-0.000 282 008 98 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 008 98(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
-0.000 282 008 98(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:

|-0.000 282 008 98| = 0.000 282 008 98


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: 0.000 282 008 98.

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) 0.000 282 008 98 × 2 = 0 + 0.000 564 017 96;
  • 2) 0.000 564 017 96 × 2 = 0 + 0.001 128 035 92;
  • 3) 0.001 128 035 92 × 2 = 0 + 0.002 256 071 84;
  • 4) 0.002 256 071 84 × 2 = 0 + 0.004 512 143 68;
  • 5) 0.004 512 143 68 × 2 = 0 + 0.009 024 287 36;
  • 6) 0.009 024 287 36 × 2 = 0 + 0.018 048 574 72;
  • 7) 0.018 048 574 72 × 2 = 0 + 0.036 097 149 44;
  • 8) 0.036 097 149 44 × 2 = 0 + 0.072 194 298 88;
  • 9) 0.072 194 298 88 × 2 = 0 + 0.144 388 597 76;
  • 10) 0.144 388 597 76 × 2 = 0 + 0.288 777 195 52;
  • 11) 0.288 777 195 52 × 2 = 0 + 0.577 554 391 04;
  • 12) 0.577 554 391 04 × 2 = 1 + 0.155 108 782 08;
  • 13) 0.155 108 782 08 × 2 = 0 + 0.310 217 564 16;
  • 14) 0.310 217 564 16 × 2 = 0 + 0.620 435 128 32;
  • 15) 0.620 435 128 32 × 2 = 1 + 0.240 870 256 64;
  • 16) 0.240 870 256 64 × 2 = 0 + 0.481 740 513 28;
  • 17) 0.481 740 513 28 × 2 = 0 + 0.963 481 026 56;
  • 18) 0.963 481 026 56 × 2 = 1 + 0.926 962 053 12;
  • 19) 0.926 962 053 12 × 2 = 1 + 0.853 924 106 24;
  • 20) 0.853 924 106 24 × 2 = 1 + 0.707 848 212 48;
  • 21) 0.707 848 212 48 × 2 = 1 + 0.415 696 424 96;
  • 22) 0.415 696 424 96 × 2 = 0 + 0.831 392 849 92;
  • 23) 0.831 392 849 92 × 2 = 1 + 0.662 785 699 84;
  • 24) 0.662 785 699 84 × 2 = 1 + 0.325 571 399 68;
  • 25) 0.325 571 399 68 × 2 = 0 + 0.651 142 799 36;
  • 26) 0.651 142 799 36 × 2 = 1 + 0.302 285 598 72;
  • 27) 0.302 285 598 72 × 2 = 0 + 0.604 571 197 44;
  • 28) 0.604 571 197 44 × 2 = 1 + 0.209 142 394 88;
  • 29) 0.209 142 394 88 × 2 = 0 + 0.418 284 789 76;
  • 30) 0.418 284 789 76 × 2 = 0 + 0.836 569 579 52;
  • 31) 0.836 569 579 52 × 2 = 1 + 0.673 139 159 04;
  • 32) 0.673 139 159 04 × 2 = 1 + 0.346 278 318 08;
  • 33) 0.346 278 318 08 × 2 = 0 + 0.692 556 636 16;
  • 34) 0.692 556 636 16 × 2 = 1 + 0.385 113 272 32;
  • 35) 0.385 113 272 32 × 2 = 0 + 0.770 226 544 64;
  • 36) 0.770 226 544 64 × 2 = 1 + 0.540 453 089 28;
  • 37) 0.540 453 089 28 × 2 = 1 + 0.080 906 178 56;
  • 38) 0.080 906 178 56 × 2 = 0 + 0.161 812 357 12;
  • 39) 0.161 812 357 12 × 2 = 0 + 0.323 624 714 24;
  • 40) 0.323 624 714 24 × 2 = 0 + 0.647 249 428 48;
  • 41) 0.647 249 428 48 × 2 = 1 + 0.294 498 856 96;
  • 42) 0.294 498 856 96 × 2 = 0 + 0.588 997 713 92;
  • 43) 0.588 997 713 92 × 2 = 1 + 0.177 995 427 84;
  • 44) 0.177 995 427 84 × 2 = 0 + 0.355 990 855 68;
  • 45) 0.355 990 855 68 × 2 = 0 + 0.711 981 711 36;
  • 46) 0.711 981 711 36 × 2 = 1 + 0.423 963 422 72;
  • 47) 0.423 963 422 72 × 2 = 0 + 0.847 926 845 44;
  • 48) 0.847 926 845 44 × 2 = 1 + 0.695 853 690 88;
  • 49) 0.695 853 690 88 × 2 = 1 + 0.391 707 381 76;
  • 50) 0.391 707 381 76 × 2 = 0 + 0.783 414 763 52;
  • 51) 0.783 414 763 52 × 2 = 1 + 0.566 829 527 04;
  • 52) 0.566 829 527 04 × 2 = 1 + 0.133 659 054 08;
  • 53) 0.133 659 054 08 × 2 = 0 + 0.267 318 108 16;
  • 54) 0.267 318 108 16 × 2 = 0 + 0.534 636 216 32;
  • 55) 0.534 636 216 32 × 2 = 1 + 0.069 272 432 64;
  • 56) 0.069 272 432 64 × 2 = 0 + 0.138 544 865 28;
  • 57) 0.138 544 865 28 × 2 = 0 + 0.277 089 730 56;
  • 58) 0.277 089 730 56 × 2 = 0 + 0.554 179 461 12;
  • 59) 0.554 179 461 12 × 2 = 1 + 0.108 358 922 24;
  • 60) 0.108 358 922 24 × 2 = 0 + 0.216 717 844 48;
  • 61) 0.216 717 844 48 × 2 = 0 + 0.433 435 688 96;
  • 62) 0.433 435 688 96 × 2 = 0 + 0.866 871 377 92;
  • 63) 0.866 871 377 92 × 2 = 1 + 0.733 742 755 84;
  • 64) 0.733 742 755 84 × 2 = 1 + 0.467 485 511 68;

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:


0.000 282 008 98(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 0011(2)

6. Positive number before normalization:

0.000 282 008 98(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 0011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 008 98(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 0011(2) =

0.0000 0000 0001 0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 0011(2) × 20 =

1.0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 0011(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 0011 =

0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 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 0011

Mantissa (52 bits) =
0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 0011


Decimal number -0.000 282 008 98 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0011 0101 1000 1010 0101 1011 0010 0010 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