-0.016 738 891 601 562 497 042 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 042(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
-0.016 738 891 601 562 497 042(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:

|-0.016 738 891 601 562 497 042| = 0.016 738 891 601 562 497 042


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: 0.016 738 891 601 562 497 042.

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) 0.016 738 891 601 562 497 042 × 2 = 0 + 0.033 477 783 203 124 994 084;
  • 2) 0.033 477 783 203 124 994 084 × 2 = 0 + 0.066 955 566 406 249 988 168;
  • 3) 0.066 955 566 406 249 988 168 × 2 = 0 + 0.133 911 132 812 499 976 336;
  • 4) 0.133 911 132 812 499 976 336 × 2 = 0 + 0.267 822 265 624 999 952 672;
  • 5) 0.267 822 265 624 999 952 672 × 2 = 0 + 0.535 644 531 249 999 905 344;
  • 6) 0.535 644 531 249 999 905 344 × 2 = 1 + 0.071 289 062 499 999 810 688;
  • 7) 0.071 289 062 499 999 810 688 × 2 = 0 + 0.142 578 124 999 999 621 376;
  • 8) 0.142 578 124 999 999 621 376 × 2 = 0 + 0.285 156 249 999 999 242 752;
  • 9) 0.285 156 249 999 999 242 752 × 2 = 0 + 0.570 312 499 999 998 485 504;
  • 10) 0.570 312 499 999 998 485 504 × 2 = 1 + 0.140 624 999 999 996 971 008;
  • 11) 0.140 624 999 999 996 971 008 × 2 = 0 + 0.281 249 999 999 993 942 016;
  • 12) 0.281 249 999 999 993 942 016 × 2 = 0 + 0.562 499 999 999 987 884 032;
  • 13) 0.562 499 999 999 987 884 032 × 2 = 1 + 0.124 999 999 999 975 768 064;
  • 14) 0.124 999 999 999 975 768 064 × 2 = 0 + 0.249 999 999 999 951 536 128;
  • 15) 0.249 999 999 999 951 536 128 × 2 = 0 + 0.499 999 999 999 903 072 256;
  • 16) 0.499 999 999 999 903 072 256 × 2 = 0 + 0.999 999 999 999 806 144 512;
  • 17) 0.999 999 999 999 806 144 512 × 2 = 1 + 0.999 999 999 999 612 289 024;
  • 18) 0.999 999 999 999 612 289 024 × 2 = 1 + 0.999 999 999 999 224 578 048;
  • 19) 0.999 999 999 999 224 578 048 × 2 = 1 + 0.999 999 999 998 449 156 096;
  • 20) 0.999 999 999 998 449 156 096 × 2 = 1 + 0.999 999 999 996 898 312 192;
  • 21) 0.999 999 999 996 898 312 192 × 2 = 1 + 0.999 999 999 993 796 624 384;
  • 22) 0.999 999 999 993 796 624 384 × 2 = 1 + 0.999 999 999 987 593 248 768;
  • 23) 0.999 999 999 987 593 248 768 × 2 = 1 + 0.999 999 999 975 186 497 536;
  • 24) 0.999 999 999 975 186 497 536 × 2 = 1 + 0.999 999 999 950 372 995 072;
  • 25) 0.999 999 999 950 372 995 072 × 2 = 1 + 0.999 999 999 900 745 990 144;
  • 26) 0.999 999 999 900 745 990 144 × 2 = 1 + 0.999 999 999 801 491 980 288;
  • 27) 0.999 999 999 801 491 980 288 × 2 = 1 + 0.999 999 999 602 983 960 576;
  • 28) 0.999 999 999 602 983 960 576 × 2 = 1 + 0.999 999 999 205 967 921 152;
  • 29) 0.999 999 999 205 967 921 152 × 2 = 1 + 0.999 999 998 411 935 842 304;
  • 30) 0.999 999 998 411 935 842 304 × 2 = 1 + 0.999 999 996 823 871 684 608;
  • 31) 0.999 999 996 823 871 684 608 × 2 = 1 + 0.999 999 993 647 743 369 216;
  • 32) 0.999 999 993 647 743 369 216 × 2 = 1 + 0.999 999 987 295 486 738 432;
  • 33) 0.999 999 987 295 486 738 432 × 2 = 1 + 0.999 999 974 590 973 476 864;
  • 34) 0.999 999 974 590 973 476 864 × 2 = 1 + 0.999 999 949 181 946 953 728;
  • 35) 0.999 999 949 181 946 953 728 × 2 = 1 + 0.999 999 898 363 893 907 456;
  • 36) 0.999 999 898 363 893 907 456 × 2 = 1 + 0.999 999 796 727 787 814 912;
  • 37) 0.999 999 796 727 787 814 912 × 2 = 1 + 0.999 999 593 455 575 629 824;
  • 38) 0.999 999 593 455 575 629 824 × 2 = 1 + 0.999 999 186 911 151 259 648;
  • 39) 0.999 999 186 911 151 259 648 × 2 = 1 + 0.999 998 373 822 302 519 296;
  • 40) 0.999 998 373 822 302 519 296 × 2 = 1 + 0.999 996 747 644 605 038 592;
  • 41) 0.999 996 747 644 605 038 592 × 2 = 1 + 0.999 993 495 289 210 077 184;
  • 42) 0.999 993 495 289 210 077 184 × 2 = 1 + 0.999 986 990 578 420 154 368;
  • 43) 0.999 986 990 578 420 154 368 × 2 = 1 + 0.999 973 981 156 840 308 736;
  • 44) 0.999 973 981 156 840 308 736 × 2 = 1 + 0.999 947 962 313 680 617 472;
  • 45) 0.999 947 962 313 680 617 472 × 2 = 1 + 0.999 895 924 627 361 234 944;
  • 46) 0.999 895 924 627 361 234 944 × 2 = 1 + 0.999 791 849 254 722 469 888;
  • 47) 0.999 791 849 254 722 469 888 × 2 = 1 + 0.999 583 698 509 444 939 776;
  • 48) 0.999 583 698 509 444 939 776 × 2 = 1 + 0.999 167 397 018 889 879 552;
  • 49) 0.999 167 397 018 889 879 552 × 2 = 1 + 0.998 334 794 037 779 759 104;
  • 50) 0.998 334 794 037 779 759 104 × 2 = 1 + 0.996 669 588 075 559 518 208;
  • 51) 0.996 669 588 075 559 518 208 × 2 = 1 + 0.993 339 176 151 119 036 416;
  • 52) 0.993 339 176 151 119 036 416 × 2 = 1 + 0.986 678 352 302 238 072 832;
  • 53) 0.986 678 352 302 238 072 832 × 2 = 1 + 0.973 356 704 604 476 145 664;
  • 54) 0.973 356 704 604 476 145 664 × 2 = 1 + 0.946 713 409 208 952 291 328;
  • 55) 0.946 713 409 208 952 291 328 × 2 = 1 + 0.893 426 818 417 904 582 656;
  • 56) 0.893 426 818 417 904 582 656 × 2 = 1 + 0.786 853 636 835 809 165 312;
  • 57) 0.786 853 636 835 809 165 312 × 2 = 1 + 0.573 707 273 671 618 330 624;
  • 58) 0.573 707 273 671 618 330 624 × 2 = 1 + 0.147 414 547 343 236 661 248;

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:


0.016 738 891 601 562 497 042(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 497 042(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:


0.016 738 891 601 562 497 042(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 -0.016 738 891 601 562 497 042 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