-0.000 282 005 862 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 862 6(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
-0.000 282 005 862 6(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:

|-0.000 282 005 862 6| = 0.000 282 005 862 6


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: 0.000 282 005 862 6.

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) 0.000 282 005 862 6 × 2 = 0 + 0.000 564 011 725 2;
  • 2) 0.000 564 011 725 2 × 2 = 0 + 0.001 128 023 450 4;
  • 3) 0.001 128 023 450 4 × 2 = 0 + 0.002 256 046 900 8;
  • 4) 0.002 256 046 900 8 × 2 = 0 + 0.004 512 093 801 6;
  • 5) 0.004 512 093 801 6 × 2 = 0 + 0.009 024 187 603 2;
  • 6) 0.009 024 187 603 2 × 2 = 0 + 0.018 048 375 206 4;
  • 7) 0.018 048 375 206 4 × 2 = 0 + 0.036 096 750 412 8;
  • 8) 0.036 096 750 412 8 × 2 = 0 + 0.072 193 500 825 6;
  • 9) 0.072 193 500 825 6 × 2 = 0 + 0.144 387 001 651 2;
  • 10) 0.144 387 001 651 2 × 2 = 0 + 0.288 774 003 302 4;
  • 11) 0.288 774 003 302 4 × 2 = 0 + 0.577 548 006 604 8;
  • 12) 0.577 548 006 604 8 × 2 = 1 + 0.155 096 013 209 6;
  • 13) 0.155 096 013 209 6 × 2 = 0 + 0.310 192 026 419 2;
  • 14) 0.310 192 026 419 2 × 2 = 0 + 0.620 384 052 838 4;
  • 15) 0.620 384 052 838 4 × 2 = 1 + 0.240 768 105 676 8;
  • 16) 0.240 768 105 676 8 × 2 = 0 + 0.481 536 211 353 6;
  • 17) 0.481 536 211 353 6 × 2 = 0 + 0.963 072 422 707 2;
  • 18) 0.963 072 422 707 2 × 2 = 1 + 0.926 144 845 414 4;
  • 19) 0.926 144 845 414 4 × 2 = 1 + 0.852 289 690 828 8;
  • 20) 0.852 289 690 828 8 × 2 = 1 + 0.704 579 381 657 6;
  • 21) 0.704 579 381 657 6 × 2 = 1 + 0.409 158 763 315 2;
  • 22) 0.409 158 763 315 2 × 2 = 0 + 0.818 317 526 630 4;
  • 23) 0.818 317 526 630 4 × 2 = 1 + 0.636 635 053 260 8;
  • 24) 0.636 635 053 260 8 × 2 = 1 + 0.273 270 106 521 6;
  • 25) 0.273 270 106 521 6 × 2 = 0 + 0.546 540 213 043 2;
  • 26) 0.546 540 213 043 2 × 2 = 1 + 0.093 080 426 086 4;
  • 27) 0.093 080 426 086 4 × 2 = 0 + 0.186 160 852 172 8;
  • 28) 0.186 160 852 172 8 × 2 = 0 + 0.372 321 704 345 6;
  • 29) 0.372 321 704 345 6 × 2 = 0 + 0.744 643 408 691 2;
  • 30) 0.744 643 408 691 2 × 2 = 1 + 0.489 286 817 382 4;
  • 31) 0.489 286 817 382 4 × 2 = 0 + 0.978 573 634 764 8;
  • 32) 0.978 573 634 764 8 × 2 = 1 + 0.957 147 269 529 6;
  • 33) 0.957 147 269 529 6 × 2 = 1 + 0.914 294 539 059 2;
  • 34) 0.914 294 539 059 2 × 2 = 1 + 0.828 589 078 118 4;
  • 35) 0.828 589 078 118 4 × 2 = 1 + 0.657 178 156 236 8;
  • 36) 0.657 178 156 236 8 × 2 = 1 + 0.314 356 312 473 6;
  • 37) 0.314 356 312 473 6 × 2 = 0 + 0.628 712 624 947 2;
  • 38) 0.628 712 624 947 2 × 2 = 1 + 0.257 425 249 894 4;
  • 39) 0.257 425 249 894 4 × 2 = 0 + 0.514 850 499 788 8;
  • 40) 0.514 850 499 788 8 × 2 = 1 + 0.029 700 999 577 6;
  • 41) 0.029 700 999 577 6 × 2 = 0 + 0.059 401 999 155 2;
  • 42) 0.059 401 999 155 2 × 2 = 0 + 0.118 803 998 310 4;
  • 43) 0.118 803 998 310 4 × 2 = 0 + 0.237 607 996 620 8;
  • 44) 0.237 607 996 620 8 × 2 = 0 + 0.475 215 993 241 6;
  • 45) 0.475 215 993 241 6 × 2 = 0 + 0.950 431 986 483 2;
  • 46) 0.950 431 986 483 2 × 2 = 1 + 0.900 863 972 966 4;
  • 47) 0.900 863 972 966 4 × 2 = 1 + 0.801 727 945 932 8;
  • 48) 0.801 727 945 932 8 × 2 = 1 + 0.603 455 891 865 6;
  • 49) 0.603 455 891 865 6 × 2 = 1 + 0.206 911 783 731 2;
  • 50) 0.206 911 783 731 2 × 2 = 0 + 0.413 823 567 462 4;
  • 51) 0.413 823 567 462 4 × 2 = 0 + 0.827 647 134 924 8;
  • 52) 0.827 647 134 924 8 × 2 = 1 + 0.655 294 269 849 6;
  • 53) 0.655 294 269 849 6 × 2 = 1 + 0.310 588 539 699 2;
  • 54) 0.310 588 539 699 2 × 2 = 0 + 0.621 177 079 398 4;
  • 55) 0.621 177 079 398 4 × 2 = 1 + 0.242 354 158 796 8;
  • 56) 0.242 354 158 796 8 × 2 = 0 + 0.484 708 317 593 6;
  • 57) 0.484 708 317 593 6 × 2 = 0 + 0.969 416 635 187 2;
  • 58) 0.969 416 635 187 2 × 2 = 1 + 0.938 833 270 374 4;
  • 59) 0.938 833 270 374 4 × 2 = 1 + 0.877 666 540 748 8;
  • 60) 0.877 666 540 748 8 × 2 = 1 + 0.755 333 081 497 6;
  • 61) 0.755 333 081 497 6 × 2 = 1 + 0.510 666 162 995 2;
  • 62) 0.510 666 162 995 2 × 2 = 1 + 0.021 332 325 990 4;
  • 63) 0.021 332 325 990 4 × 2 = 0 + 0.042 664 651 980 8;
  • 64) 0.042 664 651 980 8 × 2 = 0 + 0.085 329 303 961 6;

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:


0.000 282 005 862 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0101 0000 0111 1001 1010 0111 1100(2)

6. Positive number before normalization:

0.000 282 005 862 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0101 0000 0111 1001 1010 0111 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:


0.000 282 005 862 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0101 0000 0111 1001 1010 0111 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0101 0000 0111 1001 1010 0111 1100(2) × 20 =

1.0010 0111 1011 0100 0101 1111 0101 0000 0111 1001 1010 0111 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 0101 1111 0101 0000 0111 1001 1010 0111 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 0101 1111 0101 0000 0111 1001 1010 0111 1100 =

0010 0111 1011 0100 0101 1111 0101 0000 0111 1001 1010 0111 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 0101 1111 0101 0000 0111 1001 1010 0111 1100


Decimal number -0.000 282 005 862 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 0101 0000 0111 1001 1010 0111 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