-0.016 738 891 601 562 497 23 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 23(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 23(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 23| = 0.016 738 891 601 562 497 23


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 23.

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 23 × 2 = 0 + 0.033 477 783 203 124 994 46;
  • 2) 0.033 477 783 203 124 994 46 × 2 = 0 + 0.066 955 566 406 249 988 92;
  • 3) 0.066 955 566 406 249 988 92 × 2 = 0 + 0.133 911 132 812 499 977 84;
  • 4) 0.133 911 132 812 499 977 84 × 2 = 0 + 0.267 822 265 624 999 955 68;
  • 5) 0.267 822 265 624 999 955 68 × 2 = 0 + 0.535 644 531 249 999 911 36;
  • 6) 0.535 644 531 249 999 911 36 × 2 = 1 + 0.071 289 062 499 999 822 72;
  • 7) 0.071 289 062 499 999 822 72 × 2 = 0 + 0.142 578 124 999 999 645 44;
  • 8) 0.142 578 124 999 999 645 44 × 2 = 0 + 0.285 156 249 999 999 290 88;
  • 9) 0.285 156 249 999 999 290 88 × 2 = 0 + 0.570 312 499 999 998 581 76;
  • 10) 0.570 312 499 999 998 581 76 × 2 = 1 + 0.140 624 999 999 997 163 52;
  • 11) 0.140 624 999 999 997 163 52 × 2 = 0 + 0.281 249 999 999 994 327 04;
  • 12) 0.281 249 999 999 994 327 04 × 2 = 0 + 0.562 499 999 999 988 654 08;
  • 13) 0.562 499 999 999 988 654 08 × 2 = 1 + 0.124 999 999 999 977 308 16;
  • 14) 0.124 999 999 999 977 308 16 × 2 = 0 + 0.249 999 999 999 954 616 32;
  • 15) 0.249 999 999 999 954 616 32 × 2 = 0 + 0.499 999 999 999 909 232 64;
  • 16) 0.499 999 999 999 909 232 64 × 2 = 0 + 0.999 999 999 999 818 465 28;
  • 17) 0.999 999 999 999 818 465 28 × 2 = 1 + 0.999 999 999 999 636 930 56;
  • 18) 0.999 999 999 999 636 930 56 × 2 = 1 + 0.999 999 999 999 273 861 12;
  • 19) 0.999 999 999 999 273 861 12 × 2 = 1 + 0.999 999 999 998 547 722 24;
  • 20) 0.999 999 999 998 547 722 24 × 2 = 1 + 0.999 999 999 997 095 444 48;
  • 21) 0.999 999 999 997 095 444 48 × 2 = 1 + 0.999 999 999 994 190 888 96;
  • 22) 0.999 999 999 994 190 888 96 × 2 = 1 + 0.999 999 999 988 381 777 92;
  • 23) 0.999 999 999 988 381 777 92 × 2 = 1 + 0.999 999 999 976 763 555 84;
  • 24) 0.999 999 999 976 763 555 84 × 2 = 1 + 0.999 999 999 953 527 111 68;
  • 25) 0.999 999 999 953 527 111 68 × 2 = 1 + 0.999 999 999 907 054 223 36;
  • 26) 0.999 999 999 907 054 223 36 × 2 = 1 + 0.999 999 999 814 108 446 72;
  • 27) 0.999 999 999 814 108 446 72 × 2 = 1 + 0.999 999 999 628 216 893 44;
  • 28) 0.999 999 999 628 216 893 44 × 2 = 1 + 0.999 999 999 256 433 786 88;
  • 29) 0.999 999 999 256 433 786 88 × 2 = 1 + 0.999 999 998 512 867 573 76;
  • 30) 0.999 999 998 512 867 573 76 × 2 = 1 + 0.999 999 997 025 735 147 52;
  • 31) 0.999 999 997 025 735 147 52 × 2 = 1 + 0.999 999 994 051 470 295 04;
  • 32) 0.999 999 994 051 470 295 04 × 2 = 1 + 0.999 999 988 102 940 590 08;
  • 33) 0.999 999 988 102 940 590 08 × 2 = 1 + 0.999 999 976 205 881 180 16;
  • 34) 0.999 999 976 205 881 180 16 × 2 = 1 + 0.999 999 952 411 762 360 32;
  • 35) 0.999 999 952 411 762 360 32 × 2 = 1 + 0.999 999 904 823 524 720 64;
  • 36) 0.999 999 904 823 524 720 64 × 2 = 1 + 0.999 999 809 647 049 441 28;
  • 37) 0.999 999 809 647 049 441 28 × 2 = 1 + 0.999 999 619 294 098 882 56;
  • 38) 0.999 999 619 294 098 882 56 × 2 = 1 + 0.999 999 238 588 197 765 12;
  • 39) 0.999 999 238 588 197 765 12 × 2 = 1 + 0.999 998 477 176 395 530 24;
  • 40) 0.999 998 477 176 395 530 24 × 2 = 1 + 0.999 996 954 352 791 060 48;
  • 41) 0.999 996 954 352 791 060 48 × 2 = 1 + 0.999 993 908 705 582 120 96;
  • 42) 0.999 993 908 705 582 120 96 × 2 = 1 + 0.999 987 817 411 164 241 92;
  • 43) 0.999 987 817 411 164 241 92 × 2 = 1 + 0.999 975 634 822 328 483 84;
  • 44) 0.999 975 634 822 328 483 84 × 2 = 1 + 0.999 951 269 644 656 967 68;
  • 45) 0.999 951 269 644 656 967 68 × 2 = 1 + 0.999 902 539 289 313 935 36;
  • 46) 0.999 902 539 289 313 935 36 × 2 = 1 + 0.999 805 078 578 627 870 72;
  • 47) 0.999 805 078 578 627 870 72 × 2 = 1 + 0.999 610 157 157 255 741 44;
  • 48) 0.999 610 157 157 255 741 44 × 2 = 1 + 0.999 220 314 314 511 482 88;
  • 49) 0.999 220 314 314 511 482 88 × 2 = 1 + 0.998 440 628 629 022 965 76;
  • 50) 0.998 440 628 629 022 965 76 × 2 = 1 + 0.996 881 257 258 045 931 52;
  • 51) 0.996 881 257 258 045 931 52 × 2 = 1 + 0.993 762 514 516 091 863 04;
  • 52) 0.993 762 514 516 091 863 04 × 2 = 1 + 0.987 525 029 032 183 726 08;
  • 53) 0.987 525 029 032 183 726 08 × 2 = 1 + 0.975 050 058 064 367 452 16;
  • 54) 0.975 050 058 064 367 452 16 × 2 = 1 + 0.950 100 116 128 734 904 32;
  • 55) 0.950 100 116 128 734 904 32 × 2 = 1 + 0.900 200 232 257 469 808 64;
  • 56) 0.900 200 232 257 469 808 64 × 2 = 1 + 0.800 400 464 514 939 617 28;
  • 57) 0.800 400 464 514 939 617 28 × 2 = 1 + 0.600 800 929 029 879 234 56;
  • 58) 0.600 800 929 029 879 234 56 × 2 = 1 + 0.201 601 858 059 758 469 12;

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 23(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 23(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 23(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 23 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