-0.016 738 891 601 562 496 539 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 539(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
-0.016 738 891 601 562 496 539(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:

|-0.016 738 891 601 562 496 539| = 0.016 738 891 601 562 496 539


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: 0.016 738 891 601 562 496 539.

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) 0.016 738 891 601 562 496 539 × 2 = 0 + 0.033 477 783 203 124 993 078;
  • 2) 0.033 477 783 203 124 993 078 × 2 = 0 + 0.066 955 566 406 249 986 156;
  • 3) 0.066 955 566 406 249 986 156 × 2 = 0 + 0.133 911 132 812 499 972 312;
  • 4) 0.133 911 132 812 499 972 312 × 2 = 0 + 0.267 822 265 624 999 944 624;
  • 5) 0.267 822 265 624 999 944 624 × 2 = 0 + 0.535 644 531 249 999 889 248;
  • 6) 0.535 644 531 249 999 889 248 × 2 = 1 + 0.071 289 062 499 999 778 496;
  • 7) 0.071 289 062 499 999 778 496 × 2 = 0 + 0.142 578 124 999 999 556 992;
  • 8) 0.142 578 124 999 999 556 992 × 2 = 0 + 0.285 156 249 999 999 113 984;
  • 9) 0.285 156 249 999 999 113 984 × 2 = 0 + 0.570 312 499 999 998 227 968;
  • 10) 0.570 312 499 999 998 227 968 × 2 = 1 + 0.140 624 999 999 996 455 936;
  • 11) 0.140 624 999 999 996 455 936 × 2 = 0 + 0.281 249 999 999 992 911 872;
  • 12) 0.281 249 999 999 992 911 872 × 2 = 0 + 0.562 499 999 999 985 823 744;
  • 13) 0.562 499 999 999 985 823 744 × 2 = 1 + 0.124 999 999 999 971 647 488;
  • 14) 0.124 999 999 999 971 647 488 × 2 = 0 + 0.249 999 999 999 943 294 976;
  • 15) 0.249 999 999 999 943 294 976 × 2 = 0 + 0.499 999 999 999 886 589 952;
  • 16) 0.499 999 999 999 886 589 952 × 2 = 0 + 0.999 999 999 999 773 179 904;
  • 17) 0.999 999 999 999 773 179 904 × 2 = 1 + 0.999 999 999 999 546 359 808;
  • 18) 0.999 999 999 999 546 359 808 × 2 = 1 + 0.999 999 999 999 092 719 616;
  • 19) 0.999 999 999 999 092 719 616 × 2 = 1 + 0.999 999 999 998 185 439 232;
  • 20) 0.999 999 999 998 185 439 232 × 2 = 1 + 0.999 999 999 996 370 878 464;
  • 21) 0.999 999 999 996 370 878 464 × 2 = 1 + 0.999 999 999 992 741 756 928;
  • 22) 0.999 999 999 992 741 756 928 × 2 = 1 + 0.999 999 999 985 483 513 856;
  • 23) 0.999 999 999 985 483 513 856 × 2 = 1 + 0.999 999 999 970 967 027 712;
  • 24) 0.999 999 999 970 967 027 712 × 2 = 1 + 0.999 999 999 941 934 055 424;
  • 25) 0.999 999 999 941 934 055 424 × 2 = 1 + 0.999 999 999 883 868 110 848;
  • 26) 0.999 999 999 883 868 110 848 × 2 = 1 + 0.999 999 999 767 736 221 696;
  • 27) 0.999 999 999 767 736 221 696 × 2 = 1 + 0.999 999 999 535 472 443 392;
  • 28) 0.999 999 999 535 472 443 392 × 2 = 1 + 0.999 999 999 070 944 886 784;
  • 29) 0.999 999 999 070 944 886 784 × 2 = 1 + 0.999 999 998 141 889 773 568;
  • 30) 0.999 999 998 141 889 773 568 × 2 = 1 + 0.999 999 996 283 779 547 136;
  • 31) 0.999 999 996 283 779 547 136 × 2 = 1 + 0.999 999 992 567 559 094 272;
  • 32) 0.999 999 992 567 559 094 272 × 2 = 1 + 0.999 999 985 135 118 188 544;
  • 33) 0.999 999 985 135 118 188 544 × 2 = 1 + 0.999 999 970 270 236 377 088;
  • 34) 0.999 999 970 270 236 377 088 × 2 = 1 + 0.999 999 940 540 472 754 176;
  • 35) 0.999 999 940 540 472 754 176 × 2 = 1 + 0.999 999 881 080 945 508 352;
  • 36) 0.999 999 881 080 945 508 352 × 2 = 1 + 0.999 999 762 161 891 016 704;
  • 37) 0.999 999 762 161 891 016 704 × 2 = 1 + 0.999 999 524 323 782 033 408;
  • 38) 0.999 999 524 323 782 033 408 × 2 = 1 + 0.999 999 048 647 564 066 816;
  • 39) 0.999 999 048 647 564 066 816 × 2 = 1 + 0.999 998 097 295 128 133 632;
  • 40) 0.999 998 097 295 128 133 632 × 2 = 1 + 0.999 996 194 590 256 267 264;
  • 41) 0.999 996 194 590 256 267 264 × 2 = 1 + 0.999 992 389 180 512 534 528;
  • 42) 0.999 992 389 180 512 534 528 × 2 = 1 + 0.999 984 778 361 025 069 056;
  • 43) 0.999 984 778 361 025 069 056 × 2 = 1 + 0.999 969 556 722 050 138 112;
  • 44) 0.999 969 556 722 050 138 112 × 2 = 1 + 0.999 939 113 444 100 276 224;
  • 45) 0.999 939 113 444 100 276 224 × 2 = 1 + 0.999 878 226 888 200 552 448;
  • 46) 0.999 878 226 888 200 552 448 × 2 = 1 + 0.999 756 453 776 401 104 896;
  • 47) 0.999 756 453 776 401 104 896 × 2 = 1 + 0.999 512 907 552 802 209 792;
  • 48) 0.999 512 907 552 802 209 792 × 2 = 1 + 0.999 025 815 105 604 419 584;
  • 49) 0.999 025 815 105 604 419 584 × 2 = 1 + 0.998 051 630 211 208 839 168;
  • 50) 0.998 051 630 211 208 839 168 × 2 = 1 + 0.996 103 260 422 417 678 336;
  • 51) 0.996 103 260 422 417 678 336 × 2 = 1 + 0.992 206 520 844 835 356 672;
  • 52) 0.992 206 520 844 835 356 672 × 2 = 1 + 0.984 413 041 689 670 713 344;
  • 53) 0.984 413 041 689 670 713 344 × 2 = 1 + 0.968 826 083 379 341 426 688;
  • 54) 0.968 826 083 379 341 426 688 × 2 = 1 + 0.937 652 166 758 682 853 376;
  • 55) 0.937 652 166 758 682 853 376 × 2 = 1 + 0.875 304 333 517 365 706 752;
  • 56) 0.875 304 333 517 365 706 752 × 2 = 1 + 0.750 608 667 034 731 413 504;
  • 57) 0.750 608 667 034 731 413 504 × 2 = 1 + 0.501 217 334 069 462 827 008;
  • 58) 0.501 217 334 069 462 827 008 × 2 = 1 + 0.002 434 668 138 925 654 016;

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:


0.016 738 891 601 562 496 539(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 539(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 539(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(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 1111


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 496 539 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 1111

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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