-0.016 738 891 601 562 497 09 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 09(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 09(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 09| = 0.016 738 891 601 562 497 09


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 09.

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 09 × 2 = 0 + 0.033 477 783 203 124 994 18;
  • 2) 0.033 477 783 203 124 994 18 × 2 = 0 + 0.066 955 566 406 249 988 36;
  • 3) 0.066 955 566 406 249 988 36 × 2 = 0 + 0.133 911 132 812 499 976 72;
  • 4) 0.133 911 132 812 499 976 72 × 2 = 0 + 0.267 822 265 624 999 953 44;
  • 5) 0.267 822 265 624 999 953 44 × 2 = 0 + 0.535 644 531 249 999 906 88;
  • 6) 0.535 644 531 249 999 906 88 × 2 = 1 + 0.071 289 062 499 999 813 76;
  • 7) 0.071 289 062 499 999 813 76 × 2 = 0 + 0.142 578 124 999 999 627 52;
  • 8) 0.142 578 124 999 999 627 52 × 2 = 0 + 0.285 156 249 999 999 255 04;
  • 9) 0.285 156 249 999 999 255 04 × 2 = 0 + 0.570 312 499 999 998 510 08;
  • 10) 0.570 312 499 999 998 510 08 × 2 = 1 + 0.140 624 999 999 997 020 16;
  • 11) 0.140 624 999 999 997 020 16 × 2 = 0 + 0.281 249 999 999 994 040 32;
  • 12) 0.281 249 999 999 994 040 32 × 2 = 0 + 0.562 499 999 999 988 080 64;
  • 13) 0.562 499 999 999 988 080 64 × 2 = 1 + 0.124 999 999 999 976 161 28;
  • 14) 0.124 999 999 999 976 161 28 × 2 = 0 + 0.249 999 999 999 952 322 56;
  • 15) 0.249 999 999 999 952 322 56 × 2 = 0 + 0.499 999 999 999 904 645 12;
  • 16) 0.499 999 999 999 904 645 12 × 2 = 0 + 0.999 999 999 999 809 290 24;
  • 17) 0.999 999 999 999 809 290 24 × 2 = 1 + 0.999 999 999 999 618 580 48;
  • 18) 0.999 999 999 999 618 580 48 × 2 = 1 + 0.999 999 999 999 237 160 96;
  • 19) 0.999 999 999 999 237 160 96 × 2 = 1 + 0.999 999 999 998 474 321 92;
  • 20) 0.999 999 999 998 474 321 92 × 2 = 1 + 0.999 999 999 996 948 643 84;
  • 21) 0.999 999 999 996 948 643 84 × 2 = 1 + 0.999 999 999 993 897 287 68;
  • 22) 0.999 999 999 993 897 287 68 × 2 = 1 + 0.999 999 999 987 794 575 36;
  • 23) 0.999 999 999 987 794 575 36 × 2 = 1 + 0.999 999 999 975 589 150 72;
  • 24) 0.999 999 999 975 589 150 72 × 2 = 1 + 0.999 999 999 951 178 301 44;
  • 25) 0.999 999 999 951 178 301 44 × 2 = 1 + 0.999 999 999 902 356 602 88;
  • 26) 0.999 999 999 902 356 602 88 × 2 = 1 + 0.999 999 999 804 713 205 76;
  • 27) 0.999 999 999 804 713 205 76 × 2 = 1 + 0.999 999 999 609 426 411 52;
  • 28) 0.999 999 999 609 426 411 52 × 2 = 1 + 0.999 999 999 218 852 823 04;
  • 29) 0.999 999 999 218 852 823 04 × 2 = 1 + 0.999 999 998 437 705 646 08;
  • 30) 0.999 999 998 437 705 646 08 × 2 = 1 + 0.999 999 996 875 411 292 16;
  • 31) 0.999 999 996 875 411 292 16 × 2 = 1 + 0.999 999 993 750 822 584 32;
  • 32) 0.999 999 993 750 822 584 32 × 2 = 1 + 0.999 999 987 501 645 168 64;
  • 33) 0.999 999 987 501 645 168 64 × 2 = 1 + 0.999 999 975 003 290 337 28;
  • 34) 0.999 999 975 003 290 337 28 × 2 = 1 + 0.999 999 950 006 580 674 56;
  • 35) 0.999 999 950 006 580 674 56 × 2 = 1 + 0.999 999 900 013 161 349 12;
  • 36) 0.999 999 900 013 161 349 12 × 2 = 1 + 0.999 999 800 026 322 698 24;
  • 37) 0.999 999 800 026 322 698 24 × 2 = 1 + 0.999 999 600 052 645 396 48;
  • 38) 0.999 999 600 052 645 396 48 × 2 = 1 + 0.999 999 200 105 290 792 96;
  • 39) 0.999 999 200 105 290 792 96 × 2 = 1 + 0.999 998 400 210 581 585 92;
  • 40) 0.999 998 400 210 581 585 92 × 2 = 1 + 0.999 996 800 421 163 171 84;
  • 41) 0.999 996 800 421 163 171 84 × 2 = 1 + 0.999 993 600 842 326 343 68;
  • 42) 0.999 993 600 842 326 343 68 × 2 = 1 + 0.999 987 201 684 652 687 36;
  • 43) 0.999 987 201 684 652 687 36 × 2 = 1 + 0.999 974 403 369 305 374 72;
  • 44) 0.999 974 403 369 305 374 72 × 2 = 1 + 0.999 948 806 738 610 749 44;
  • 45) 0.999 948 806 738 610 749 44 × 2 = 1 + 0.999 897 613 477 221 498 88;
  • 46) 0.999 897 613 477 221 498 88 × 2 = 1 + 0.999 795 226 954 442 997 76;
  • 47) 0.999 795 226 954 442 997 76 × 2 = 1 + 0.999 590 453 908 885 995 52;
  • 48) 0.999 590 453 908 885 995 52 × 2 = 1 + 0.999 180 907 817 771 991 04;
  • 49) 0.999 180 907 817 771 991 04 × 2 = 1 + 0.998 361 815 635 543 982 08;
  • 50) 0.998 361 815 635 543 982 08 × 2 = 1 + 0.996 723 631 271 087 964 16;
  • 51) 0.996 723 631 271 087 964 16 × 2 = 1 + 0.993 447 262 542 175 928 32;
  • 52) 0.993 447 262 542 175 928 32 × 2 = 1 + 0.986 894 525 084 351 856 64;
  • 53) 0.986 894 525 084 351 856 64 × 2 = 1 + 0.973 789 050 168 703 713 28;
  • 54) 0.973 789 050 168 703 713 28 × 2 = 1 + 0.947 578 100 337 407 426 56;
  • 55) 0.947 578 100 337 407 426 56 × 2 = 1 + 0.895 156 200 674 814 853 12;
  • 56) 0.895 156 200 674 814 853 12 × 2 = 1 + 0.790 312 401 349 629 706 24;
  • 57) 0.790 312 401 349 629 706 24 × 2 = 1 + 0.580 624 802 699 259 412 48;
  • 58) 0.580 624 802 699 259 412 48 × 2 = 1 + 0.161 249 605 398 518 824 96;

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 09(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 09(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 09(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 09 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