-0.033 343 988 072 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.033 343 988 072 41(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
-0.033 343 988 072 41(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:

|-0.033 343 988 072 41| = 0.033 343 988 072 41


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: 0.033 343 988 072 41.

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) 0.033 343 988 072 41 × 2 = 0 + 0.066 687 976 144 82;
  • 2) 0.066 687 976 144 82 × 2 = 0 + 0.133 375 952 289 64;
  • 3) 0.133 375 952 289 64 × 2 = 0 + 0.266 751 904 579 28;
  • 4) 0.266 751 904 579 28 × 2 = 0 + 0.533 503 809 158 56;
  • 5) 0.533 503 809 158 56 × 2 = 1 + 0.067 007 618 317 12;
  • 6) 0.067 007 618 317 12 × 2 = 0 + 0.134 015 236 634 24;
  • 7) 0.134 015 236 634 24 × 2 = 0 + 0.268 030 473 268 48;
  • 8) 0.268 030 473 268 48 × 2 = 0 + 0.536 060 946 536 96;
  • 9) 0.536 060 946 536 96 × 2 = 1 + 0.072 121 893 073 92;
  • 10) 0.072 121 893 073 92 × 2 = 0 + 0.144 243 786 147 84;
  • 11) 0.144 243 786 147 84 × 2 = 0 + 0.288 487 572 295 68;
  • 12) 0.288 487 572 295 68 × 2 = 0 + 0.576 975 144 591 36;
  • 13) 0.576 975 144 591 36 × 2 = 1 + 0.153 950 289 182 72;
  • 14) 0.153 950 289 182 72 × 2 = 0 + 0.307 900 578 365 44;
  • 15) 0.307 900 578 365 44 × 2 = 0 + 0.615 801 156 730 88;
  • 16) 0.615 801 156 730 88 × 2 = 1 + 0.231 602 313 461 76;
  • 17) 0.231 602 313 461 76 × 2 = 0 + 0.463 204 626 923 52;
  • 18) 0.463 204 626 923 52 × 2 = 0 + 0.926 409 253 847 04;
  • 19) 0.926 409 253 847 04 × 2 = 1 + 0.852 818 507 694 08;
  • 20) 0.852 818 507 694 08 × 2 = 1 + 0.705 637 015 388 16;
  • 21) 0.705 637 015 388 16 × 2 = 1 + 0.411 274 030 776 32;
  • 22) 0.411 274 030 776 32 × 2 = 0 + 0.822 548 061 552 64;
  • 23) 0.822 548 061 552 64 × 2 = 1 + 0.645 096 123 105 28;
  • 24) 0.645 096 123 105 28 × 2 = 1 + 0.290 192 246 210 56;
  • 25) 0.290 192 246 210 56 × 2 = 0 + 0.580 384 492 421 12;
  • 26) 0.580 384 492 421 12 × 2 = 1 + 0.160 768 984 842 24;
  • 27) 0.160 768 984 842 24 × 2 = 0 + 0.321 537 969 684 48;
  • 28) 0.321 537 969 684 48 × 2 = 0 + 0.643 075 939 368 96;
  • 29) 0.643 075 939 368 96 × 2 = 1 + 0.286 151 878 737 92;
  • 30) 0.286 151 878 737 92 × 2 = 0 + 0.572 303 757 475 84;
  • 31) 0.572 303 757 475 84 × 2 = 1 + 0.144 607 514 951 68;
  • 32) 0.144 607 514 951 68 × 2 = 0 + 0.289 215 029 903 36;
  • 33) 0.289 215 029 903 36 × 2 = 0 + 0.578 430 059 806 72;
  • 34) 0.578 430 059 806 72 × 2 = 1 + 0.156 860 119 613 44;
  • 35) 0.156 860 119 613 44 × 2 = 0 + 0.313 720 239 226 88;
  • 36) 0.313 720 239 226 88 × 2 = 0 + 0.627 440 478 453 76;
  • 37) 0.627 440 478 453 76 × 2 = 1 + 0.254 880 956 907 52;
  • 38) 0.254 880 956 907 52 × 2 = 0 + 0.509 761 913 815 04;
  • 39) 0.509 761 913 815 04 × 2 = 1 + 0.019 523 827 630 08;
  • 40) 0.019 523 827 630 08 × 2 = 0 + 0.039 047 655 260 16;
  • 41) 0.039 047 655 260 16 × 2 = 0 + 0.078 095 310 520 32;
  • 42) 0.078 095 310 520 32 × 2 = 0 + 0.156 190 621 040 64;
  • 43) 0.156 190 621 040 64 × 2 = 0 + 0.312 381 242 081 28;
  • 44) 0.312 381 242 081 28 × 2 = 0 + 0.624 762 484 162 56;
  • 45) 0.624 762 484 162 56 × 2 = 1 + 0.249 524 968 325 12;
  • 46) 0.249 524 968 325 12 × 2 = 0 + 0.499 049 936 650 24;
  • 47) 0.499 049 936 650 24 × 2 = 0 + 0.998 099 873 300 48;
  • 48) 0.998 099 873 300 48 × 2 = 1 + 0.996 199 746 600 96;
  • 49) 0.996 199 746 600 96 × 2 = 1 + 0.992 399 493 201 92;
  • 50) 0.992 399 493 201 92 × 2 = 1 + 0.984 798 986 403 84;
  • 51) 0.984 798 986 403 84 × 2 = 1 + 0.969 597 972 807 68;
  • 52) 0.969 597 972 807 68 × 2 = 1 + 0.939 195 945 615 36;
  • 53) 0.939 195 945 615 36 × 2 = 1 + 0.878 391 891 230 72;
  • 54) 0.878 391 891 230 72 × 2 = 1 + 0.756 783 782 461 44;
  • 55) 0.756 783 782 461 44 × 2 = 1 + 0.513 567 564 922 88;
  • 56) 0.513 567 564 922 88 × 2 = 1 + 0.027 135 129 845 76;
  • 57) 0.027 135 129 845 76 × 2 = 0 + 0.054 270 259 691 52;

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:


0.033 343 988 072 41(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1010 0000 1001 1111 1111 0(2)

6. Positive number before normalization:

0.033 343 988 072 41(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1010 0000 1001 1111 1111 0(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:


0.033 343 988 072 41(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1010 0000 1001 1111 1111 0(2) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1010 0000 1001 1111 1111 0(2) × 20 =

1.0001 0001 0010 0111 0110 1001 0100 1001 0100 0001 0011 1111 1110(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 0100 0001 0011 1111 1110


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 0100 0001 0011 1111 1110 =

0001 0001 0010 0111 0110 1001 0100 1001 0100 0001 0011 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 1010

Mantissa (52 bits) =
0001 0001 0010 0111 0110 1001 0100 1001 0100 0001 0011 1111 1110


Decimal number -0.033 343 988 072 41 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0001 0001 0010 0111 0110 1001 0100 1001 0100 0001 0011 1111 1110

» Convert decimal -0.033 343 988 071 76 to 64 bit double precision IEEE 754 binary floating point representation standard

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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