-0.016 738 891 601 562 496 811 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 811(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
-0.016 738 891 601 562 496 811(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:

|-0.016 738 891 601 562 496 811| = 0.016 738 891 601 562 496 811


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: 0.016 738 891 601 562 496 811.

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) 0.016 738 891 601 562 496 811 × 2 = 0 + 0.033 477 783 203 124 993 622;
  • 2) 0.033 477 783 203 124 993 622 × 2 = 0 + 0.066 955 566 406 249 987 244;
  • 3) 0.066 955 566 406 249 987 244 × 2 = 0 + 0.133 911 132 812 499 974 488;
  • 4) 0.133 911 132 812 499 974 488 × 2 = 0 + 0.267 822 265 624 999 948 976;
  • 5) 0.267 822 265 624 999 948 976 × 2 = 0 + 0.535 644 531 249 999 897 952;
  • 6) 0.535 644 531 249 999 897 952 × 2 = 1 + 0.071 289 062 499 999 795 904;
  • 7) 0.071 289 062 499 999 795 904 × 2 = 0 + 0.142 578 124 999 999 591 808;
  • 8) 0.142 578 124 999 999 591 808 × 2 = 0 + 0.285 156 249 999 999 183 616;
  • 9) 0.285 156 249 999 999 183 616 × 2 = 0 + 0.570 312 499 999 998 367 232;
  • 10) 0.570 312 499 999 998 367 232 × 2 = 1 + 0.140 624 999 999 996 734 464;
  • 11) 0.140 624 999 999 996 734 464 × 2 = 0 + 0.281 249 999 999 993 468 928;
  • 12) 0.281 249 999 999 993 468 928 × 2 = 0 + 0.562 499 999 999 986 937 856;
  • 13) 0.562 499 999 999 986 937 856 × 2 = 1 + 0.124 999 999 999 973 875 712;
  • 14) 0.124 999 999 999 973 875 712 × 2 = 0 + 0.249 999 999 999 947 751 424;
  • 15) 0.249 999 999 999 947 751 424 × 2 = 0 + 0.499 999 999 999 895 502 848;
  • 16) 0.499 999 999 999 895 502 848 × 2 = 0 + 0.999 999 999 999 791 005 696;
  • 17) 0.999 999 999 999 791 005 696 × 2 = 1 + 0.999 999 999 999 582 011 392;
  • 18) 0.999 999 999 999 582 011 392 × 2 = 1 + 0.999 999 999 999 164 022 784;
  • 19) 0.999 999 999 999 164 022 784 × 2 = 1 + 0.999 999 999 998 328 045 568;
  • 20) 0.999 999 999 998 328 045 568 × 2 = 1 + 0.999 999 999 996 656 091 136;
  • 21) 0.999 999 999 996 656 091 136 × 2 = 1 + 0.999 999 999 993 312 182 272;
  • 22) 0.999 999 999 993 312 182 272 × 2 = 1 + 0.999 999 999 986 624 364 544;
  • 23) 0.999 999 999 986 624 364 544 × 2 = 1 + 0.999 999 999 973 248 729 088;
  • 24) 0.999 999 999 973 248 729 088 × 2 = 1 + 0.999 999 999 946 497 458 176;
  • 25) 0.999 999 999 946 497 458 176 × 2 = 1 + 0.999 999 999 892 994 916 352;
  • 26) 0.999 999 999 892 994 916 352 × 2 = 1 + 0.999 999 999 785 989 832 704;
  • 27) 0.999 999 999 785 989 832 704 × 2 = 1 + 0.999 999 999 571 979 665 408;
  • 28) 0.999 999 999 571 979 665 408 × 2 = 1 + 0.999 999 999 143 959 330 816;
  • 29) 0.999 999 999 143 959 330 816 × 2 = 1 + 0.999 999 998 287 918 661 632;
  • 30) 0.999 999 998 287 918 661 632 × 2 = 1 + 0.999 999 996 575 837 323 264;
  • 31) 0.999 999 996 575 837 323 264 × 2 = 1 + 0.999 999 993 151 674 646 528;
  • 32) 0.999 999 993 151 674 646 528 × 2 = 1 + 0.999 999 986 303 349 293 056;
  • 33) 0.999 999 986 303 349 293 056 × 2 = 1 + 0.999 999 972 606 698 586 112;
  • 34) 0.999 999 972 606 698 586 112 × 2 = 1 + 0.999 999 945 213 397 172 224;
  • 35) 0.999 999 945 213 397 172 224 × 2 = 1 + 0.999 999 890 426 794 344 448;
  • 36) 0.999 999 890 426 794 344 448 × 2 = 1 + 0.999 999 780 853 588 688 896;
  • 37) 0.999 999 780 853 588 688 896 × 2 = 1 + 0.999 999 561 707 177 377 792;
  • 38) 0.999 999 561 707 177 377 792 × 2 = 1 + 0.999 999 123 414 354 755 584;
  • 39) 0.999 999 123 414 354 755 584 × 2 = 1 + 0.999 998 246 828 709 511 168;
  • 40) 0.999 998 246 828 709 511 168 × 2 = 1 + 0.999 996 493 657 419 022 336;
  • 41) 0.999 996 493 657 419 022 336 × 2 = 1 + 0.999 992 987 314 838 044 672;
  • 42) 0.999 992 987 314 838 044 672 × 2 = 1 + 0.999 985 974 629 676 089 344;
  • 43) 0.999 985 974 629 676 089 344 × 2 = 1 + 0.999 971 949 259 352 178 688;
  • 44) 0.999 971 949 259 352 178 688 × 2 = 1 + 0.999 943 898 518 704 357 376;
  • 45) 0.999 943 898 518 704 357 376 × 2 = 1 + 0.999 887 797 037 408 714 752;
  • 46) 0.999 887 797 037 408 714 752 × 2 = 1 + 0.999 775 594 074 817 429 504;
  • 47) 0.999 775 594 074 817 429 504 × 2 = 1 + 0.999 551 188 149 634 859 008;
  • 48) 0.999 551 188 149 634 859 008 × 2 = 1 + 0.999 102 376 299 269 718 016;
  • 49) 0.999 102 376 299 269 718 016 × 2 = 1 + 0.998 204 752 598 539 436 032;
  • 50) 0.998 204 752 598 539 436 032 × 2 = 1 + 0.996 409 505 197 078 872 064;
  • 51) 0.996 409 505 197 078 872 064 × 2 = 1 + 0.992 819 010 394 157 744 128;
  • 52) 0.992 819 010 394 157 744 128 × 2 = 1 + 0.985 638 020 788 315 488 256;
  • 53) 0.985 638 020 788 315 488 256 × 2 = 1 + 0.971 276 041 576 630 976 512;
  • 54) 0.971 276 041 576 630 976 512 × 2 = 1 + 0.942 552 083 153 261 953 024;
  • 55) 0.942 552 083 153 261 953 024 × 2 = 1 + 0.885 104 166 306 523 906 048;
  • 56) 0.885 104 166 306 523 906 048 × 2 = 1 + 0.770 208 332 613 047 812 096;
  • 57) 0.770 208 332 613 047 812 096 × 2 = 1 + 0.540 416 665 226 095 624 192;
  • 58) 0.540 416 665 226 095 624 192 × 2 = 1 + 0.080 833 330 452 191 248 384;

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:


0.016 738 891 601 562 496 811(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 811(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:


0.016 738 891 601 562 496 811(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 -0.016 738 891 601 562 496 811 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