0.000 000 000 013 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 013 2(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
0.000 000 000 013 2(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: 0.000 000 000 013 2.

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) 0.000 000 000 013 2 × 2 = 0 + 0.000 000 000 026 4;
  • 2) 0.000 000 000 026 4 × 2 = 0 + 0.000 000 000 052 8;
  • 3) 0.000 000 000 052 8 × 2 = 0 + 0.000 000 000 105 6;
  • 4) 0.000 000 000 105 6 × 2 = 0 + 0.000 000 000 211 2;
  • 5) 0.000 000 000 211 2 × 2 = 0 + 0.000 000 000 422 4;
  • 6) 0.000 000 000 422 4 × 2 = 0 + 0.000 000 000 844 8;
  • 7) 0.000 000 000 844 8 × 2 = 0 + 0.000 000 001 689 6;
  • 8) 0.000 000 001 689 6 × 2 = 0 + 0.000 000 003 379 2;
  • 9) 0.000 000 003 379 2 × 2 = 0 + 0.000 000 006 758 4;
  • 10) 0.000 000 006 758 4 × 2 = 0 + 0.000 000 013 516 8;
  • 11) 0.000 000 013 516 8 × 2 = 0 + 0.000 000 027 033 6;
  • 12) 0.000 000 027 033 6 × 2 = 0 + 0.000 000 054 067 2;
  • 13) 0.000 000 054 067 2 × 2 = 0 + 0.000 000 108 134 4;
  • 14) 0.000 000 108 134 4 × 2 = 0 + 0.000 000 216 268 8;
  • 15) 0.000 000 216 268 8 × 2 = 0 + 0.000 000 432 537 6;
  • 16) 0.000 000 432 537 6 × 2 = 0 + 0.000 000 865 075 2;
  • 17) 0.000 000 865 075 2 × 2 = 0 + 0.000 001 730 150 4;
  • 18) 0.000 001 730 150 4 × 2 = 0 + 0.000 003 460 300 8;
  • 19) 0.000 003 460 300 8 × 2 = 0 + 0.000 006 920 601 6;
  • 20) 0.000 006 920 601 6 × 2 = 0 + 0.000 013 841 203 2;
  • 21) 0.000 013 841 203 2 × 2 = 0 + 0.000 027 682 406 4;
  • 22) 0.000 027 682 406 4 × 2 = 0 + 0.000 055 364 812 8;
  • 23) 0.000 055 364 812 8 × 2 = 0 + 0.000 110 729 625 6;
  • 24) 0.000 110 729 625 6 × 2 = 0 + 0.000 221 459 251 2;
  • 25) 0.000 221 459 251 2 × 2 = 0 + 0.000 442 918 502 4;
  • 26) 0.000 442 918 502 4 × 2 = 0 + 0.000 885 837 004 8;
  • 27) 0.000 885 837 004 8 × 2 = 0 + 0.001 771 674 009 6;
  • 28) 0.001 771 674 009 6 × 2 = 0 + 0.003 543 348 019 2;
  • 29) 0.003 543 348 019 2 × 2 = 0 + 0.007 086 696 038 4;
  • 30) 0.007 086 696 038 4 × 2 = 0 + 0.014 173 392 076 8;
  • 31) 0.014 173 392 076 8 × 2 = 0 + 0.028 346 784 153 6;
  • 32) 0.028 346 784 153 6 × 2 = 0 + 0.056 693 568 307 2;
  • 33) 0.056 693 568 307 2 × 2 = 0 + 0.113 387 136 614 4;
  • 34) 0.113 387 136 614 4 × 2 = 0 + 0.226 774 273 228 8;
  • 35) 0.226 774 273 228 8 × 2 = 0 + 0.453 548 546 457 6;
  • 36) 0.453 548 546 457 6 × 2 = 0 + 0.907 097 092 915 2;
  • 37) 0.907 097 092 915 2 × 2 = 1 + 0.814 194 185 830 4;
  • 38) 0.814 194 185 830 4 × 2 = 1 + 0.628 388 371 660 8;
  • 39) 0.628 388 371 660 8 × 2 = 1 + 0.256 776 743 321 6;
  • 40) 0.256 776 743 321 6 × 2 = 0 + 0.513 553 486 643 2;
  • 41) 0.513 553 486 643 2 × 2 = 1 + 0.027 106 973 286 4;
  • 42) 0.027 106 973 286 4 × 2 = 0 + 0.054 213 946 572 8;
  • 43) 0.054 213 946 572 8 × 2 = 0 + 0.108 427 893 145 6;
  • 44) 0.108 427 893 145 6 × 2 = 0 + 0.216 855 786 291 2;
  • 45) 0.216 855 786 291 2 × 2 = 0 + 0.433 711 572 582 4;
  • 46) 0.433 711 572 582 4 × 2 = 0 + 0.867 423 145 164 8;
  • 47) 0.867 423 145 164 8 × 2 = 1 + 0.734 846 290 329 6;
  • 48) 0.734 846 290 329 6 × 2 = 1 + 0.469 692 580 659 2;
  • 49) 0.469 692 580 659 2 × 2 = 0 + 0.939 385 161 318 4;
  • 50) 0.939 385 161 318 4 × 2 = 1 + 0.878 770 322 636 8;
  • 51) 0.878 770 322 636 8 × 2 = 1 + 0.757 540 645 273 6;
  • 52) 0.757 540 645 273 6 × 2 = 1 + 0.515 081 290 547 2;
  • 53) 0.515 081 290 547 2 × 2 = 1 + 0.030 162 581 094 4;
  • 54) 0.030 162 581 094 4 × 2 = 0 + 0.060 325 162 188 8;
  • 55) 0.060 325 162 188 8 × 2 = 0 + 0.120 650 324 377 6;
  • 56) 0.120 650 324 377 6 × 2 = 0 + 0.241 300 648 755 2;
  • 57) 0.241 300 648 755 2 × 2 = 0 + 0.482 601 297 510 4;
  • 58) 0.482 601 297 510 4 × 2 = 0 + 0.965 202 595 020 8;
  • 59) 0.965 202 595 020 8 × 2 = 1 + 0.930 405 190 041 6;
  • 60) 0.930 405 190 041 6 × 2 = 1 + 0.860 810 380 083 2;
  • 61) 0.860 810 380 083 2 × 2 = 1 + 0.721 620 760 166 4;
  • 62) 0.721 620 760 166 4 × 2 = 1 + 0.443 241 520 332 8;
  • 63) 0.443 241 520 332 8 × 2 = 0 + 0.886 483 040 665 6;
  • 64) 0.886 483 040 665 6 × 2 = 1 + 0.772 966 081 331 2;
  • 65) 0.772 966 081 331 2 × 2 = 1 + 0.545 932 162 662 4;
  • 66) 0.545 932 162 662 4 × 2 = 1 + 0.091 864 325 324 8;
  • 67) 0.091 864 325 324 8 × 2 = 0 + 0.183 728 650 649 6;
  • 68) 0.183 728 650 649 6 × 2 = 0 + 0.367 457 301 299 2;
  • 69) 0.367 457 301 299 2 × 2 = 0 + 0.734 914 602 598 4;
  • 70) 0.734 914 602 598 4 × 2 = 1 + 0.469 829 205 196 8;
  • 71) 0.469 829 205 196 8 × 2 = 0 + 0.939 658 410 393 6;
  • 72) 0.939 658 410 393 6 × 2 = 1 + 0.879 316 820 787 2;
  • 73) 0.879 316 820 787 2 × 2 = 1 + 0.758 633 641 574 4;
  • 74) 0.758 633 641 574 4 × 2 = 1 + 0.517 267 283 148 8;
  • 75) 0.517 267 283 148 8 × 2 = 1 + 0.034 534 566 297 6;
  • 76) 0.034 534 566 297 6 × 2 = 0 + 0.069 069 132 595 2;
  • 77) 0.069 069 132 595 2 × 2 = 0 + 0.138 138 265 190 4;
  • 78) 0.138 138 265 190 4 × 2 = 0 + 0.276 276 530 380 8;
  • 79) 0.276 276 530 380 8 × 2 = 0 + 0.552 553 060 761 6;
  • 80) 0.552 553 060 761 6 × 2 = 1 + 0.105 106 121 523 2;
  • 81) 0.105 106 121 523 2 × 2 = 0 + 0.210 212 243 046 4;
  • 82) 0.210 212 243 046 4 × 2 = 0 + 0.420 424 486 092 8;
  • 83) 0.420 424 486 092 8 × 2 = 0 + 0.840 848 972 185 6;
  • 84) 0.840 848 972 185 6 × 2 = 1 + 0.681 697 944 371 2;
  • 85) 0.681 697 944 371 2 × 2 = 1 + 0.363 395 888 742 4;
  • 86) 0.363 395 888 742 4 × 2 = 0 + 0.726 791 777 484 8;
  • 87) 0.726 791 777 484 8 × 2 = 1 + 0.453 583 554 969 6;
  • 88) 0.453 583 554 969 6 × 2 = 0 + 0.907 167 109 939 2;
  • 89) 0.907 167 109 939 2 × 2 = 1 + 0.814 334 219 878 4;

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:


0.000 000 000 013 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1110 1000 0011 0111 1000 0011 1101 1100 0101 1110 0001 0001 1010 1(2)

5. Positive number before normalization:

0.000 000 000 013 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1110 1000 0011 0111 1000 0011 1101 1100 0101 1110 0001 0001 1010 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 013 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1110 1000 0011 0111 1000 0011 1101 1100 0101 1110 0001 0001 1010 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1110 1000 0011 0111 1000 0011 1101 1100 0101 1110 0001 0001 1010 1(2) × 20 =

1.1101 0000 0110 1111 0000 0111 1011 1000 1011 1100 0010 0011 0101(2) × 2-37


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): -37

Mantissa (not normalized):
1.1101 0000 0110 1111 0000 0111 1011 1000 1011 1100 0010 0011 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-37 + 2(11-1) - 1 =

(-37 + 1 023)(10) =

986(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;
  • 986 ÷ 2 = 493 + 0;
  • 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) =

986(10) =

011 1101 1010(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. 1101 0000 0110 1111 0000 0111 1011 1000 1011 1100 0010 0011 0101 =

1101 0000 0110 1111 0000 0111 1011 1000 1011 1100 0010 0011 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 1101 1010

Mantissa (52 bits) =
1101 0000 0110 1111 0000 0111 1011 1000 1011 1100 0010 0011 0101


Decimal number 0.000 000 000 013 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 1010 - 1101 0000 0110 1111 0000 0111 1011 1000 1011 1100 0010 0011 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