-0.023 219 428 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.023 219 428 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
-0.023 219 428 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:

|-0.023 219 428 1| = 0.023 219 428 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: 0.023 219 428 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) 0.023 219 428 1 × 2 = 0 + 0.046 438 856 2;
  • 2) 0.046 438 856 2 × 2 = 0 + 0.092 877 712 4;
  • 3) 0.092 877 712 4 × 2 = 0 + 0.185 755 424 8;
  • 4) 0.185 755 424 8 × 2 = 0 + 0.371 510 849 6;
  • 5) 0.371 510 849 6 × 2 = 0 + 0.743 021 699 2;
  • 6) 0.743 021 699 2 × 2 = 1 + 0.486 043 398 4;
  • 7) 0.486 043 398 4 × 2 = 0 + 0.972 086 796 8;
  • 8) 0.972 086 796 8 × 2 = 1 + 0.944 173 593 6;
  • 9) 0.944 173 593 6 × 2 = 1 + 0.888 347 187 2;
  • 10) 0.888 347 187 2 × 2 = 1 + 0.776 694 374 4;
  • 11) 0.776 694 374 4 × 2 = 1 + 0.553 388 748 8;
  • 12) 0.553 388 748 8 × 2 = 1 + 0.106 777 497 6;
  • 13) 0.106 777 497 6 × 2 = 0 + 0.213 554 995 2;
  • 14) 0.213 554 995 2 × 2 = 0 + 0.427 109 990 4;
  • 15) 0.427 109 990 4 × 2 = 0 + 0.854 219 980 8;
  • 16) 0.854 219 980 8 × 2 = 1 + 0.708 439 961 6;
  • 17) 0.708 439 961 6 × 2 = 1 + 0.416 879 923 2;
  • 18) 0.416 879 923 2 × 2 = 0 + 0.833 759 846 4;
  • 19) 0.833 759 846 4 × 2 = 1 + 0.667 519 692 8;
  • 20) 0.667 519 692 8 × 2 = 1 + 0.335 039 385 6;
  • 21) 0.335 039 385 6 × 2 = 0 + 0.670 078 771 2;
  • 22) 0.670 078 771 2 × 2 = 1 + 0.340 157 542 4;
  • 23) 0.340 157 542 4 × 2 = 0 + 0.680 315 084 8;
  • 24) 0.680 315 084 8 × 2 = 1 + 0.360 630 169 6;
  • 25) 0.360 630 169 6 × 2 = 0 + 0.721 260 339 2;
  • 26) 0.721 260 339 2 × 2 = 1 + 0.442 520 678 4;
  • 27) 0.442 520 678 4 × 2 = 0 + 0.885 041 356 8;
  • 28) 0.885 041 356 8 × 2 = 1 + 0.770 082 713 6;
  • 29) 0.770 082 713 6 × 2 = 1 + 0.540 165 427 2;
  • 30) 0.540 165 427 2 × 2 = 1 + 0.080 330 854 4;
  • 31) 0.080 330 854 4 × 2 = 0 + 0.160 661 708 8;
  • 32) 0.160 661 708 8 × 2 = 0 + 0.321 323 417 6;
  • 33) 0.321 323 417 6 × 2 = 0 + 0.642 646 835 2;
  • 34) 0.642 646 835 2 × 2 = 1 + 0.285 293 670 4;
  • 35) 0.285 293 670 4 × 2 = 0 + 0.570 587 340 8;
  • 36) 0.570 587 340 8 × 2 = 1 + 0.141 174 681 6;
  • 37) 0.141 174 681 6 × 2 = 0 + 0.282 349 363 2;
  • 38) 0.282 349 363 2 × 2 = 0 + 0.564 698 726 4;
  • 39) 0.564 698 726 4 × 2 = 1 + 0.129 397 452 8;
  • 40) 0.129 397 452 8 × 2 = 0 + 0.258 794 905 6;
  • 41) 0.258 794 905 6 × 2 = 0 + 0.517 589 811 2;
  • 42) 0.517 589 811 2 × 2 = 1 + 0.035 179 622 4;
  • 43) 0.035 179 622 4 × 2 = 0 + 0.070 359 244 8;
  • 44) 0.070 359 244 8 × 2 = 0 + 0.140 718 489 6;
  • 45) 0.140 718 489 6 × 2 = 0 + 0.281 436 979 2;
  • 46) 0.281 436 979 2 × 2 = 0 + 0.562 873 958 4;
  • 47) 0.562 873 958 4 × 2 = 1 + 0.125 747 916 8;
  • 48) 0.125 747 916 8 × 2 = 0 + 0.251 495 833 6;
  • 49) 0.251 495 833 6 × 2 = 0 + 0.502 991 667 2;
  • 50) 0.502 991 667 2 × 2 = 1 + 0.005 983 334 4;
  • 51) 0.005 983 334 4 × 2 = 0 + 0.011 966 668 8;
  • 52) 0.011 966 668 8 × 2 = 0 + 0.023 933 337 6;
  • 53) 0.023 933 337 6 × 2 = 0 + 0.047 866 675 2;
  • 54) 0.047 866 675 2 × 2 = 0 + 0.095 733 350 4;
  • 55) 0.095 733 350 4 × 2 = 0 + 0.191 466 700 8;
  • 56) 0.191 466 700 8 × 2 = 0 + 0.382 933 401 6;
  • 57) 0.382 933 401 6 × 2 = 0 + 0.765 866 803 2;
  • 58) 0.765 866 803 2 × 2 = 1 + 0.531 733 606 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:


0.023 219 428 1(10) =

0.0000 0101 1111 0001 1011 0101 0101 1100 0101 0010 0100 0010 0100 0000 01(2)

6. Positive number before normalization:

0.023 219 428 1(10) =

0.0000 0101 1111 0001 1011 0101 0101 1100 0101 0010 0100 0010 0100 0000 01(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:


0.023 219 428 1(10) =

0.0000 0101 1111 0001 1011 0101 0101 1100 0101 0010 0100 0010 0100 0000 01(2) =

0.0000 0101 1111 0001 1011 0101 0101 1100 0101 0010 0100 0010 0100 0000 01(2) × 20 =

1.0111 1100 0110 1101 0101 0111 0001 0100 1001 0000 1001 0000 0001(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):
1.0111 1100 0110 1101 0101 0111 0001 0100 1001 0000 1001 0000 0001


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. 0111 1100 0110 1101 0101 0111 0001 0100 1001 0000 1001 0000 0001 =

0111 1100 0110 1101 0101 0111 0001 0100 1001 0000 1001 0000 0001


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) =
0111 1100 0110 1101 0101 0111 0001 0100 1001 0000 1001 0000 0001


Decimal number -0.023 219 428 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0111 1100 0110 1101 0101 0111 0001 0100 1001 0000 1001 0000 0001

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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