-0.016 738 891 601 562 494 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 494 4(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
-0.016 738 891 601 562 494 4(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:

|-0.016 738 891 601 562 494 4| = 0.016 738 891 601 562 494 4


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: 0.016 738 891 601 562 494 4.

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) 0.016 738 891 601 562 494 4 × 2 = 0 + 0.033 477 783 203 124 988 8;
  • 2) 0.033 477 783 203 124 988 8 × 2 = 0 + 0.066 955 566 406 249 977 6;
  • 3) 0.066 955 566 406 249 977 6 × 2 = 0 + 0.133 911 132 812 499 955 2;
  • 4) 0.133 911 132 812 499 955 2 × 2 = 0 + 0.267 822 265 624 999 910 4;
  • 5) 0.267 822 265 624 999 910 4 × 2 = 0 + 0.535 644 531 249 999 820 8;
  • 6) 0.535 644 531 249 999 820 8 × 2 = 1 + 0.071 289 062 499 999 641 6;
  • 7) 0.071 289 062 499 999 641 6 × 2 = 0 + 0.142 578 124 999 999 283 2;
  • 8) 0.142 578 124 999 999 283 2 × 2 = 0 + 0.285 156 249 999 998 566 4;
  • 9) 0.285 156 249 999 998 566 4 × 2 = 0 + 0.570 312 499 999 997 132 8;
  • 10) 0.570 312 499 999 997 132 8 × 2 = 1 + 0.140 624 999 999 994 265 6;
  • 11) 0.140 624 999 999 994 265 6 × 2 = 0 + 0.281 249 999 999 988 531 2;
  • 12) 0.281 249 999 999 988 531 2 × 2 = 0 + 0.562 499 999 999 977 062 4;
  • 13) 0.562 499 999 999 977 062 4 × 2 = 1 + 0.124 999 999 999 954 124 8;
  • 14) 0.124 999 999 999 954 124 8 × 2 = 0 + 0.249 999 999 999 908 249 6;
  • 15) 0.249 999 999 999 908 249 6 × 2 = 0 + 0.499 999 999 999 816 499 2;
  • 16) 0.499 999 999 999 816 499 2 × 2 = 0 + 0.999 999 999 999 632 998 4;
  • 17) 0.999 999 999 999 632 998 4 × 2 = 1 + 0.999 999 999 999 265 996 8;
  • 18) 0.999 999 999 999 265 996 8 × 2 = 1 + 0.999 999 999 998 531 993 6;
  • 19) 0.999 999 999 998 531 993 6 × 2 = 1 + 0.999 999 999 997 063 987 2;
  • 20) 0.999 999 999 997 063 987 2 × 2 = 1 + 0.999 999 999 994 127 974 4;
  • 21) 0.999 999 999 994 127 974 4 × 2 = 1 + 0.999 999 999 988 255 948 8;
  • 22) 0.999 999 999 988 255 948 8 × 2 = 1 + 0.999 999 999 976 511 897 6;
  • 23) 0.999 999 999 976 511 897 6 × 2 = 1 + 0.999 999 999 953 023 795 2;
  • 24) 0.999 999 999 953 023 795 2 × 2 = 1 + 0.999 999 999 906 047 590 4;
  • 25) 0.999 999 999 906 047 590 4 × 2 = 1 + 0.999 999 999 812 095 180 8;
  • 26) 0.999 999 999 812 095 180 8 × 2 = 1 + 0.999 999 999 624 190 361 6;
  • 27) 0.999 999 999 624 190 361 6 × 2 = 1 + 0.999 999 999 248 380 723 2;
  • 28) 0.999 999 999 248 380 723 2 × 2 = 1 + 0.999 999 998 496 761 446 4;
  • 29) 0.999 999 998 496 761 446 4 × 2 = 1 + 0.999 999 996 993 522 892 8;
  • 30) 0.999 999 996 993 522 892 8 × 2 = 1 + 0.999 999 993 987 045 785 6;
  • 31) 0.999 999 993 987 045 785 6 × 2 = 1 + 0.999 999 987 974 091 571 2;
  • 32) 0.999 999 987 974 091 571 2 × 2 = 1 + 0.999 999 975 948 183 142 4;
  • 33) 0.999 999 975 948 183 142 4 × 2 = 1 + 0.999 999 951 896 366 284 8;
  • 34) 0.999 999 951 896 366 284 8 × 2 = 1 + 0.999 999 903 792 732 569 6;
  • 35) 0.999 999 903 792 732 569 6 × 2 = 1 + 0.999 999 807 585 465 139 2;
  • 36) 0.999 999 807 585 465 139 2 × 2 = 1 + 0.999 999 615 170 930 278 4;
  • 37) 0.999 999 615 170 930 278 4 × 2 = 1 + 0.999 999 230 341 860 556 8;
  • 38) 0.999 999 230 341 860 556 8 × 2 = 1 + 0.999 998 460 683 721 113 6;
  • 39) 0.999 998 460 683 721 113 6 × 2 = 1 + 0.999 996 921 367 442 227 2;
  • 40) 0.999 996 921 367 442 227 2 × 2 = 1 + 0.999 993 842 734 884 454 4;
  • 41) 0.999 993 842 734 884 454 4 × 2 = 1 + 0.999 987 685 469 768 908 8;
  • 42) 0.999 987 685 469 768 908 8 × 2 = 1 + 0.999 975 370 939 537 817 6;
  • 43) 0.999 975 370 939 537 817 6 × 2 = 1 + 0.999 950 741 879 075 635 2;
  • 44) 0.999 950 741 879 075 635 2 × 2 = 1 + 0.999 901 483 758 151 270 4;
  • 45) 0.999 901 483 758 151 270 4 × 2 = 1 + 0.999 802 967 516 302 540 8;
  • 46) 0.999 802 967 516 302 540 8 × 2 = 1 + 0.999 605 935 032 605 081 6;
  • 47) 0.999 605 935 032 605 081 6 × 2 = 1 + 0.999 211 870 065 210 163 2;
  • 48) 0.999 211 870 065 210 163 2 × 2 = 1 + 0.998 423 740 130 420 326 4;
  • 49) 0.998 423 740 130 420 326 4 × 2 = 1 + 0.996 847 480 260 840 652 8;
  • 50) 0.996 847 480 260 840 652 8 × 2 = 1 + 0.993 694 960 521 681 305 6;
  • 51) 0.993 694 960 521 681 305 6 × 2 = 1 + 0.987 389 921 043 362 611 2;
  • 52) 0.987 389 921 043 362 611 2 × 2 = 1 + 0.974 779 842 086 725 222 4;
  • 53) 0.974 779 842 086 725 222 4 × 2 = 1 + 0.949 559 684 173 450 444 8;
  • 54) 0.949 559 684 173 450 444 8 × 2 = 1 + 0.899 119 368 346 900 889 6;
  • 55) 0.899 119 368 346 900 889 6 × 2 = 1 + 0.798 238 736 693 801 779 2;
  • 56) 0.798 238 736 693 801 779 2 × 2 = 1 + 0.596 477 473 387 603 558 4;
  • 57) 0.596 477 473 387 603 558 4 × 2 = 1 + 0.192 954 946 775 207 116 8;
  • 58) 0.192 954 946 775 207 116 8 × 2 = 0 + 0.385 909 893 550 414 233 6;

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:


0.016 738 891 601 562 494 4(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 494 4(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 494 4(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 494 4 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 1110

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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