-0.000 001 323 71 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 323 71(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
-0.000 001 323 71(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:

|-0.000 001 323 71| = 0.000 001 323 71


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: 0.000 001 323 71.

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) 0.000 001 323 71 × 2 = 0 + 0.000 002 647 42;
  • 2) 0.000 002 647 42 × 2 = 0 + 0.000 005 294 84;
  • 3) 0.000 005 294 84 × 2 = 0 + 0.000 010 589 68;
  • 4) 0.000 010 589 68 × 2 = 0 + 0.000 021 179 36;
  • 5) 0.000 021 179 36 × 2 = 0 + 0.000 042 358 72;
  • 6) 0.000 042 358 72 × 2 = 0 + 0.000 084 717 44;
  • 7) 0.000 084 717 44 × 2 = 0 + 0.000 169 434 88;
  • 8) 0.000 169 434 88 × 2 = 0 + 0.000 338 869 76;
  • 9) 0.000 338 869 76 × 2 = 0 + 0.000 677 739 52;
  • 10) 0.000 677 739 52 × 2 = 0 + 0.001 355 479 04;
  • 11) 0.001 355 479 04 × 2 = 0 + 0.002 710 958 08;
  • 12) 0.002 710 958 08 × 2 = 0 + 0.005 421 916 16;
  • 13) 0.005 421 916 16 × 2 = 0 + 0.010 843 832 32;
  • 14) 0.010 843 832 32 × 2 = 0 + 0.021 687 664 64;
  • 15) 0.021 687 664 64 × 2 = 0 + 0.043 375 329 28;
  • 16) 0.043 375 329 28 × 2 = 0 + 0.086 750 658 56;
  • 17) 0.086 750 658 56 × 2 = 0 + 0.173 501 317 12;
  • 18) 0.173 501 317 12 × 2 = 0 + 0.347 002 634 24;
  • 19) 0.347 002 634 24 × 2 = 0 + 0.694 005 268 48;
  • 20) 0.694 005 268 48 × 2 = 1 + 0.388 010 536 96;
  • 21) 0.388 010 536 96 × 2 = 0 + 0.776 021 073 92;
  • 22) 0.776 021 073 92 × 2 = 1 + 0.552 042 147 84;
  • 23) 0.552 042 147 84 × 2 = 1 + 0.104 084 295 68;
  • 24) 0.104 084 295 68 × 2 = 0 + 0.208 168 591 36;
  • 25) 0.208 168 591 36 × 2 = 0 + 0.416 337 182 72;
  • 26) 0.416 337 182 72 × 2 = 0 + 0.832 674 365 44;
  • 27) 0.832 674 365 44 × 2 = 1 + 0.665 348 730 88;
  • 28) 0.665 348 730 88 × 2 = 1 + 0.330 697 461 76;
  • 29) 0.330 697 461 76 × 2 = 0 + 0.661 394 923 52;
  • 30) 0.661 394 923 52 × 2 = 1 + 0.322 789 847 04;
  • 31) 0.322 789 847 04 × 2 = 0 + 0.645 579 694 08;
  • 32) 0.645 579 694 08 × 2 = 1 + 0.291 159 388 16;
  • 33) 0.291 159 388 16 × 2 = 0 + 0.582 318 776 32;
  • 34) 0.582 318 776 32 × 2 = 1 + 0.164 637 552 64;
  • 35) 0.164 637 552 64 × 2 = 0 + 0.329 275 105 28;
  • 36) 0.329 275 105 28 × 2 = 0 + 0.658 550 210 56;
  • 37) 0.658 550 210 56 × 2 = 1 + 0.317 100 421 12;
  • 38) 0.317 100 421 12 × 2 = 0 + 0.634 200 842 24;
  • 39) 0.634 200 842 24 × 2 = 1 + 0.268 401 684 48;
  • 40) 0.268 401 684 48 × 2 = 0 + 0.536 803 368 96;
  • 41) 0.536 803 368 96 × 2 = 1 + 0.073 606 737 92;
  • 42) 0.073 606 737 92 × 2 = 0 + 0.147 213 475 84;
  • 43) 0.147 213 475 84 × 2 = 0 + 0.294 426 951 68;
  • 44) 0.294 426 951 68 × 2 = 0 + 0.588 853 903 36;
  • 45) 0.588 853 903 36 × 2 = 1 + 0.177 707 806 72;
  • 46) 0.177 707 806 72 × 2 = 0 + 0.355 415 613 44;
  • 47) 0.355 415 613 44 × 2 = 0 + 0.710 831 226 88;
  • 48) 0.710 831 226 88 × 2 = 1 + 0.421 662 453 76;
  • 49) 0.421 662 453 76 × 2 = 0 + 0.843 324 907 52;
  • 50) 0.843 324 907 52 × 2 = 1 + 0.686 649 815 04;
  • 51) 0.686 649 815 04 × 2 = 1 + 0.373 299 630 08;
  • 52) 0.373 299 630 08 × 2 = 0 + 0.746 599 260 16;
  • 53) 0.746 599 260 16 × 2 = 1 + 0.493 198 520 32;
  • 54) 0.493 198 520 32 × 2 = 0 + 0.986 397 040 64;
  • 55) 0.986 397 040 64 × 2 = 1 + 0.972 794 081 28;
  • 56) 0.972 794 081 28 × 2 = 1 + 0.945 588 162 56;
  • 57) 0.945 588 162 56 × 2 = 1 + 0.891 176 325 12;
  • 58) 0.891 176 325 12 × 2 = 1 + 0.782 352 650 24;
  • 59) 0.782 352 650 24 × 2 = 1 + 0.564 705 300 48;
  • 60) 0.564 705 300 48 × 2 = 1 + 0.129 410 600 96;
  • 61) 0.129 410 600 96 × 2 = 0 + 0.258 821 201 92;
  • 62) 0.258 821 201 92 × 2 = 0 + 0.517 642 403 84;
  • 63) 0.517 642 403 84 × 2 = 1 + 0.035 284 807 68;
  • 64) 0.035 284 807 68 × 2 = 0 + 0.070 569 615 36;
  • 65) 0.070 569 615 36 × 2 = 0 + 0.141 139 230 72;
  • 66) 0.141 139 230 72 × 2 = 0 + 0.282 278 461 44;
  • 67) 0.282 278 461 44 × 2 = 0 + 0.564 556 922 88;
  • 68) 0.564 556 922 88 × 2 = 1 + 0.129 113 845 76;
  • 69) 0.129 113 845 76 × 2 = 0 + 0.258 227 691 52;
  • 70) 0.258 227 691 52 × 2 = 0 + 0.516 455 383 04;
  • 71) 0.516 455 383 04 × 2 = 1 + 0.032 910 766 08;
  • 72) 0.032 910 766 08 × 2 = 0 + 0.065 821 532 16;

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:


0.000 001 323 71(10) =

0.0000 0000 0000 0000 0001 0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010(2)

6. Positive number before normalization:

0.000 001 323 71(10) =

0.0000 0000 0000 0000 0001 0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 323 71(10) =

0.0000 0000 0000 0000 0001 0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010(2) =

0.0000 0000 0000 0000 0001 0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010(2) × 20 =

1.0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010(2) × 2-20


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): -20

Mantissa (not normalized):
1.0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1003(10) =

011 1110 1011(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. 0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010 =

0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010


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 1110 1011

Mantissa (52 bits) =
0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010


Decimal number -0.000 001 323 71 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1011 - 0110 0011 0101 0100 1010 1000 1001 0110 1011 1111 0010 0001 0010

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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