-0.000 282 008 67 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 008 67(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 67(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 67| = 0.000 282 008 67


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 67.

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 67 × 2 = 0 + 0.000 564 017 34;
  • 2) 0.000 564 017 34 × 2 = 0 + 0.001 128 034 68;
  • 3) 0.001 128 034 68 × 2 = 0 + 0.002 256 069 36;
  • 4) 0.002 256 069 36 × 2 = 0 + 0.004 512 138 72;
  • 5) 0.004 512 138 72 × 2 = 0 + 0.009 024 277 44;
  • 6) 0.009 024 277 44 × 2 = 0 + 0.018 048 554 88;
  • 7) 0.018 048 554 88 × 2 = 0 + 0.036 097 109 76;
  • 8) 0.036 097 109 76 × 2 = 0 + 0.072 194 219 52;
  • 9) 0.072 194 219 52 × 2 = 0 + 0.144 388 439 04;
  • 10) 0.144 388 439 04 × 2 = 0 + 0.288 776 878 08;
  • 11) 0.288 776 878 08 × 2 = 0 + 0.577 553 756 16;
  • 12) 0.577 553 756 16 × 2 = 1 + 0.155 107 512 32;
  • 13) 0.155 107 512 32 × 2 = 0 + 0.310 215 024 64;
  • 14) 0.310 215 024 64 × 2 = 0 + 0.620 430 049 28;
  • 15) 0.620 430 049 28 × 2 = 1 + 0.240 860 098 56;
  • 16) 0.240 860 098 56 × 2 = 0 + 0.481 720 197 12;
  • 17) 0.481 720 197 12 × 2 = 0 + 0.963 440 394 24;
  • 18) 0.963 440 394 24 × 2 = 1 + 0.926 880 788 48;
  • 19) 0.926 880 788 48 × 2 = 1 + 0.853 761 576 96;
  • 20) 0.853 761 576 96 × 2 = 1 + 0.707 523 153 92;
  • 21) 0.707 523 153 92 × 2 = 1 + 0.415 046 307 84;
  • 22) 0.415 046 307 84 × 2 = 0 + 0.830 092 615 68;
  • 23) 0.830 092 615 68 × 2 = 1 + 0.660 185 231 36;
  • 24) 0.660 185 231 36 × 2 = 1 + 0.320 370 462 72;
  • 25) 0.320 370 462 72 × 2 = 0 + 0.640 740 925 44;
  • 26) 0.640 740 925 44 × 2 = 1 + 0.281 481 850 88;
  • 27) 0.281 481 850 88 × 2 = 0 + 0.562 963 701 76;
  • 28) 0.562 963 701 76 × 2 = 1 + 0.125 927 403 52;
  • 29) 0.125 927 403 52 × 2 = 0 + 0.251 854 807 04;
  • 30) 0.251 854 807 04 × 2 = 0 + 0.503 709 614 08;
  • 31) 0.503 709 614 08 × 2 = 1 + 0.007 419 228 16;
  • 32) 0.007 419 228 16 × 2 = 0 + 0.014 838 456 32;
  • 33) 0.014 838 456 32 × 2 = 0 + 0.029 676 912 64;
  • 34) 0.029 676 912 64 × 2 = 0 + 0.059 353 825 28;
  • 35) 0.059 353 825 28 × 2 = 0 + 0.118 707 650 56;
  • 36) 0.118 707 650 56 × 2 = 0 + 0.237 415 301 12;
  • 37) 0.237 415 301 12 × 2 = 0 + 0.474 830 602 24;
  • 38) 0.474 830 602 24 × 2 = 0 + 0.949 661 204 48;
  • 39) 0.949 661 204 48 × 2 = 1 + 0.899 322 408 96;
  • 40) 0.899 322 408 96 × 2 = 1 + 0.798 644 817 92;
  • 41) 0.798 644 817 92 × 2 = 1 + 0.597 289 635 84;
  • 42) 0.597 289 635 84 × 2 = 1 + 0.194 579 271 68;
  • 43) 0.194 579 271 68 × 2 = 0 + 0.389 158 543 36;
  • 44) 0.389 158 543 36 × 2 = 0 + 0.778 317 086 72;
  • 45) 0.778 317 086 72 × 2 = 1 + 0.556 634 173 44;
  • 46) 0.556 634 173 44 × 2 = 1 + 0.113 268 346 88;
  • 47) 0.113 268 346 88 × 2 = 0 + 0.226 536 693 76;
  • 48) 0.226 536 693 76 × 2 = 0 + 0.453 073 387 52;
  • 49) 0.453 073 387 52 × 2 = 0 + 0.906 146 775 04;
  • 50) 0.906 146 775 04 × 2 = 1 + 0.812 293 550 08;
  • 51) 0.812 293 550 08 × 2 = 1 + 0.624 587 100 16;
  • 52) 0.624 587 100 16 × 2 = 1 + 0.249 174 200 32;
  • 53) 0.249 174 200 32 × 2 = 0 + 0.498 348 400 64;
  • 54) 0.498 348 400 64 × 2 = 0 + 0.996 696 801 28;
  • 55) 0.996 696 801 28 × 2 = 1 + 0.993 393 602 56;
  • 56) 0.993 393 602 56 × 2 = 1 + 0.986 787 205 12;
  • 57) 0.986 787 205 12 × 2 = 1 + 0.973 574 410 24;
  • 58) 0.973 574 410 24 × 2 = 1 + 0.947 148 820 48;
  • 59) 0.947 148 820 48 × 2 = 1 + 0.894 297 640 96;
  • 60) 0.894 297 640 96 × 2 = 1 + 0.788 595 281 92;
  • 61) 0.788 595 281 92 × 2 = 1 + 0.577 190 563 84;
  • 62) 0.577 190 563 84 × 2 = 1 + 0.154 381 127 68;
  • 63) 0.154 381 127 68 × 2 = 0 + 0.308 762 255 36;
  • 64) 0.308 762 255 36 × 2 = 0 + 0.617 524 510 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:


0.000 282 008 67(10) =

0.0000 0000 0001 0010 0111 1011 0101 0010 0000 0011 1100 1100 0111 0011 1111 1100(2)

6. Positive number before normalization:

0.000 282 008 67(10) =

0.0000 0000 0001 0010 0111 1011 0101 0010 0000 0011 1100 1100 0111 0011 1111 1100(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 67(10) =

0.0000 0000 0001 0010 0111 1011 0101 0010 0000 0011 1100 1100 0111 0011 1111 1100(2) =

0.0000 0000 0001 0010 0111 1011 0101 0010 0000 0011 1100 1100 0111 0011 1111 1100(2) × 20 =

1.0010 0111 1011 0101 0010 0000 0011 1100 1100 0111 0011 1111 1100(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 0010 0000 0011 1100 1100 0111 0011 1111 1100


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 0010 0000 0011 1100 1100 0111 0011 1111 1100 =

0010 0111 1011 0101 0010 0000 0011 1100 1100 0111 0011 1111 1100


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 0010 0000 0011 1100 1100 0111 0011 1111 1100


Decimal number -0.000 282 008 67 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0010 0000 0011 1100 1100 0111 0011 1111 1100

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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