-0.016 738 891 601 562 496 385 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 385(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
-0.016 738 891 601 562 496 385(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:

|-0.016 738 891 601 562 496 385| = 0.016 738 891 601 562 496 385


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: 0.016 738 891 601 562 496 385.

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) 0.016 738 891 601 562 496 385 × 2 = 0 + 0.033 477 783 203 124 992 77;
  • 2) 0.033 477 783 203 124 992 77 × 2 = 0 + 0.066 955 566 406 249 985 54;
  • 3) 0.066 955 566 406 249 985 54 × 2 = 0 + 0.133 911 132 812 499 971 08;
  • 4) 0.133 911 132 812 499 971 08 × 2 = 0 + 0.267 822 265 624 999 942 16;
  • 5) 0.267 822 265 624 999 942 16 × 2 = 0 + 0.535 644 531 249 999 884 32;
  • 6) 0.535 644 531 249 999 884 32 × 2 = 1 + 0.071 289 062 499 999 768 64;
  • 7) 0.071 289 062 499 999 768 64 × 2 = 0 + 0.142 578 124 999 999 537 28;
  • 8) 0.142 578 124 999 999 537 28 × 2 = 0 + 0.285 156 249 999 999 074 56;
  • 9) 0.285 156 249 999 999 074 56 × 2 = 0 + 0.570 312 499 999 998 149 12;
  • 10) 0.570 312 499 999 998 149 12 × 2 = 1 + 0.140 624 999 999 996 298 24;
  • 11) 0.140 624 999 999 996 298 24 × 2 = 0 + 0.281 249 999 999 992 596 48;
  • 12) 0.281 249 999 999 992 596 48 × 2 = 0 + 0.562 499 999 999 985 192 96;
  • 13) 0.562 499 999 999 985 192 96 × 2 = 1 + 0.124 999 999 999 970 385 92;
  • 14) 0.124 999 999 999 970 385 92 × 2 = 0 + 0.249 999 999 999 940 771 84;
  • 15) 0.249 999 999 999 940 771 84 × 2 = 0 + 0.499 999 999 999 881 543 68;
  • 16) 0.499 999 999 999 881 543 68 × 2 = 0 + 0.999 999 999 999 763 087 36;
  • 17) 0.999 999 999 999 763 087 36 × 2 = 1 + 0.999 999 999 999 526 174 72;
  • 18) 0.999 999 999 999 526 174 72 × 2 = 1 + 0.999 999 999 999 052 349 44;
  • 19) 0.999 999 999 999 052 349 44 × 2 = 1 + 0.999 999 999 998 104 698 88;
  • 20) 0.999 999 999 998 104 698 88 × 2 = 1 + 0.999 999 999 996 209 397 76;
  • 21) 0.999 999 999 996 209 397 76 × 2 = 1 + 0.999 999 999 992 418 795 52;
  • 22) 0.999 999 999 992 418 795 52 × 2 = 1 + 0.999 999 999 984 837 591 04;
  • 23) 0.999 999 999 984 837 591 04 × 2 = 1 + 0.999 999 999 969 675 182 08;
  • 24) 0.999 999 999 969 675 182 08 × 2 = 1 + 0.999 999 999 939 350 364 16;
  • 25) 0.999 999 999 939 350 364 16 × 2 = 1 + 0.999 999 999 878 700 728 32;
  • 26) 0.999 999 999 878 700 728 32 × 2 = 1 + 0.999 999 999 757 401 456 64;
  • 27) 0.999 999 999 757 401 456 64 × 2 = 1 + 0.999 999 999 514 802 913 28;
  • 28) 0.999 999 999 514 802 913 28 × 2 = 1 + 0.999 999 999 029 605 826 56;
  • 29) 0.999 999 999 029 605 826 56 × 2 = 1 + 0.999 999 998 059 211 653 12;
  • 30) 0.999 999 998 059 211 653 12 × 2 = 1 + 0.999 999 996 118 423 306 24;
  • 31) 0.999 999 996 118 423 306 24 × 2 = 1 + 0.999 999 992 236 846 612 48;
  • 32) 0.999 999 992 236 846 612 48 × 2 = 1 + 0.999 999 984 473 693 224 96;
  • 33) 0.999 999 984 473 693 224 96 × 2 = 1 + 0.999 999 968 947 386 449 92;
  • 34) 0.999 999 968 947 386 449 92 × 2 = 1 + 0.999 999 937 894 772 899 84;
  • 35) 0.999 999 937 894 772 899 84 × 2 = 1 + 0.999 999 875 789 545 799 68;
  • 36) 0.999 999 875 789 545 799 68 × 2 = 1 + 0.999 999 751 579 091 599 36;
  • 37) 0.999 999 751 579 091 599 36 × 2 = 1 + 0.999 999 503 158 183 198 72;
  • 38) 0.999 999 503 158 183 198 72 × 2 = 1 + 0.999 999 006 316 366 397 44;
  • 39) 0.999 999 006 316 366 397 44 × 2 = 1 + 0.999 998 012 632 732 794 88;
  • 40) 0.999 998 012 632 732 794 88 × 2 = 1 + 0.999 996 025 265 465 589 76;
  • 41) 0.999 996 025 265 465 589 76 × 2 = 1 + 0.999 992 050 530 931 179 52;
  • 42) 0.999 992 050 530 931 179 52 × 2 = 1 + 0.999 984 101 061 862 359 04;
  • 43) 0.999 984 101 061 862 359 04 × 2 = 1 + 0.999 968 202 123 724 718 08;
  • 44) 0.999 968 202 123 724 718 08 × 2 = 1 + 0.999 936 404 247 449 436 16;
  • 45) 0.999 936 404 247 449 436 16 × 2 = 1 + 0.999 872 808 494 898 872 32;
  • 46) 0.999 872 808 494 898 872 32 × 2 = 1 + 0.999 745 616 989 797 744 64;
  • 47) 0.999 745 616 989 797 744 64 × 2 = 1 + 0.999 491 233 979 595 489 28;
  • 48) 0.999 491 233 979 595 489 28 × 2 = 1 + 0.998 982 467 959 190 978 56;
  • 49) 0.998 982 467 959 190 978 56 × 2 = 1 + 0.997 964 935 918 381 957 12;
  • 50) 0.997 964 935 918 381 957 12 × 2 = 1 + 0.995 929 871 836 763 914 24;
  • 51) 0.995 929 871 836 763 914 24 × 2 = 1 + 0.991 859 743 673 527 828 48;
  • 52) 0.991 859 743 673 527 828 48 × 2 = 1 + 0.983 719 487 347 055 656 96;
  • 53) 0.983 719 487 347 055 656 96 × 2 = 1 + 0.967 438 974 694 111 313 92;
  • 54) 0.967 438 974 694 111 313 92 × 2 = 1 + 0.934 877 949 388 222 627 84;
  • 55) 0.934 877 949 388 222 627 84 × 2 = 1 + 0.869 755 898 776 445 255 68;
  • 56) 0.869 755 898 776 445 255 68 × 2 = 1 + 0.739 511 797 552 890 511 36;
  • 57) 0.739 511 797 552 890 511 36 × 2 = 1 + 0.479 023 595 105 781 022 72;
  • 58) 0.479 023 595 105 781 022 72 × 2 = 0 + 0.958 047 190 211 562 045 44;

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:


0.016 738 891 601 562 496 385(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 385(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:


0.016 738 891 601 562 496 385(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 -0.016 738 891 601 562 496 385 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