-0.000 282 040 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 040 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
-0.000 282 040 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:

|-0.000 282 040 2| = 0.000 282 040 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: 0.000 282 040 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) 0.000 282 040 2 × 2 = 0 + 0.000 564 080 4;
  • 2) 0.000 564 080 4 × 2 = 0 + 0.001 128 160 8;
  • 3) 0.001 128 160 8 × 2 = 0 + 0.002 256 321 6;
  • 4) 0.002 256 321 6 × 2 = 0 + 0.004 512 643 2;
  • 5) 0.004 512 643 2 × 2 = 0 + 0.009 025 286 4;
  • 6) 0.009 025 286 4 × 2 = 0 + 0.018 050 572 8;
  • 7) 0.018 050 572 8 × 2 = 0 + 0.036 101 145 6;
  • 8) 0.036 101 145 6 × 2 = 0 + 0.072 202 291 2;
  • 9) 0.072 202 291 2 × 2 = 0 + 0.144 404 582 4;
  • 10) 0.144 404 582 4 × 2 = 0 + 0.288 809 164 8;
  • 11) 0.288 809 164 8 × 2 = 0 + 0.577 618 329 6;
  • 12) 0.577 618 329 6 × 2 = 1 + 0.155 236 659 2;
  • 13) 0.155 236 659 2 × 2 = 0 + 0.310 473 318 4;
  • 14) 0.310 473 318 4 × 2 = 0 + 0.620 946 636 8;
  • 15) 0.620 946 636 8 × 2 = 1 + 0.241 893 273 6;
  • 16) 0.241 893 273 6 × 2 = 0 + 0.483 786 547 2;
  • 17) 0.483 786 547 2 × 2 = 0 + 0.967 573 094 4;
  • 18) 0.967 573 094 4 × 2 = 1 + 0.935 146 188 8;
  • 19) 0.935 146 188 8 × 2 = 1 + 0.870 292 377 6;
  • 20) 0.870 292 377 6 × 2 = 1 + 0.740 584 755 2;
  • 21) 0.740 584 755 2 × 2 = 1 + 0.481 169 510 4;
  • 22) 0.481 169 510 4 × 2 = 0 + 0.962 339 020 8;
  • 23) 0.962 339 020 8 × 2 = 1 + 0.924 678 041 6;
  • 24) 0.924 678 041 6 × 2 = 1 + 0.849 356 083 2;
  • 25) 0.849 356 083 2 × 2 = 1 + 0.698 712 166 4;
  • 26) 0.698 712 166 4 × 2 = 1 + 0.397 424 332 8;
  • 27) 0.397 424 332 8 × 2 = 0 + 0.794 848 665 6;
  • 28) 0.794 848 665 6 × 2 = 1 + 0.589 697 331 2;
  • 29) 0.589 697 331 2 × 2 = 1 + 0.179 394 662 4;
  • 30) 0.179 394 662 4 × 2 = 0 + 0.358 789 324 8;
  • 31) 0.358 789 324 8 × 2 = 0 + 0.717 578 649 6;
  • 32) 0.717 578 649 6 × 2 = 1 + 0.435 157 299 2;
  • 33) 0.435 157 299 2 × 2 = 0 + 0.870 314 598 4;
  • 34) 0.870 314 598 4 × 2 = 1 + 0.740 629 196 8;
  • 35) 0.740 629 196 8 × 2 = 1 + 0.481 258 393 6;
  • 36) 0.481 258 393 6 × 2 = 0 + 0.962 516 787 2;
  • 37) 0.962 516 787 2 × 2 = 1 + 0.925 033 574 4;
  • 38) 0.925 033 574 4 × 2 = 1 + 0.850 067 148 8;
  • 39) 0.850 067 148 8 × 2 = 1 + 0.700 134 297 6;
  • 40) 0.700 134 297 6 × 2 = 1 + 0.400 268 595 2;
  • 41) 0.400 268 595 2 × 2 = 0 + 0.800 537 190 4;
  • 42) 0.800 537 190 4 × 2 = 1 + 0.601 074 380 8;
  • 43) 0.601 074 380 8 × 2 = 1 + 0.202 148 761 6;
  • 44) 0.202 148 761 6 × 2 = 0 + 0.404 297 523 2;
  • 45) 0.404 297 523 2 × 2 = 0 + 0.808 595 046 4;
  • 46) 0.808 595 046 4 × 2 = 1 + 0.617 190 092 8;
  • 47) 0.617 190 092 8 × 2 = 1 + 0.234 380 185 6;
  • 48) 0.234 380 185 6 × 2 = 0 + 0.468 760 371 2;
  • 49) 0.468 760 371 2 × 2 = 0 + 0.937 520 742 4;
  • 50) 0.937 520 742 4 × 2 = 1 + 0.875 041 484 8;
  • 51) 0.875 041 484 8 × 2 = 1 + 0.750 082 969 6;
  • 52) 0.750 082 969 6 × 2 = 1 + 0.500 165 939 2;
  • 53) 0.500 165 939 2 × 2 = 1 + 0.000 331 878 4;
  • 54) 0.000 331 878 4 × 2 = 0 + 0.000 663 756 8;
  • 55) 0.000 663 756 8 × 2 = 0 + 0.001 327 513 6;
  • 56) 0.001 327 513 6 × 2 = 0 + 0.002 655 027 2;
  • 57) 0.002 655 027 2 × 2 = 0 + 0.005 310 054 4;
  • 58) 0.005 310 054 4 × 2 = 0 + 0.010 620 108 8;
  • 59) 0.010 620 108 8 × 2 = 0 + 0.021 240 217 6;
  • 60) 0.021 240 217 6 × 2 = 0 + 0.042 480 435 2;
  • 61) 0.042 480 435 2 × 2 = 0 + 0.084 960 870 4;
  • 62) 0.084 960 870 4 × 2 = 0 + 0.169 921 740 8;
  • 63) 0.169 921 740 8 × 2 = 0 + 0.339 843 481 6;
  • 64) 0.339 843 481 6 × 2 = 0 + 0.679 686 963 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:


0.000 282 040 2(10) =

0.0000 0000 0001 0010 0111 1011 1101 1001 0110 1111 0110 0110 0111 1000 0000 0000(2)

6. Positive number before normalization:

0.000 282 040 2(10) =

0.0000 0000 0001 0010 0111 1011 1101 1001 0110 1111 0110 0110 0111 1000 0000 0000(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:


0.000 282 040 2(10) =

0.0000 0000 0001 0010 0111 1011 1101 1001 0110 1111 0110 0110 0111 1000 0000 0000(2) =

0.0000 0000 0001 0010 0111 1011 1101 1001 0110 1111 0110 0110 0111 1000 0000 0000(2) × 20 =

1.0010 0111 1011 1101 1001 0110 1111 0110 0110 0111 1000 0000 0000(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 1001 0110 1111 0110 0110 0111 1000 0000 0000


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 1001 0110 1111 0110 0110 0111 1000 0000 0000 =

0010 0111 1011 1101 1001 0110 1111 0110 0110 0111 1000 0000 0000


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 1001 0110 1111 0110 0110 0111 1000 0000 0000


Decimal number -0.000 282 040 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1101 1001 0110 1111 0110 0110 0111 1000 0000 0000

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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