-0.214 999 999 999 701 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.214 999 999 999 701(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
-0.214 999 999 999 701(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:

|-0.214 999 999 999 701| = 0.214 999 999 999 701


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: 0.214 999 999 999 701.

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) 0.214 999 999 999 701 × 2 = 0 + 0.429 999 999 999 402;
  • 2) 0.429 999 999 999 402 × 2 = 0 + 0.859 999 999 998 804;
  • 3) 0.859 999 999 998 804 × 2 = 1 + 0.719 999 999 997 608;
  • 4) 0.719 999 999 997 608 × 2 = 1 + 0.439 999 999 995 216;
  • 5) 0.439 999 999 995 216 × 2 = 0 + 0.879 999 999 990 432;
  • 6) 0.879 999 999 990 432 × 2 = 1 + 0.759 999 999 980 864;
  • 7) 0.759 999 999 980 864 × 2 = 1 + 0.519 999 999 961 728;
  • 8) 0.519 999 999 961 728 × 2 = 1 + 0.039 999 999 923 456;
  • 9) 0.039 999 999 923 456 × 2 = 0 + 0.079 999 999 846 912;
  • 10) 0.079 999 999 846 912 × 2 = 0 + 0.159 999 999 693 824;
  • 11) 0.159 999 999 693 824 × 2 = 0 + 0.319 999 999 387 648;
  • 12) 0.319 999 999 387 648 × 2 = 0 + 0.639 999 998 775 296;
  • 13) 0.639 999 998 775 296 × 2 = 1 + 0.279 999 997 550 592;
  • 14) 0.279 999 997 550 592 × 2 = 0 + 0.559 999 995 101 184;
  • 15) 0.559 999 995 101 184 × 2 = 1 + 0.119 999 990 202 368;
  • 16) 0.119 999 990 202 368 × 2 = 0 + 0.239 999 980 404 736;
  • 17) 0.239 999 980 404 736 × 2 = 0 + 0.479 999 960 809 472;
  • 18) 0.479 999 960 809 472 × 2 = 0 + 0.959 999 921 618 944;
  • 19) 0.959 999 921 618 944 × 2 = 1 + 0.919 999 843 237 888;
  • 20) 0.919 999 843 237 888 × 2 = 1 + 0.839 999 686 475 776;
  • 21) 0.839 999 686 475 776 × 2 = 1 + 0.679 999 372 951 552;
  • 22) 0.679 999 372 951 552 × 2 = 1 + 0.359 998 745 903 104;
  • 23) 0.359 998 745 903 104 × 2 = 0 + 0.719 997 491 806 208;
  • 24) 0.719 997 491 806 208 × 2 = 1 + 0.439 994 983 612 416;
  • 25) 0.439 994 983 612 416 × 2 = 0 + 0.879 989 967 224 832;
  • 26) 0.879 989 967 224 832 × 2 = 1 + 0.759 979 934 449 664;
  • 27) 0.759 979 934 449 664 × 2 = 1 + 0.519 959 868 899 328;
  • 28) 0.519 959 868 899 328 × 2 = 1 + 0.039 919 737 798 656;
  • 29) 0.039 919 737 798 656 × 2 = 0 + 0.079 839 475 597 312;
  • 30) 0.079 839 475 597 312 × 2 = 0 + 0.159 678 951 194 624;
  • 31) 0.159 678 951 194 624 × 2 = 0 + 0.319 357 902 389 248;
  • 32) 0.319 357 902 389 248 × 2 = 0 + 0.638 715 804 778 496;
  • 33) 0.638 715 804 778 496 × 2 = 1 + 0.277 431 609 556 992;
  • 34) 0.277 431 609 556 992 × 2 = 0 + 0.554 863 219 113 984;
  • 35) 0.554 863 219 113 984 × 2 = 1 + 0.109 726 438 227 968;
  • 36) 0.109 726 438 227 968 × 2 = 0 + 0.219 452 876 455 936;
  • 37) 0.219 452 876 455 936 × 2 = 0 + 0.438 905 752 911 872;
  • 38) 0.438 905 752 911 872 × 2 = 0 + 0.877 811 505 823 744;
  • 39) 0.877 811 505 823 744 × 2 = 1 + 0.755 623 011 647 488;
  • 40) 0.755 623 011 647 488 × 2 = 1 + 0.511 246 023 294 976;
  • 41) 0.511 246 023 294 976 × 2 = 1 + 0.022 492 046 589 952;
  • 42) 0.022 492 046 589 952 × 2 = 0 + 0.044 984 093 179 904;
  • 43) 0.044 984 093 179 904 × 2 = 0 + 0.089 968 186 359 808;
  • 44) 0.089 968 186 359 808 × 2 = 0 + 0.179 936 372 719 616;
  • 45) 0.179 936 372 719 616 × 2 = 0 + 0.359 872 745 439 232;
  • 46) 0.359 872 745 439 232 × 2 = 0 + 0.719 745 490 878 464;
  • 47) 0.719 745 490 878 464 × 2 = 1 + 0.439 490 981 756 928;
  • 48) 0.439 490 981 756 928 × 2 = 0 + 0.878 981 963 513 856;
  • 49) 0.878 981 963 513 856 × 2 = 1 + 0.757 963 927 027 712;
  • 50) 0.757 963 927 027 712 × 2 = 1 + 0.515 927 854 055 424;
  • 51) 0.515 927 854 055 424 × 2 = 1 + 0.031 855 708 110 848;
  • 52) 0.031 855 708 110 848 × 2 = 0 + 0.063 711 416 221 696;
  • 53) 0.063 711 416 221 696 × 2 = 0 + 0.127 422 832 443 392;
  • 54) 0.127 422 832 443 392 × 2 = 0 + 0.254 845 664 886 784;
  • 55) 0.254 845 664 886 784 × 2 = 0 + 0.509 691 329 773 568;

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:


0.214 999 999 999 701(10) =

0.0011 0111 0000 1010 0011 1101 0111 0000 1010 0011 1000 0010 1110 000(2)

6. Positive number before normalization:

0.214 999 999 999 701(10) =

0.0011 0111 0000 1010 0011 1101 0111 0000 1010 0011 1000 0010 1110 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.214 999 999 999 701(10) =

0.0011 0111 0000 1010 0011 1101 0111 0000 1010 0011 1000 0010 1110 000(2) =

0.0011 0111 0000 1010 0011 1101 0111 0000 1010 0011 1000 0010 1110 000(2) × 20 =

1.1011 1000 0101 0001 1110 1011 1000 0101 0001 1100 0001 0111 0000(2) × 2-3


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): -3

Mantissa (not normalized):
1.1011 1000 0101 0001 1110 1011 1000 0101 0001 1100 0001 0111 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1020(10) =

011 1111 1100(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. 1011 1000 0101 0001 1110 1011 1000 0101 0001 1100 0001 0111 0000 =

1011 1000 0101 0001 1110 1011 1000 0101 0001 1100 0001 0111 0000


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 1100

Mantissa (52 bits) =
1011 1000 0101 0001 1110 1011 1000 0101 0001 1100 0001 0111 0000


Decimal number -0.214 999 999 999 701 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 1011 1000 0101 0001 1110 1011 1000 0101 0001 1100 0001 0111 0000

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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