0.000 000 000 016 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 016 4(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
0.000 000 000 016 4(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: 0.000 000 000 016 4.

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) 0.000 000 000 016 4 × 2 = 0 + 0.000 000 000 032 8;
  • 2) 0.000 000 000 032 8 × 2 = 0 + 0.000 000 000 065 6;
  • 3) 0.000 000 000 065 6 × 2 = 0 + 0.000 000 000 131 2;
  • 4) 0.000 000 000 131 2 × 2 = 0 + 0.000 000 000 262 4;
  • 5) 0.000 000 000 262 4 × 2 = 0 + 0.000 000 000 524 8;
  • 6) 0.000 000 000 524 8 × 2 = 0 + 0.000 000 001 049 6;
  • 7) 0.000 000 001 049 6 × 2 = 0 + 0.000 000 002 099 2;
  • 8) 0.000 000 002 099 2 × 2 = 0 + 0.000 000 004 198 4;
  • 9) 0.000 000 004 198 4 × 2 = 0 + 0.000 000 008 396 8;
  • 10) 0.000 000 008 396 8 × 2 = 0 + 0.000 000 016 793 6;
  • 11) 0.000 000 016 793 6 × 2 = 0 + 0.000 000 033 587 2;
  • 12) 0.000 000 033 587 2 × 2 = 0 + 0.000 000 067 174 4;
  • 13) 0.000 000 067 174 4 × 2 = 0 + 0.000 000 134 348 8;
  • 14) 0.000 000 134 348 8 × 2 = 0 + 0.000 000 268 697 6;
  • 15) 0.000 000 268 697 6 × 2 = 0 + 0.000 000 537 395 2;
  • 16) 0.000 000 537 395 2 × 2 = 0 + 0.000 001 074 790 4;
  • 17) 0.000 001 074 790 4 × 2 = 0 + 0.000 002 149 580 8;
  • 18) 0.000 002 149 580 8 × 2 = 0 + 0.000 004 299 161 6;
  • 19) 0.000 004 299 161 6 × 2 = 0 + 0.000 008 598 323 2;
  • 20) 0.000 008 598 323 2 × 2 = 0 + 0.000 017 196 646 4;
  • 21) 0.000 017 196 646 4 × 2 = 0 + 0.000 034 393 292 8;
  • 22) 0.000 034 393 292 8 × 2 = 0 + 0.000 068 786 585 6;
  • 23) 0.000 068 786 585 6 × 2 = 0 + 0.000 137 573 171 2;
  • 24) 0.000 137 573 171 2 × 2 = 0 + 0.000 275 146 342 4;
  • 25) 0.000 275 146 342 4 × 2 = 0 + 0.000 550 292 684 8;
  • 26) 0.000 550 292 684 8 × 2 = 0 + 0.001 100 585 369 6;
  • 27) 0.001 100 585 369 6 × 2 = 0 + 0.002 201 170 739 2;
  • 28) 0.002 201 170 739 2 × 2 = 0 + 0.004 402 341 478 4;
  • 29) 0.004 402 341 478 4 × 2 = 0 + 0.008 804 682 956 8;
  • 30) 0.008 804 682 956 8 × 2 = 0 + 0.017 609 365 913 6;
  • 31) 0.017 609 365 913 6 × 2 = 0 + 0.035 218 731 827 2;
  • 32) 0.035 218 731 827 2 × 2 = 0 + 0.070 437 463 654 4;
  • 33) 0.070 437 463 654 4 × 2 = 0 + 0.140 874 927 308 8;
  • 34) 0.140 874 927 308 8 × 2 = 0 + 0.281 749 854 617 6;
  • 35) 0.281 749 854 617 6 × 2 = 0 + 0.563 499 709 235 2;
  • 36) 0.563 499 709 235 2 × 2 = 1 + 0.126 999 418 470 4;
  • 37) 0.126 999 418 470 4 × 2 = 0 + 0.253 998 836 940 8;
  • 38) 0.253 998 836 940 8 × 2 = 0 + 0.507 997 673 881 6;
  • 39) 0.507 997 673 881 6 × 2 = 1 + 0.015 995 347 763 2;
  • 40) 0.015 995 347 763 2 × 2 = 0 + 0.031 990 695 526 4;
  • 41) 0.031 990 695 526 4 × 2 = 0 + 0.063 981 391 052 8;
  • 42) 0.063 981 391 052 8 × 2 = 0 + 0.127 962 782 105 6;
  • 43) 0.127 962 782 105 6 × 2 = 0 + 0.255 925 564 211 2;
  • 44) 0.255 925 564 211 2 × 2 = 0 + 0.511 851 128 422 4;
  • 45) 0.511 851 128 422 4 × 2 = 1 + 0.023 702 256 844 8;
  • 46) 0.023 702 256 844 8 × 2 = 0 + 0.047 404 513 689 6;
  • 47) 0.047 404 513 689 6 × 2 = 0 + 0.094 809 027 379 2;
  • 48) 0.094 809 027 379 2 × 2 = 0 + 0.189 618 054 758 4;
  • 49) 0.189 618 054 758 4 × 2 = 0 + 0.379 236 109 516 8;
  • 50) 0.379 236 109 516 8 × 2 = 0 + 0.758 472 219 033 6;
  • 51) 0.758 472 219 033 6 × 2 = 1 + 0.516 944 438 067 2;
  • 52) 0.516 944 438 067 2 × 2 = 1 + 0.033 888 876 134 4;
  • 53) 0.033 888 876 134 4 × 2 = 0 + 0.067 777 752 268 8;
  • 54) 0.067 777 752 268 8 × 2 = 0 + 0.135 555 504 537 6;
  • 55) 0.135 555 504 537 6 × 2 = 0 + 0.271 111 009 075 2;
  • 56) 0.271 111 009 075 2 × 2 = 0 + 0.542 222 018 150 4;
  • 57) 0.542 222 018 150 4 × 2 = 1 + 0.084 444 036 300 8;
  • 58) 0.084 444 036 300 8 × 2 = 0 + 0.168 888 072 601 6;
  • 59) 0.168 888 072 601 6 × 2 = 0 + 0.337 776 145 203 2;
  • 60) 0.337 776 145 203 2 × 2 = 0 + 0.675 552 290 406 4;
  • 61) 0.675 552 290 406 4 × 2 = 1 + 0.351 104 580 812 8;
  • 62) 0.351 104 580 812 8 × 2 = 0 + 0.702 209 161 625 6;
  • 63) 0.702 209 161 625 6 × 2 = 1 + 0.404 418 323 251 2;
  • 64) 0.404 418 323 251 2 × 2 = 0 + 0.808 836 646 502 4;
  • 65) 0.808 836 646 502 4 × 2 = 1 + 0.617 673 293 004 8;
  • 66) 0.617 673 293 004 8 × 2 = 1 + 0.235 346 586 009 6;
  • 67) 0.235 346 586 009 6 × 2 = 0 + 0.470 693 172 019 2;
  • 68) 0.470 693 172 019 2 × 2 = 0 + 0.941 386 344 038 4;
  • 69) 0.941 386 344 038 4 × 2 = 1 + 0.882 772 688 076 8;
  • 70) 0.882 772 688 076 8 × 2 = 1 + 0.765 545 376 153 6;
  • 71) 0.765 545 376 153 6 × 2 = 1 + 0.531 090 752 307 2;
  • 72) 0.531 090 752 307 2 × 2 = 1 + 0.062 181 504 614 4;
  • 73) 0.062 181 504 614 4 × 2 = 0 + 0.124 363 009 228 8;
  • 74) 0.124 363 009 228 8 × 2 = 0 + 0.248 726 018 457 6;
  • 75) 0.248 726 018 457 6 × 2 = 0 + 0.497 452 036 915 2;
  • 76) 0.497 452 036 915 2 × 2 = 0 + 0.994 904 073 830 4;
  • 77) 0.994 904 073 830 4 × 2 = 1 + 0.989 808 147 660 8;
  • 78) 0.989 808 147 660 8 × 2 = 1 + 0.979 616 295 321 6;
  • 79) 0.979 616 295 321 6 × 2 = 1 + 0.959 232 590 643 2;
  • 80) 0.959 232 590 643 2 × 2 = 1 + 0.918 465 181 286 4;
  • 81) 0.918 465 181 286 4 × 2 = 1 + 0.836 930 362 572 8;
  • 82) 0.836 930 362 572 8 × 2 = 1 + 0.673 860 725 145 6;
  • 83) 0.673 860 725 145 6 × 2 = 1 + 0.347 721 450 291 2;
  • 84) 0.347 721 450 291 2 × 2 = 0 + 0.695 442 900 582 4;
  • 85) 0.695 442 900 582 4 × 2 = 1 + 0.390 885 801 164 8;
  • 86) 0.390 885 801 164 8 × 2 = 0 + 0.781 771 602 329 6;
  • 87) 0.781 771 602 329 6 × 2 = 1 + 0.563 543 204 659 2;
  • 88) 0.563 543 204 659 2 × 2 = 1 + 0.127 086 409 318 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:


0.000 000 000 016 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011(2)

5. Positive number before normalization:

0.000 000 000 016 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011(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:


0.000 000 000 016 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011(2) × 20 =

1.0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011(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):
1.0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011


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. 0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011 =

0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011


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) =
0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011


Decimal number 0.000 000 000 016 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 1011 - 0010 0000 1000 0011 0000 1000 1010 1100 1111 0000 1111 1110 1011

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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