-0.016 738 891 601 562 496 558 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 558 1(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
-0.016 738 891 601 562 496 558 1(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:

|-0.016 738 891 601 562 496 558 1| = 0.016 738 891 601 562 496 558 1


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: 0.016 738 891 601 562 496 558 1.

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) 0.016 738 891 601 562 496 558 1 × 2 = 0 + 0.033 477 783 203 124 993 116 2;
  • 2) 0.033 477 783 203 124 993 116 2 × 2 = 0 + 0.066 955 566 406 249 986 232 4;
  • 3) 0.066 955 566 406 249 986 232 4 × 2 = 0 + 0.133 911 132 812 499 972 464 8;
  • 4) 0.133 911 132 812 499 972 464 8 × 2 = 0 + 0.267 822 265 624 999 944 929 6;
  • 5) 0.267 822 265 624 999 944 929 6 × 2 = 0 + 0.535 644 531 249 999 889 859 2;
  • 6) 0.535 644 531 249 999 889 859 2 × 2 = 1 + 0.071 289 062 499 999 779 718 4;
  • 7) 0.071 289 062 499 999 779 718 4 × 2 = 0 + 0.142 578 124 999 999 559 436 8;
  • 8) 0.142 578 124 999 999 559 436 8 × 2 = 0 + 0.285 156 249 999 999 118 873 6;
  • 9) 0.285 156 249 999 999 118 873 6 × 2 = 0 + 0.570 312 499 999 998 237 747 2;
  • 10) 0.570 312 499 999 998 237 747 2 × 2 = 1 + 0.140 624 999 999 996 475 494 4;
  • 11) 0.140 624 999 999 996 475 494 4 × 2 = 0 + 0.281 249 999 999 992 950 988 8;
  • 12) 0.281 249 999 999 992 950 988 8 × 2 = 0 + 0.562 499 999 999 985 901 977 6;
  • 13) 0.562 499 999 999 985 901 977 6 × 2 = 1 + 0.124 999 999 999 971 803 955 2;
  • 14) 0.124 999 999 999 971 803 955 2 × 2 = 0 + 0.249 999 999 999 943 607 910 4;
  • 15) 0.249 999 999 999 943 607 910 4 × 2 = 0 + 0.499 999 999 999 887 215 820 8;
  • 16) 0.499 999 999 999 887 215 820 8 × 2 = 0 + 0.999 999 999 999 774 431 641 6;
  • 17) 0.999 999 999 999 774 431 641 6 × 2 = 1 + 0.999 999 999 999 548 863 283 2;
  • 18) 0.999 999 999 999 548 863 283 2 × 2 = 1 + 0.999 999 999 999 097 726 566 4;
  • 19) 0.999 999 999 999 097 726 566 4 × 2 = 1 + 0.999 999 999 998 195 453 132 8;
  • 20) 0.999 999 999 998 195 453 132 8 × 2 = 1 + 0.999 999 999 996 390 906 265 6;
  • 21) 0.999 999 999 996 390 906 265 6 × 2 = 1 + 0.999 999 999 992 781 812 531 2;
  • 22) 0.999 999 999 992 781 812 531 2 × 2 = 1 + 0.999 999 999 985 563 625 062 4;
  • 23) 0.999 999 999 985 563 625 062 4 × 2 = 1 + 0.999 999 999 971 127 250 124 8;
  • 24) 0.999 999 999 971 127 250 124 8 × 2 = 1 + 0.999 999 999 942 254 500 249 6;
  • 25) 0.999 999 999 942 254 500 249 6 × 2 = 1 + 0.999 999 999 884 509 000 499 2;
  • 26) 0.999 999 999 884 509 000 499 2 × 2 = 1 + 0.999 999 999 769 018 000 998 4;
  • 27) 0.999 999 999 769 018 000 998 4 × 2 = 1 + 0.999 999 999 538 036 001 996 8;
  • 28) 0.999 999 999 538 036 001 996 8 × 2 = 1 + 0.999 999 999 076 072 003 993 6;
  • 29) 0.999 999 999 076 072 003 993 6 × 2 = 1 + 0.999 999 998 152 144 007 987 2;
  • 30) 0.999 999 998 152 144 007 987 2 × 2 = 1 + 0.999 999 996 304 288 015 974 4;
  • 31) 0.999 999 996 304 288 015 974 4 × 2 = 1 + 0.999 999 992 608 576 031 948 8;
  • 32) 0.999 999 992 608 576 031 948 8 × 2 = 1 + 0.999 999 985 217 152 063 897 6;
  • 33) 0.999 999 985 217 152 063 897 6 × 2 = 1 + 0.999 999 970 434 304 127 795 2;
  • 34) 0.999 999 970 434 304 127 795 2 × 2 = 1 + 0.999 999 940 868 608 255 590 4;
  • 35) 0.999 999 940 868 608 255 590 4 × 2 = 1 + 0.999 999 881 737 216 511 180 8;
  • 36) 0.999 999 881 737 216 511 180 8 × 2 = 1 + 0.999 999 763 474 433 022 361 6;
  • 37) 0.999 999 763 474 433 022 361 6 × 2 = 1 + 0.999 999 526 948 866 044 723 2;
  • 38) 0.999 999 526 948 866 044 723 2 × 2 = 1 + 0.999 999 053 897 732 089 446 4;
  • 39) 0.999 999 053 897 732 089 446 4 × 2 = 1 + 0.999 998 107 795 464 178 892 8;
  • 40) 0.999 998 107 795 464 178 892 8 × 2 = 1 + 0.999 996 215 590 928 357 785 6;
  • 41) 0.999 996 215 590 928 357 785 6 × 2 = 1 + 0.999 992 431 181 856 715 571 2;
  • 42) 0.999 992 431 181 856 715 571 2 × 2 = 1 + 0.999 984 862 363 713 431 142 4;
  • 43) 0.999 984 862 363 713 431 142 4 × 2 = 1 + 0.999 969 724 727 426 862 284 8;
  • 44) 0.999 969 724 727 426 862 284 8 × 2 = 1 + 0.999 939 449 454 853 724 569 6;
  • 45) 0.999 939 449 454 853 724 569 6 × 2 = 1 + 0.999 878 898 909 707 449 139 2;
  • 46) 0.999 878 898 909 707 449 139 2 × 2 = 1 + 0.999 757 797 819 414 898 278 4;
  • 47) 0.999 757 797 819 414 898 278 4 × 2 = 1 + 0.999 515 595 638 829 796 556 8;
  • 48) 0.999 515 595 638 829 796 556 8 × 2 = 1 + 0.999 031 191 277 659 593 113 6;
  • 49) 0.999 031 191 277 659 593 113 6 × 2 = 1 + 0.998 062 382 555 319 186 227 2;
  • 50) 0.998 062 382 555 319 186 227 2 × 2 = 1 + 0.996 124 765 110 638 372 454 4;
  • 51) 0.996 124 765 110 638 372 454 4 × 2 = 1 + 0.992 249 530 221 276 744 908 8;
  • 52) 0.992 249 530 221 276 744 908 8 × 2 = 1 + 0.984 499 060 442 553 489 817 6;
  • 53) 0.984 499 060 442 553 489 817 6 × 2 = 1 + 0.968 998 120 885 106 979 635 2;
  • 54) 0.968 998 120 885 106 979 635 2 × 2 = 1 + 0.937 996 241 770 213 959 270 4;
  • 55) 0.937 996 241 770 213 959 270 4 × 2 = 1 + 0.875 992 483 540 427 918 540 8;
  • 56) 0.875 992 483 540 427 918 540 8 × 2 = 1 + 0.751 984 967 080 855 837 081 6;
  • 57) 0.751 984 967 080 855 837 081 6 × 2 = 1 + 0.503 969 934 161 711 674 163 2;
  • 58) 0.503 969 934 161 711 674 163 2 × 2 = 1 + 0.007 939 868 323 423 348 326 4;

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:


0.016 738 891 601 562 496 558 1(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 558 1(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:


0.016 738 891 601 562 496 558 1(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 -0.016 738 891 601 562 496 558 1 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