-0.033 343 988 073 34 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.033 343 988 073 34(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
-0.033 343 988 073 34(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:

|-0.033 343 988 073 34| = 0.033 343 988 073 34


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: 0.033 343 988 073 34.

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) 0.033 343 988 073 34 × 2 = 0 + 0.066 687 976 146 68;
  • 2) 0.066 687 976 146 68 × 2 = 0 + 0.133 375 952 293 36;
  • 3) 0.133 375 952 293 36 × 2 = 0 + 0.266 751 904 586 72;
  • 4) 0.266 751 904 586 72 × 2 = 0 + 0.533 503 809 173 44;
  • 5) 0.533 503 809 173 44 × 2 = 1 + 0.067 007 618 346 88;
  • 6) 0.067 007 618 346 88 × 2 = 0 + 0.134 015 236 693 76;
  • 7) 0.134 015 236 693 76 × 2 = 0 + 0.268 030 473 387 52;
  • 8) 0.268 030 473 387 52 × 2 = 0 + 0.536 060 946 775 04;
  • 9) 0.536 060 946 775 04 × 2 = 1 + 0.072 121 893 550 08;
  • 10) 0.072 121 893 550 08 × 2 = 0 + 0.144 243 787 100 16;
  • 11) 0.144 243 787 100 16 × 2 = 0 + 0.288 487 574 200 32;
  • 12) 0.288 487 574 200 32 × 2 = 0 + 0.576 975 148 400 64;
  • 13) 0.576 975 148 400 64 × 2 = 1 + 0.153 950 296 801 28;
  • 14) 0.153 950 296 801 28 × 2 = 0 + 0.307 900 593 602 56;
  • 15) 0.307 900 593 602 56 × 2 = 0 + 0.615 801 187 205 12;
  • 16) 0.615 801 187 205 12 × 2 = 1 + 0.231 602 374 410 24;
  • 17) 0.231 602 374 410 24 × 2 = 0 + 0.463 204 748 820 48;
  • 18) 0.463 204 748 820 48 × 2 = 0 + 0.926 409 497 640 96;
  • 19) 0.926 409 497 640 96 × 2 = 1 + 0.852 818 995 281 92;
  • 20) 0.852 818 995 281 92 × 2 = 1 + 0.705 637 990 563 84;
  • 21) 0.705 637 990 563 84 × 2 = 1 + 0.411 275 981 127 68;
  • 22) 0.411 275 981 127 68 × 2 = 0 + 0.822 551 962 255 36;
  • 23) 0.822 551 962 255 36 × 2 = 1 + 0.645 103 924 510 72;
  • 24) 0.645 103 924 510 72 × 2 = 1 + 0.290 207 849 021 44;
  • 25) 0.290 207 849 021 44 × 2 = 0 + 0.580 415 698 042 88;
  • 26) 0.580 415 698 042 88 × 2 = 1 + 0.160 831 396 085 76;
  • 27) 0.160 831 396 085 76 × 2 = 0 + 0.321 662 792 171 52;
  • 28) 0.321 662 792 171 52 × 2 = 0 + 0.643 325 584 343 04;
  • 29) 0.643 325 584 343 04 × 2 = 1 + 0.286 651 168 686 08;
  • 30) 0.286 651 168 686 08 × 2 = 0 + 0.573 302 337 372 16;
  • 31) 0.573 302 337 372 16 × 2 = 1 + 0.146 604 674 744 32;
  • 32) 0.146 604 674 744 32 × 2 = 0 + 0.293 209 349 488 64;
  • 33) 0.293 209 349 488 64 × 2 = 0 + 0.586 418 698 977 28;
  • 34) 0.586 418 698 977 28 × 2 = 1 + 0.172 837 397 954 56;
  • 35) 0.172 837 397 954 56 × 2 = 0 + 0.345 674 795 909 12;
  • 36) 0.345 674 795 909 12 × 2 = 0 + 0.691 349 591 818 24;
  • 37) 0.691 349 591 818 24 × 2 = 1 + 0.382 699 183 636 48;
  • 38) 0.382 699 183 636 48 × 2 = 0 + 0.765 398 367 272 96;
  • 39) 0.765 398 367 272 96 × 2 = 1 + 0.530 796 734 545 92;
  • 40) 0.530 796 734 545 92 × 2 = 1 + 0.061 593 469 091 84;
  • 41) 0.061 593 469 091 84 × 2 = 0 + 0.123 186 938 183 68;
  • 42) 0.123 186 938 183 68 × 2 = 0 + 0.246 373 876 367 36;
  • 43) 0.246 373 876 367 36 × 2 = 0 + 0.492 747 752 734 72;
  • 44) 0.492 747 752 734 72 × 2 = 0 + 0.985 495 505 469 44;
  • 45) 0.985 495 505 469 44 × 2 = 1 + 0.970 991 010 938 88;
  • 46) 0.970 991 010 938 88 × 2 = 1 + 0.941 982 021 877 76;
  • 47) 0.941 982 021 877 76 × 2 = 1 + 0.883 964 043 755 52;
  • 48) 0.883 964 043 755 52 × 2 = 1 + 0.767 928 087 511 04;
  • 49) 0.767 928 087 511 04 × 2 = 1 + 0.535 856 175 022 08;
  • 50) 0.535 856 175 022 08 × 2 = 1 + 0.071 712 350 044 16;
  • 51) 0.071 712 350 044 16 × 2 = 0 + 0.143 424 700 088 32;
  • 52) 0.143 424 700 088 32 × 2 = 0 + 0.286 849 400 176 64;
  • 53) 0.286 849 400 176 64 × 2 = 0 + 0.573 698 800 353 28;
  • 54) 0.573 698 800 353 28 × 2 = 1 + 0.147 397 600 706 56;
  • 55) 0.147 397 600 706 56 × 2 = 0 + 0.294 795 201 413 12;
  • 56) 0.294 795 201 413 12 × 2 = 0 + 0.589 590 402 826 24;
  • 57) 0.589 590 402 826 24 × 2 = 1 + 0.179 180 805 652 48;

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:


0.033 343 988 073 34(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1011 0000 1111 1100 0100 1(2)

6. Positive number before normalization:

0.033 343 988 073 34(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1011 0000 1111 1100 0100 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.033 343 988 073 34(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1011 0000 1111 1100 0100 1(2) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1011 0000 1111 1100 0100 1(2) × 20 =

1.0001 0001 0010 0111 0110 1001 0100 1001 0110 0001 1111 1000 1001(2) × 2-5


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): -5

Mantissa (not normalized):
1.0001 0001 0010 0111 0110 1001 0100 1001 0110 0001 1111 1000 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 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) =

1018(10) =

011 1111 1010(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 0001 0010 0111 0110 1001 0100 1001 0110 0001 1111 1000 1001 =

0001 0001 0010 0111 0110 1001 0100 1001 0110 0001 1111 1000 1001


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 1010

Mantissa (52 bits) =
0001 0001 0010 0111 0110 1001 0100 1001 0110 0001 1111 1000 1001


Decimal number -0.033 343 988 073 34 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0001 0001 0010 0111 0110 1001 0100 1001 0110 0001 1111 1000 1001

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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