-0.000 282 006 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 06(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
-0.000 282 006 06(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:

|-0.000 282 006 06| = 0.000 282 006 06


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: 0.000 282 006 06.

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) 0.000 282 006 06 × 2 = 0 + 0.000 564 012 12;
  • 2) 0.000 564 012 12 × 2 = 0 + 0.001 128 024 24;
  • 3) 0.001 128 024 24 × 2 = 0 + 0.002 256 048 48;
  • 4) 0.002 256 048 48 × 2 = 0 + 0.004 512 096 96;
  • 5) 0.004 512 096 96 × 2 = 0 + 0.009 024 193 92;
  • 6) 0.009 024 193 92 × 2 = 0 + 0.018 048 387 84;
  • 7) 0.018 048 387 84 × 2 = 0 + 0.036 096 775 68;
  • 8) 0.036 096 775 68 × 2 = 0 + 0.072 193 551 36;
  • 9) 0.072 193 551 36 × 2 = 0 + 0.144 387 102 72;
  • 10) 0.144 387 102 72 × 2 = 0 + 0.288 774 205 44;
  • 11) 0.288 774 205 44 × 2 = 0 + 0.577 548 410 88;
  • 12) 0.577 548 410 88 × 2 = 1 + 0.155 096 821 76;
  • 13) 0.155 096 821 76 × 2 = 0 + 0.310 193 643 52;
  • 14) 0.310 193 643 52 × 2 = 0 + 0.620 387 287 04;
  • 15) 0.620 387 287 04 × 2 = 1 + 0.240 774 574 08;
  • 16) 0.240 774 574 08 × 2 = 0 + 0.481 549 148 16;
  • 17) 0.481 549 148 16 × 2 = 0 + 0.963 098 296 32;
  • 18) 0.963 098 296 32 × 2 = 1 + 0.926 196 592 64;
  • 19) 0.926 196 592 64 × 2 = 1 + 0.852 393 185 28;
  • 20) 0.852 393 185 28 × 2 = 1 + 0.704 786 370 56;
  • 21) 0.704 786 370 56 × 2 = 1 + 0.409 572 741 12;
  • 22) 0.409 572 741 12 × 2 = 0 + 0.819 145 482 24;
  • 23) 0.819 145 482 24 × 2 = 1 + 0.638 290 964 48;
  • 24) 0.638 290 964 48 × 2 = 1 + 0.276 581 928 96;
  • 25) 0.276 581 928 96 × 2 = 0 + 0.553 163 857 92;
  • 26) 0.553 163 857 92 × 2 = 1 + 0.106 327 715 84;
  • 27) 0.106 327 715 84 × 2 = 0 + 0.212 655 431 68;
  • 28) 0.212 655 431 68 × 2 = 0 + 0.425 310 863 36;
  • 29) 0.425 310 863 36 × 2 = 0 + 0.850 621 726 72;
  • 30) 0.850 621 726 72 × 2 = 1 + 0.701 243 453 44;
  • 31) 0.701 243 453 44 × 2 = 1 + 0.402 486 906 88;
  • 32) 0.402 486 906 88 × 2 = 0 + 0.804 973 813 76;
  • 33) 0.804 973 813 76 × 2 = 1 + 0.609 947 627 52;
  • 34) 0.609 947 627 52 × 2 = 1 + 0.219 895 255 04;
  • 35) 0.219 895 255 04 × 2 = 0 + 0.439 790 510 08;
  • 36) 0.439 790 510 08 × 2 = 0 + 0.879 581 020 16;
  • 37) 0.879 581 020 16 × 2 = 1 + 0.759 162 040 32;
  • 38) 0.759 162 040 32 × 2 = 1 + 0.518 324 080 64;
  • 39) 0.518 324 080 64 × 2 = 1 + 0.036 648 161 28;
  • 40) 0.036 648 161 28 × 2 = 0 + 0.073 296 322 56;
  • 41) 0.073 296 322 56 × 2 = 0 + 0.146 592 645 12;
  • 42) 0.146 592 645 12 × 2 = 0 + 0.293 185 290 24;
  • 43) 0.293 185 290 24 × 2 = 0 + 0.586 370 580 48;
  • 44) 0.586 370 580 48 × 2 = 1 + 0.172 741 160 96;
  • 45) 0.172 741 160 96 × 2 = 0 + 0.345 482 321 92;
  • 46) 0.345 482 321 92 × 2 = 0 + 0.690 964 643 84;
  • 47) 0.690 964 643 84 × 2 = 1 + 0.381 929 287 68;
  • 48) 0.381 929 287 68 × 2 = 0 + 0.763 858 575 36;
  • 49) 0.763 858 575 36 × 2 = 1 + 0.527 717 150 72;
  • 50) 0.527 717 150 72 × 2 = 1 + 0.055 434 301 44;
  • 51) 0.055 434 301 44 × 2 = 0 + 0.110 868 602 88;
  • 52) 0.110 868 602 88 × 2 = 0 + 0.221 737 205 76;
  • 53) 0.221 737 205 76 × 2 = 0 + 0.443 474 411 52;
  • 54) 0.443 474 411 52 × 2 = 0 + 0.886 948 823 04;
  • 55) 0.886 948 823 04 × 2 = 1 + 0.773 897 646 08;
  • 56) 0.773 897 646 08 × 2 = 1 + 0.547 795 292 16;
  • 57) 0.547 795 292 16 × 2 = 1 + 0.095 590 584 32;
  • 58) 0.095 590 584 32 × 2 = 0 + 0.191 181 168 64;
  • 59) 0.191 181 168 64 × 2 = 0 + 0.382 362 337 28;
  • 60) 0.382 362 337 28 × 2 = 0 + 0.764 724 674 56;
  • 61) 0.764 724 674 56 × 2 = 1 + 0.529 449 349 12;
  • 62) 0.529 449 349 12 × 2 = 1 + 0.058 898 698 24;
  • 63) 0.058 898 698 24 × 2 = 0 + 0.117 797 396 48;
  • 64) 0.117 797 396 48 × 2 = 0 + 0.235 594 792 96;

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:


0.000 282 006 06(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1100 1110 0001 0010 1100 0011 1000 1100(2)

6. Positive number before normalization:

0.000 282 006 06(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1100 1110 0001 0010 1100 0011 1000 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:


0.000 282 006 06(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1100 1110 0001 0010 1100 0011 1000 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1100 1110 0001 0010 1100 0011 1000 1100(2) × 20 =

1.0010 0111 1011 0100 0110 1100 1110 0001 0010 1100 0011 1000 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 0100 0110 1100 1110 0001 0010 1100 0011 1000 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 0100 0110 1100 1110 0001 0010 1100 0011 1000 1100 =

0010 0111 1011 0100 0110 1100 1110 0001 0010 1100 0011 1000 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 0100 0110 1100 1110 0001 0010 1100 0011 1000 1100


Decimal number -0.000 282 006 06 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1100 1110 0001 0010 1100 0011 1000 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