0.905 571 769 548 8 × 2 = 1 + 0.811 143 539 097 6;
  • 74) 0.811 143 539 097 6 × 2 = 1 + 0.622 287 078 195 2;
  • 75) 0.622 287 078 195 2 × 2 = 1 + 0.244 574 156 390 4;
  • 76) 0.244 574 156 390 4 × 2 = 0 + 0.489 148 312 780 8;
  • 77) 0.489 148 312 780 8 × 2 = 0 + 0.978 296 625 561 6;
  • 78) 0.978 296 625 561 6 × 2 = 1 + 0.956 593 251 123 2;
  • 79) 0.956 593 251 123 2 × 2 = 1 + 0.913 186 502 246 4;
  • 80) 0.913 186 502 246 4 × 2 = 1 + 0.826 373 004 492 8;
  • 81) 0.826 373 004 492 8 × 2 = 1 + 0.652 746 008 985 6;
  • 82) 0.652 746 008 985 6 × 2 = 1 + 0.305 492 017 971 2;
  • 83) 0.305 492 017 971 2 × 2 = 0 + 0.610 984 035 942 4;
  • 84) 0.610 984 035 942 4 × 2 = 1 + 0.221 968 071 884 8;
  • 85) 0.221 968 071 884 8 × 2 = 0 + 0.443 936 143 769 6;
  • 86) 0.443 936 143 769 6 × 2 = 0 + 0.887 872 287 539 2;
  • 87) 0.887 872 287 539 2 × 2 = 1 + 0.775 744 575 078 4;
  • 88) 0.775 744 575 078 4 × 2 = 1 + 0.551 489 150 156 8;

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:


0.000 000 000 015 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 0011(2)

5. Positive number before normalization:

0.000 000 000 015 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 0011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 36 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 015 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 0011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 0011(2) × 20 =

1.0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 0011(2) × 2-36


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): -36

Mantissa (not normalized):
1.0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-36 + 2(11-1) - 1 =

(-36 + 1 023)(10) =

987(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;
  • 987 ÷ 2 = 493 + 1;
  • 493 ÷ 2 = 246 + 1;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 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) =

987(10) =

011 1101 1011(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 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 0011 =

0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 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 1101 1011

Mantissa (52 bits) =
0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 0011


Decimal number 0.000 000 000 015 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 1011 - 0000 1101 0010 1001 0001 0011 0000 0101 0011 1110 0111 1101 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