-0.000 282 038 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 038 2(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
-0.000 282 038 2(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:

|-0.000 282 038 2| = 0.000 282 038 2


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: 0.000 282 038 2.

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) 0.000 282 038 2 × 2 = 0 + 0.000 564 076 4;
  • 2) 0.000 564 076 4 × 2 = 0 + 0.001 128 152 8;
  • 3) 0.001 128 152 8 × 2 = 0 + 0.002 256 305 6;
  • 4) 0.002 256 305 6 × 2 = 0 + 0.004 512 611 2;
  • 5) 0.004 512 611 2 × 2 = 0 + 0.009 025 222 4;
  • 6) 0.009 025 222 4 × 2 = 0 + 0.018 050 444 8;
  • 7) 0.018 050 444 8 × 2 = 0 + 0.036 100 889 6;
  • 8) 0.036 100 889 6 × 2 = 0 + 0.072 201 779 2;
  • 9) 0.072 201 779 2 × 2 = 0 + 0.144 403 558 4;
  • 10) 0.144 403 558 4 × 2 = 0 + 0.288 807 116 8;
  • 11) 0.288 807 116 8 × 2 = 0 + 0.577 614 233 6;
  • 12) 0.577 614 233 6 × 2 = 1 + 0.155 228 467 2;
  • 13) 0.155 228 467 2 × 2 = 0 + 0.310 456 934 4;
  • 14) 0.310 456 934 4 × 2 = 0 + 0.620 913 868 8;
  • 15) 0.620 913 868 8 × 2 = 1 + 0.241 827 737 6;
  • 16) 0.241 827 737 6 × 2 = 0 + 0.483 655 475 2;
  • 17) 0.483 655 475 2 × 2 = 0 + 0.967 310 950 4;
  • 18) 0.967 310 950 4 × 2 = 1 + 0.934 621 900 8;
  • 19) 0.934 621 900 8 × 2 = 1 + 0.869 243 801 6;
  • 20) 0.869 243 801 6 × 2 = 1 + 0.738 487 603 2;
  • 21) 0.738 487 603 2 × 2 = 1 + 0.476 975 206 4;
  • 22) 0.476 975 206 4 × 2 = 0 + 0.953 950 412 8;
  • 23) 0.953 950 412 8 × 2 = 1 + 0.907 900 825 6;
  • 24) 0.907 900 825 6 × 2 = 1 + 0.815 801 651 2;
  • 25) 0.815 801 651 2 × 2 = 1 + 0.631 603 302 4;
  • 26) 0.631 603 302 4 × 2 = 1 + 0.263 206 604 8;
  • 27) 0.263 206 604 8 × 2 = 0 + 0.526 413 209 6;
  • 28) 0.526 413 209 6 × 2 = 1 + 0.052 826 419 2;
  • 29) 0.052 826 419 2 × 2 = 0 + 0.105 652 838 4;
  • 30) 0.105 652 838 4 × 2 = 0 + 0.211 305 676 8;
  • 31) 0.211 305 676 8 × 2 = 0 + 0.422 611 353 6;
  • 32) 0.422 611 353 6 × 2 = 0 + 0.845 222 707 2;
  • 33) 0.845 222 707 2 × 2 = 1 + 0.690 445 414 4;
  • 34) 0.690 445 414 4 × 2 = 1 + 0.380 890 828 8;
  • 35) 0.380 890 828 8 × 2 = 0 + 0.761 781 657 6;
  • 36) 0.761 781 657 6 × 2 = 1 + 0.523 563 315 2;
  • 37) 0.523 563 315 2 × 2 = 1 + 0.047 126 630 4;
  • 38) 0.047 126 630 4 × 2 = 0 + 0.094 253 260 8;
  • 39) 0.094 253 260 8 × 2 = 0 + 0.188 506 521 6;
  • 40) 0.188 506 521 6 × 2 = 0 + 0.377 013 043 2;
  • 41) 0.377 013 043 2 × 2 = 0 + 0.754 026 086 4;
  • 42) 0.754 026 086 4 × 2 = 1 + 0.508 052 172 8;
  • 43) 0.508 052 172 8 × 2 = 1 + 0.016 104 345 6;
  • 44) 0.016 104 345 6 × 2 = 0 + 0.032 208 691 2;
  • 45) 0.032 208 691 2 × 2 = 0 + 0.064 417 382 4;
  • 46) 0.064 417 382 4 × 2 = 0 + 0.128 834 764 8;
  • 47) 0.128 834 764 8 × 2 = 0 + 0.257 669 529 6;
  • 48) 0.257 669 529 6 × 2 = 0 + 0.515 339 059 2;
  • 49) 0.515 339 059 2 × 2 = 1 + 0.030 678 118 4;
  • 50) 0.030 678 118 4 × 2 = 0 + 0.061 356 236 8;
  • 51) 0.061 356 236 8 × 2 = 0 + 0.122 712 473 6;
  • 52) 0.122 712 473 6 × 2 = 0 + 0.245 424 947 2;
  • 53) 0.245 424 947 2 × 2 = 0 + 0.490 849 894 4;
  • 54) 0.490 849 894 4 × 2 = 0 + 0.981 699 788 8;
  • 55) 0.981 699 788 8 × 2 = 1 + 0.963 399 577 6;
  • 56) 0.963 399 577 6 × 2 = 1 + 0.926 799 155 2;
  • 57) 0.926 799 155 2 × 2 = 1 + 0.853 598 310 4;
  • 58) 0.853 598 310 4 × 2 = 1 + 0.707 196 620 8;
  • 59) 0.707 196 620 8 × 2 = 1 + 0.414 393 241 6;
  • 60) 0.414 393 241 6 × 2 = 0 + 0.828 786 483 2;
  • 61) 0.828 786 483 2 × 2 = 1 + 0.657 572 966 4;
  • 62) 0.657 572 966 4 × 2 = 1 + 0.315 145 932 8;
  • 63) 0.315 145 932 8 × 2 = 0 + 0.630 291 865 6;
  • 64) 0.630 291 865 6 × 2 = 1 + 0.260 583 731 2;

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:


0.000 282 038 2(10) =

0.0000 0000 0001 0010 0111 1011 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101(2)

6. Positive number before normalization:

0.000 282 038 2(10) =

0.0000 0000 0001 0010 0111 1011 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101(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:


0.000 282 038 2(10) =

0.0000 0000 0001 0010 0111 1011 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101(2) =

0.0000 0000 0001 0010 0111 1011 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101(2) × 20 =

1.0010 0111 1011 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101(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 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101


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 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101 =

0010 0111 1011 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101


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 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101


Decimal number -0.000 282 038 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1101 0000 1101 1000 0110 0000 1000 0011 1110 1101

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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