-0.000 282 003 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 62(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
-0.000 282 003 62(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:

|-0.000 282 003 62| = 0.000 282 003 62


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: 0.000 282 003 62.

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) 0.000 282 003 62 × 2 = 0 + 0.000 564 007 24;
  • 2) 0.000 564 007 24 × 2 = 0 + 0.001 128 014 48;
  • 3) 0.001 128 014 48 × 2 = 0 + 0.002 256 028 96;
  • 4) 0.002 256 028 96 × 2 = 0 + 0.004 512 057 92;
  • 5) 0.004 512 057 92 × 2 = 0 + 0.009 024 115 84;
  • 6) 0.009 024 115 84 × 2 = 0 + 0.018 048 231 68;
  • 7) 0.018 048 231 68 × 2 = 0 + 0.036 096 463 36;
  • 8) 0.036 096 463 36 × 2 = 0 + 0.072 192 926 72;
  • 9) 0.072 192 926 72 × 2 = 0 + 0.144 385 853 44;
  • 10) 0.144 385 853 44 × 2 = 0 + 0.288 771 706 88;
  • 11) 0.288 771 706 88 × 2 = 0 + 0.577 543 413 76;
  • 12) 0.577 543 413 76 × 2 = 1 + 0.155 086 827 52;
  • 13) 0.155 086 827 52 × 2 = 0 + 0.310 173 655 04;
  • 14) 0.310 173 655 04 × 2 = 0 + 0.620 347 310 08;
  • 15) 0.620 347 310 08 × 2 = 1 + 0.240 694 620 16;
  • 16) 0.240 694 620 16 × 2 = 0 + 0.481 389 240 32;
  • 17) 0.481 389 240 32 × 2 = 0 + 0.962 778 480 64;
  • 18) 0.962 778 480 64 × 2 = 1 + 0.925 556 961 28;
  • 19) 0.925 556 961 28 × 2 = 1 + 0.851 113 922 56;
  • 20) 0.851 113 922 56 × 2 = 1 + 0.702 227 845 12;
  • 21) 0.702 227 845 12 × 2 = 1 + 0.404 455 690 24;
  • 22) 0.404 455 690 24 × 2 = 0 + 0.808 911 380 48;
  • 23) 0.808 911 380 48 × 2 = 1 + 0.617 822 760 96;
  • 24) 0.617 822 760 96 × 2 = 1 + 0.235 645 521 92;
  • 25) 0.235 645 521 92 × 2 = 0 + 0.471 291 043 84;
  • 26) 0.471 291 043 84 × 2 = 0 + 0.942 582 087 68;
  • 27) 0.942 582 087 68 × 2 = 1 + 0.885 164 175 36;
  • 28) 0.885 164 175 36 × 2 = 1 + 0.770 328 350 72;
  • 29) 0.770 328 350 72 × 2 = 1 + 0.540 656 701 44;
  • 30) 0.540 656 701 44 × 2 = 1 + 0.081 313 402 88;
  • 31) 0.081 313 402 88 × 2 = 0 + 0.162 626 805 76;
  • 32) 0.162 626 805 76 × 2 = 0 + 0.325 253 611 52;
  • 33) 0.325 253 611 52 × 2 = 0 + 0.650 507 223 04;
  • 34) 0.650 507 223 04 × 2 = 1 + 0.301 014 446 08;
  • 35) 0.301 014 446 08 × 2 = 0 + 0.602 028 892 16;
  • 36) 0.602 028 892 16 × 2 = 1 + 0.204 057 784 32;
  • 37) 0.204 057 784 32 × 2 = 0 + 0.408 115 568 64;
  • 38) 0.408 115 568 64 × 2 = 0 + 0.816 231 137 28;
  • 39) 0.816 231 137 28 × 2 = 1 + 0.632 462 274 56;
  • 40) 0.632 462 274 56 × 2 = 1 + 0.264 924 549 12;
  • 41) 0.264 924 549 12 × 2 = 0 + 0.529 849 098 24;
  • 42) 0.529 849 098 24 × 2 = 1 + 0.059 698 196 48;
  • 43) 0.059 698 196 48 × 2 = 0 + 0.119 396 392 96;
  • 44) 0.119 396 392 96 × 2 = 0 + 0.238 792 785 92;
  • 45) 0.238 792 785 92 × 2 = 0 + 0.477 585 571 84;
  • 46) 0.477 585 571 84 × 2 = 0 + 0.955 171 143 68;
  • 47) 0.955 171 143 68 × 2 = 1 + 0.910 342 287 36;
  • 48) 0.910 342 287 36 × 2 = 1 + 0.820 684 574 72;
  • 49) 0.820 684 574 72 × 2 = 1 + 0.641 369 149 44;
  • 50) 0.641 369 149 44 × 2 = 1 + 0.282 738 298 88;
  • 51) 0.282 738 298 88 × 2 = 0 + 0.565 476 597 76;
  • 52) 0.565 476 597 76 × 2 = 1 + 0.130 953 195 52;
  • 53) 0.130 953 195 52 × 2 = 0 + 0.261 906 391 04;
  • 54) 0.261 906 391 04 × 2 = 0 + 0.523 812 782 08;
  • 55) 0.523 812 782 08 × 2 = 1 + 0.047 625 564 16;
  • 56) 0.047 625 564 16 × 2 = 0 + 0.095 251 128 32;
  • 57) 0.095 251 128 32 × 2 = 0 + 0.190 502 256 64;
  • 58) 0.190 502 256 64 × 2 = 0 + 0.381 004 513 28;
  • 59) 0.381 004 513 28 × 2 = 0 + 0.762 009 026 56;
  • 60) 0.762 009 026 56 × 2 = 1 + 0.524 018 053 12;
  • 61) 0.524 018 053 12 × 2 = 1 + 0.048 036 106 24;
  • 62) 0.048 036 106 24 × 2 = 0 + 0.096 072 212 48;
  • 63) 0.096 072 212 48 × 2 = 0 + 0.192 144 424 96;
  • 64) 0.192 144 424 96 × 2 = 0 + 0.384 288 849 92;

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:


0.000 282 003 62(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000(2)

6. Positive number before normalization:

0.000 282 003 62(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000(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:


0.000 282 003 62(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000(2) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000(2) × 20 =

1.0010 0111 1011 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000(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 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000


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 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000 =

0010 0111 1011 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000


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 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000


Decimal number -0.000 282 003 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1100 0101 0011 0100 0011 1101 0010 0001 1000

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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