-0.000 282 002 88 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 002 88(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
-0.000 282 002 88(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:

|-0.000 282 002 88| = 0.000 282 002 88


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: 0.000 282 002 88.

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) 0.000 282 002 88 × 2 = 0 + 0.000 564 005 76;
  • 2) 0.000 564 005 76 × 2 = 0 + 0.001 128 011 52;
  • 3) 0.001 128 011 52 × 2 = 0 + 0.002 256 023 04;
  • 4) 0.002 256 023 04 × 2 = 0 + 0.004 512 046 08;
  • 5) 0.004 512 046 08 × 2 = 0 + 0.009 024 092 16;
  • 6) 0.009 024 092 16 × 2 = 0 + 0.018 048 184 32;
  • 7) 0.018 048 184 32 × 2 = 0 + 0.036 096 368 64;
  • 8) 0.036 096 368 64 × 2 = 0 + 0.072 192 737 28;
  • 9) 0.072 192 737 28 × 2 = 0 + 0.144 385 474 56;
  • 10) 0.144 385 474 56 × 2 = 0 + 0.288 770 949 12;
  • 11) 0.288 770 949 12 × 2 = 0 + 0.577 541 898 24;
  • 12) 0.577 541 898 24 × 2 = 1 + 0.155 083 796 48;
  • 13) 0.155 083 796 48 × 2 = 0 + 0.310 167 592 96;
  • 14) 0.310 167 592 96 × 2 = 0 + 0.620 335 185 92;
  • 15) 0.620 335 185 92 × 2 = 1 + 0.240 670 371 84;
  • 16) 0.240 670 371 84 × 2 = 0 + 0.481 340 743 68;
  • 17) 0.481 340 743 68 × 2 = 0 + 0.962 681 487 36;
  • 18) 0.962 681 487 36 × 2 = 1 + 0.925 362 974 72;
  • 19) 0.925 362 974 72 × 2 = 1 + 0.850 725 949 44;
  • 20) 0.850 725 949 44 × 2 = 1 + 0.701 451 898 88;
  • 21) 0.701 451 898 88 × 2 = 1 + 0.402 903 797 76;
  • 22) 0.402 903 797 76 × 2 = 0 + 0.805 807 595 52;
  • 23) 0.805 807 595 52 × 2 = 1 + 0.611 615 191 04;
  • 24) 0.611 615 191 04 × 2 = 1 + 0.223 230 382 08;
  • 25) 0.223 230 382 08 × 2 = 0 + 0.446 460 764 16;
  • 26) 0.446 460 764 16 × 2 = 0 + 0.892 921 528 32;
  • 27) 0.892 921 528 32 × 2 = 1 + 0.785 843 056 64;
  • 28) 0.785 843 056 64 × 2 = 1 + 0.571 686 113 28;
  • 29) 0.571 686 113 28 × 2 = 1 + 0.143 372 226 56;
  • 30) 0.143 372 226 56 × 2 = 0 + 0.286 744 453 12;
  • 31) 0.286 744 453 12 × 2 = 0 + 0.573 488 906 24;
  • 32) 0.573 488 906 24 × 2 = 1 + 0.146 977 812 48;
  • 33) 0.146 977 812 48 × 2 = 0 + 0.293 955 624 96;
  • 34) 0.293 955 624 96 × 2 = 0 + 0.587 911 249 92;
  • 35) 0.587 911 249 92 × 2 = 1 + 0.175 822 499 84;
  • 36) 0.175 822 499 84 × 2 = 0 + 0.351 644 999 68;
  • 37) 0.351 644 999 68 × 2 = 0 + 0.703 289 999 36;
  • 38) 0.703 289 999 36 × 2 = 1 + 0.406 579 998 72;
  • 39) 0.406 579 998 72 × 2 = 0 + 0.813 159 997 44;
  • 40) 0.813 159 997 44 × 2 = 1 + 0.626 319 994 88;
  • 41) 0.626 319 994 88 × 2 = 1 + 0.252 639 989 76;
  • 42) 0.252 639 989 76 × 2 = 0 + 0.505 279 979 52;
  • 43) 0.505 279 979 52 × 2 = 1 + 0.010 559 959 04;
  • 44) 0.010 559 959 04 × 2 = 0 + 0.021 119 918 08;
  • 45) 0.021 119 918 08 × 2 = 0 + 0.042 239 836 16;
  • 46) 0.042 239 836 16 × 2 = 0 + 0.084 479 672 32;
  • 47) 0.084 479 672 32 × 2 = 0 + 0.168 959 344 64;
  • 48) 0.168 959 344 64 × 2 = 0 + 0.337 918 689 28;
  • 49) 0.337 918 689 28 × 2 = 0 + 0.675 837 378 56;
  • 50) 0.675 837 378 56 × 2 = 1 + 0.351 674 757 12;
  • 51) 0.351 674 757 12 × 2 = 0 + 0.703 349 514 24;
  • 52) 0.703 349 514 24 × 2 = 1 + 0.406 699 028 48;
  • 53) 0.406 699 028 48 × 2 = 0 + 0.813 398 056 96;
  • 54) 0.813 398 056 96 × 2 = 1 + 0.626 796 113 92;
  • 55) 0.626 796 113 92 × 2 = 1 + 0.253 592 227 84;
  • 56) 0.253 592 227 84 × 2 = 0 + 0.507 184 455 68;
  • 57) 0.507 184 455 68 × 2 = 1 + 0.014 368 911 36;
  • 58) 0.014 368 911 36 × 2 = 0 + 0.028 737 822 72;
  • 59) 0.028 737 822 72 × 2 = 0 + 0.057 475 645 44;
  • 60) 0.057 475 645 44 × 2 = 0 + 0.114 951 290 88;
  • 61) 0.114 951 290 88 × 2 = 0 + 0.229 902 581 76;
  • 62) 0.229 902 581 76 × 2 = 0 + 0.459 805 163 52;
  • 63) 0.459 805 163 52 × 2 = 0 + 0.919 610 327 04;
  • 64) 0.919 610 327 04 × 2 = 1 + 0.839 220 654 08;

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:


0.000 282 002 88(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0010 0101 1010 0000 0101 0110 1000 0001(2)

6. Positive number before normalization:

0.000 282 002 88(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0010 0101 1010 0000 0101 0110 1000 0001(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:


0.000 282 002 88(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0010 0101 1010 0000 0101 0110 1000 0001(2) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0010 0101 1010 0000 0101 0110 1000 0001(2) × 20 =

1.0010 0111 1011 0011 1001 0010 0101 1010 0000 0101 0110 1000 0001(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 1001 0010 0101 1010 0000 0101 0110 1000 0001


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 1001 0010 0101 1010 0000 0101 0110 1000 0001 =

0010 0111 1011 0011 1001 0010 0101 1010 0000 0101 0110 1000 0001


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 1001 0010 0101 1010 0000 0101 0110 1000 0001


Decimal number -0.000 282 002 88 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1001 0010 0101 1010 0000 0101 0110 1000 0001

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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