-0.033 343 988 071 76 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.033 343 988 071 76(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
-0.033 343 988 071 76(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:

|-0.033 343 988 071 76| = 0.033 343 988 071 76


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: 0.033 343 988 071 76.

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) 0.033 343 988 071 76 × 2 = 0 + 0.066 687 976 143 52;
  • 2) 0.066 687 976 143 52 × 2 = 0 + 0.133 375 952 287 04;
  • 3) 0.133 375 952 287 04 × 2 = 0 + 0.266 751 904 574 08;
  • 4) 0.266 751 904 574 08 × 2 = 0 + 0.533 503 809 148 16;
  • 5) 0.533 503 809 148 16 × 2 = 1 + 0.067 007 618 296 32;
  • 6) 0.067 007 618 296 32 × 2 = 0 + 0.134 015 236 592 64;
  • 7) 0.134 015 236 592 64 × 2 = 0 + 0.268 030 473 185 28;
  • 8) 0.268 030 473 185 28 × 2 = 0 + 0.536 060 946 370 56;
  • 9) 0.536 060 946 370 56 × 2 = 1 + 0.072 121 892 741 12;
  • 10) 0.072 121 892 741 12 × 2 = 0 + 0.144 243 785 482 24;
  • 11) 0.144 243 785 482 24 × 2 = 0 + 0.288 487 570 964 48;
  • 12) 0.288 487 570 964 48 × 2 = 0 + 0.576 975 141 928 96;
  • 13) 0.576 975 141 928 96 × 2 = 1 + 0.153 950 283 857 92;
  • 14) 0.153 950 283 857 92 × 2 = 0 + 0.307 900 567 715 84;
  • 15) 0.307 900 567 715 84 × 2 = 0 + 0.615 801 135 431 68;
  • 16) 0.615 801 135 431 68 × 2 = 1 + 0.231 602 270 863 36;
  • 17) 0.231 602 270 863 36 × 2 = 0 + 0.463 204 541 726 72;
  • 18) 0.463 204 541 726 72 × 2 = 0 + 0.926 409 083 453 44;
  • 19) 0.926 409 083 453 44 × 2 = 1 + 0.852 818 166 906 88;
  • 20) 0.852 818 166 906 88 × 2 = 1 + 0.705 636 333 813 76;
  • 21) 0.705 636 333 813 76 × 2 = 1 + 0.411 272 667 627 52;
  • 22) 0.411 272 667 627 52 × 2 = 0 + 0.822 545 335 255 04;
  • 23) 0.822 545 335 255 04 × 2 = 1 + 0.645 090 670 510 08;
  • 24) 0.645 090 670 510 08 × 2 = 1 + 0.290 181 341 020 16;
  • 25) 0.290 181 341 020 16 × 2 = 0 + 0.580 362 682 040 32;
  • 26) 0.580 362 682 040 32 × 2 = 1 + 0.160 725 364 080 64;
  • 27) 0.160 725 364 080 64 × 2 = 0 + 0.321 450 728 161 28;
  • 28) 0.321 450 728 161 28 × 2 = 0 + 0.642 901 456 322 56;
  • 29) 0.642 901 456 322 56 × 2 = 1 + 0.285 802 912 645 12;
  • 30) 0.285 802 912 645 12 × 2 = 0 + 0.571 605 825 290 24;
  • 31) 0.571 605 825 290 24 × 2 = 1 + 0.143 211 650 580 48;
  • 32) 0.143 211 650 580 48 × 2 = 0 + 0.286 423 301 160 96;
  • 33) 0.286 423 301 160 96 × 2 = 0 + 0.572 846 602 321 92;
  • 34) 0.572 846 602 321 92 × 2 = 1 + 0.145 693 204 643 84;
  • 35) 0.145 693 204 643 84 × 2 = 0 + 0.291 386 409 287 68;
  • 36) 0.291 386 409 287 68 × 2 = 0 + 0.582 772 818 575 36;
  • 37) 0.582 772 818 575 36 × 2 = 1 + 0.165 545 637 150 72;
  • 38) 0.165 545 637 150 72 × 2 = 0 + 0.331 091 274 301 44;
  • 39) 0.331 091 274 301 44 × 2 = 0 + 0.662 182 548 602 88;
  • 40) 0.662 182 548 602 88 × 2 = 1 + 0.324 365 097 205 76;
  • 41) 0.324 365 097 205 76 × 2 = 0 + 0.648 730 194 411 52;
  • 42) 0.648 730 194 411 52 × 2 = 1 + 0.297 460 388 823 04;
  • 43) 0.297 460 388 823 04 × 2 = 0 + 0.594 920 777 646 08;
  • 44) 0.594 920 777 646 08 × 2 = 1 + 0.189 841 555 292 16;
  • 45) 0.189 841 555 292 16 × 2 = 0 + 0.379 683 110 584 32;
  • 46) 0.379 683 110 584 32 × 2 = 0 + 0.759 366 221 168 64;
  • 47) 0.759 366 221 168 64 × 2 = 1 + 0.518 732 442 337 28;
  • 48) 0.518 732 442 337 28 × 2 = 1 + 0.037 464 884 674 56;
  • 49) 0.037 464 884 674 56 × 2 = 0 + 0.074 929 769 349 12;
  • 50) 0.074 929 769 349 12 × 2 = 0 + 0.149 859 538 698 24;
  • 51) 0.149 859 538 698 24 × 2 = 0 + 0.299 719 077 396 48;
  • 52) 0.299 719 077 396 48 × 2 = 0 + 0.599 438 154 792 96;
  • 53) 0.599 438 154 792 96 × 2 = 1 + 0.198 876 309 585 92;
  • 54) 0.198 876 309 585 92 × 2 = 0 + 0.397 752 619 171 84;
  • 55) 0.397 752 619 171 84 × 2 = 0 + 0.795 505 238 343 68;
  • 56) 0.795 505 238 343 68 × 2 = 1 + 0.591 010 476 687 36;
  • 57) 0.591 010 476 687 36 × 2 = 1 + 0.182 020 953 374 72;

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:


0.033 343 988 071 76(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1001 0101 0011 0000 1001 1(2)

6. Positive number before normalization:

0.033 343 988 071 76(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1001 0101 0011 0000 1001 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:


0.033 343 988 071 76(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1001 0101 0011 0000 1001 1(2) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1001 0101 0011 0000 1001 1(2) × 20 =

1.0001 0001 0010 0111 0110 1001 0100 1001 0010 1010 0110 0001 0011(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 0010 1010 0110 0001 0011


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 0010 1010 0110 0001 0011 =

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


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 0010 1010 0110 0001 0011


Decimal number -0.033 343 988 071 76 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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