-0.016 738 891 601 562 496 367 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 367(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
-0.016 738 891 601 562 496 367(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. Start with the positive version of the number:

|-0.016 738 891 601 562 496 367| = 0.016 738 891 601 562 496 367


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 496 367.

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) 0.016 738 891 601 562 496 367 × 2 = 0 + 0.033 477 783 203 124 992 734;
  • 2) 0.033 477 783 203 124 992 734 × 2 = 0 + 0.066 955 566 406 249 985 468;
  • 3) 0.066 955 566 406 249 985 468 × 2 = 0 + 0.133 911 132 812 499 970 936;
  • 4) 0.133 911 132 812 499 970 936 × 2 = 0 + 0.267 822 265 624 999 941 872;
  • 5) 0.267 822 265 624 999 941 872 × 2 = 0 + 0.535 644 531 249 999 883 744;
  • 6) 0.535 644 531 249 999 883 744 × 2 = 1 + 0.071 289 062 499 999 767 488;
  • 7) 0.071 289 062 499 999 767 488 × 2 = 0 + 0.142 578 124 999 999 534 976;
  • 8) 0.142 578 124 999 999 534 976 × 2 = 0 + 0.285 156 249 999 999 069 952;
  • 9) 0.285 156 249 999 999 069 952 × 2 = 0 + 0.570 312 499 999 998 139 904;
  • 10) 0.570 312 499 999 998 139 904 × 2 = 1 + 0.140 624 999 999 996 279 808;
  • 11) 0.140 624 999 999 996 279 808 × 2 = 0 + 0.281 249 999 999 992 559 616;
  • 12) 0.281 249 999 999 992 559 616 × 2 = 0 + 0.562 499 999 999 985 119 232;
  • 13) 0.562 499 999 999 985 119 232 × 2 = 1 + 0.124 999 999 999 970 238 464;
  • 14) 0.124 999 999 999 970 238 464 × 2 = 0 + 0.249 999 999 999 940 476 928;
  • 15) 0.249 999 999 999 940 476 928 × 2 = 0 + 0.499 999 999 999 880 953 856;
  • 16) 0.499 999 999 999 880 953 856 × 2 = 0 + 0.999 999 999 999 761 907 712;
  • 17) 0.999 999 999 999 761 907 712 × 2 = 1 + 0.999 999 999 999 523 815 424;
  • 18) 0.999 999 999 999 523 815 424 × 2 = 1 + 0.999 999 999 999 047 630 848;
  • 19) 0.999 999 999 999 047 630 848 × 2 = 1 + 0.999 999 999 998 095 261 696;
  • 20) 0.999 999 999 998 095 261 696 × 2 = 1 + 0.999 999 999 996 190 523 392;
  • 21) 0.999 999 999 996 190 523 392 × 2 = 1 + 0.999 999 999 992 381 046 784;
  • 22) 0.999 999 999 992 381 046 784 × 2 = 1 + 0.999 999 999 984 762 093 568;
  • 23) 0.999 999 999 984 762 093 568 × 2 = 1 + 0.999 999 999 969 524 187 136;
  • 24) 0.999 999 999 969 524 187 136 × 2 = 1 + 0.999 999 999 939 048 374 272;
  • 25) 0.999 999 999 939 048 374 272 × 2 = 1 + 0.999 999 999 878 096 748 544;
  • 26) 0.999 999 999 878 096 748 544 × 2 = 1 + 0.999 999 999 756 193 497 088;
  • 27) 0.999 999 999 756 193 497 088 × 2 = 1 + 0.999 999 999 512 386 994 176;
  • 28) 0.999 999 999 512 386 994 176 × 2 = 1 + 0.999 999 999 024 773 988 352;
  • 29) 0.999 999 999 024 773 988 352 × 2 = 1 + 0.999 999 998 049 547 976 704;
  • 30) 0.999 999 998 049 547 976 704 × 2 = 1 + 0.999 999 996 099 095 953 408;
  • 31) 0.999 999 996 099 095 953 408 × 2 = 1 + 0.999 999 992 198 191 906 816;
  • 32) 0.999 999 992 198 191 906 816 × 2 = 1 + 0.999 999 984 396 383 813 632;
  • 33) 0.999 999 984 396 383 813 632 × 2 = 1 + 0.999 999 968 792 767 627 264;
  • 34) 0.999 999 968 792 767 627 264 × 2 = 1 + 0.999 999 937 585 535 254 528;
  • 35) 0.999 999 937 585 535 254 528 × 2 = 1 + 0.999 999 875 171 070 509 056;
  • 36) 0.999 999 875 171 070 509 056 × 2 = 1 + 0.999 999 750 342 141 018 112;
  • 37) 0.999 999 750 342 141 018 112 × 2 = 1 + 0.999 999 500 684 282 036 224;
  • 38) 0.999 999 500 684 282 036 224 × 2 = 1 + 0.999 999 001 368 564 072 448;
  • 39) 0.999 999 001 368 564 072 448 × 2 = 1 + 0.999 998 002 737 128 144 896;
  • 40) 0.999 998 002 737 128 144 896 × 2 = 1 + 0.999 996 005 474 256 289 792;
  • 41) 0.999 996 005 474 256 289 792 × 2 = 1 + 0.999 992 010 948 512 579 584;
  • 42) 0.999 992 010 948 512 579 584 × 2 = 1 + 0.999 984 021 897 025 159 168;
  • 43) 0.999 984 021 897 025 159 168 × 2 = 1 + 0.999 968 043 794 050 318 336;
  • 44) 0.999 968 043 794 050 318 336 × 2 = 1 + 0.999 936 087 588 100 636 672;
  • 45) 0.999 936 087 588 100 636 672 × 2 = 1 + 0.999 872 175 176 201 273 344;
  • 46) 0.999 872 175 176 201 273 344 × 2 = 1 + 0.999 744 350 352 402 546 688;
  • 47) 0.999 744 350 352 402 546 688 × 2 = 1 + 0.999 488 700 704 805 093 376;
  • 48) 0.999 488 700 704 805 093 376 × 2 = 1 + 0.998 977 401 409 610 186 752;
  • 49) 0.998 977 401 409 610 186 752 × 2 = 1 + 0.997 954 802 819 220 373 504;
  • 50) 0.997 954 802 819 220 373 504 × 2 = 1 + 0.995 909 605 638 440 747 008;
  • 51) 0.995 909 605 638 440 747 008 × 2 = 1 + 0.991 819 211 276 881 494 016;
  • 52) 0.991 819 211 276 881 494 016 × 2 = 1 + 0.983 638 422 553 762 988 032;
  • 53) 0.983 638 422 553 762 988 032 × 2 = 1 + 0.967 276 845 107 525 976 064;
  • 54) 0.967 276 845 107 525 976 064 × 2 = 1 + 0.934 553 690 215 051 952 128;
  • 55) 0.934 553 690 215 051 952 128 × 2 = 1 + 0.869 107 380 430 103 904 256;
  • 56) 0.869 107 380 430 103 904 256 × 2 = 1 + 0.738 214 760 860 207 808 512;
  • 57) 0.738 214 760 860 207 808 512 × 2 = 1 + 0.476 429 521 720 415 617 024;
  • 58) 0.476 429 521 720 415 617 024 × 2 = 0 + 0.952 859 043 440 831 234 048;

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).


5. 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:


0.016 738 891 601 562 496 367(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 367(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 367(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


10. 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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


12. 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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 367 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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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