-0.000 282 058 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 058 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
-0.000 282 058 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:

|-0.000 282 058 2| = 0.000 282 058 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: 0.000 282 058 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) 0.000 282 058 2 × 2 = 0 + 0.000 564 116 4;
  • 2) 0.000 564 116 4 × 2 = 0 + 0.001 128 232 8;
  • 3) 0.001 128 232 8 × 2 = 0 + 0.002 256 465 6;
  • 4) 0.002 256 465 6 × 2 = 0 + 0.004 512 931 2;
  • 5) 0.004 512 931 2 × 2 = 0 + 0.009 025 862 4;
  • 6) 0.009 025 862 4 × 2 = 0 + 0.018 051 724 8;
  • 7) 0.018 051 724 8 × 2 = 0 + 0.036 103 449 6;
  • 8) 0.036 103 449 6 × 2 = 0 + 0.072 206 899 2;
  • 9) 0.072 206 899 2 × 2 = 0 + 0.144 413 798 4;
  • 10) 0.144 413 798 4 × 2 = 0 + 0.288 827 596 8;
  • 11) 0.288 827 596 8 × 2 = 0 + 0.577 655 193 6;
  • 12) 0.577 655 193 6 × 2 = 1 + 0.155 310 387 2;
  • 13) 0.155 310 387 2 × 2 = 0 + 0.310 620 774 4;
  • 14) 0.310 620 774 4 × 2 = 0 + 0.621 241 548 8;
  • 15) 0.621 241 548 8 × 2 = 1 + 0.242 483 097 6;
  • 16) 0.242 483 097 6 × 2 = 0 + 0.484 966 195 2;
  • 17) 0.484 966 195 2 × 2 = 0 + 0.969 932 390 4;
  • 18) 0.969 932 390 4 × 2 = 1 + 0.939 864 780 8;
  • 19) 0.939 864 780 8 × 2 = 1 + 0.879 729 561 6;
  • 20) 0.879 729 561 6 × 2 = 1 + 0.759 459 123 2;
  • 21) 0.759 459 123 2 × 2 = 1 + 0.518 918 246 4;
  • 22) 0.518 918 246 4 × 2 = 1 + 0.037 836 492 8;
  • 23) 0.037 836 492 8 × 2 = 0 + 0.075 672 985 6;
  • 24) 0.075 672 985 6 × 2 = 0 + 0.151 345 971 2;
  • 25) 0.151 345 971 2 × 2 = 0 + 0.302 691 942 4;
  • 26) 0.302 691 942 4 × 2 = 0 + 0.605 383 884 8;
  • 27) 0.605 383 884 8 × 2 = 1 + 0.210 767 769 6;
  • 28) 0.210 767 769 6 × 2 = 0 + 0.421 535 539 2;
  • 29) 0.421 535 539 2 × 2 = 0 + 0.843 071 078 4;
  • 30) 0.843 071 078 4 × 2 = 1 + 0.686 142 156 8;
  • 31) 0.686 142 156 8 × 2 = 1 + 0.372 284 313 6;
  • 32) 0.372 284 313 6 × 2 = 0 + 0.744 568 627 2;
  • 33) 0.744 568 627 2 × 2 = 1 + 0.489 137 254 4;
  • 34) 0.489 137 254 4 × 2 = 0 + 0.978 274 508 8;
  • 35) 0.978 274 508 8 × 2 = 1 + 0.956 549 017 6;
  • 36) 0.956 549 017 6 × 2 = 1 + 0.913 098 035 2;
  • 37) 0.913 098 035 2 × 2 = 1 + 0.826 196 070 4;
  • 38) 0.826 196 070 4 × 2 = 1 + 0.652 392 140 8;
  • 39) 0.652 392 140 8 × 2 = 1 + 0.304 784 281 6;
  • 40) 0.304 784 281 6 × 2 = 0 + 0.609 568 563 2;
  • 41) 0.609 568 563 2 × 2 = 1 + 0.219 137 126 4;
  • 42) 0.219 137 126 4 × 2 = 0 + 0.438 274 252 8;
  • 43) 0.438 274 252 8 × 2 = 0 + 0.876 548 505 6;
  • 44) 0.876 548 505 6 × 2 = 1 + 0.753 097 011 2;
  • 45) 0.753 097 011 2 × 2 = 1 + 0.506 194 022 4;
  • 46) 0.506 194 022 4 × 2 = 1 + 0.012 388 044 8;
  • 47) 0.012 388 044 8 × 2 = 0 + 0.024 776 089 6;
  • 48) 0.024 776 089 6 × 2 = 0 + 0.049 552 179 2;
  • 49) 0.049 552 179 2 × 2 = 0 + 0.099 104 358 4;
  • 50) 0.099 104 358 4 × 2 = 0 + 0.198 208 716 8;
  • 51) 0.198 208 716 8 × 2 = 0 + 0.396 417 433 6;
  • 52) 0.396 417 433 6 × 2 = 0 + 0.792 834 867 2;
  • 53) 0.792 834 867 2 × 2 = 1 + 0.585 669 734 4;
  • 54) 0.585 669 734 4 × 2 = 1 + 0.171 339 468 8;
  • 55) 0.171 339 468 8 × 2 = 0 + 0.342 678 937 6;
  • 56) 0.342 678 937 6 × 2 = 0 + 0.685 357 875 2;
  • 57) 0.685 357 875 2 × 2 = 1 + 0.370 715 750 4;
  • 58) 0.370 715 750 4 × 2 = 0 + 0.741 431 500 8;
  • 59) 0.741 431 500 8 × 2 = 1 + 0.482 863 001 6;
  • 60) 0.482 863 001 6 × 2 = 0 + 0.965 726 003 2;
  • 61) 0.965 726 003 2 × 2 = 1 + 0.931 452 006 4;
  • 62) 0.931 452 006 4 × 2 = 1 + 0.862 904 012 8;
  • 63) 0.862 904 012 8 × 2 = 1 + 0.725 808 025 6;
  • 64) 0.725 808 025 6 × 2 = 1 + 0.451 616 051 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:


0.000 282 058 2(10) =

0.0000 0000 0001 0010 0111 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111(2)

6. Positive number before normalization:

0.000 282 058 2(10) =

0.0000 0000 0001 0010 0111 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111(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:


0.000 282 058 2(10) =

0.0000 0000 0001 0010 0111 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111(2) =

0.0000 0000 0001 0010 0111 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111(2) × 20 =

1.0010 0111 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111(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 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111


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 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111 =

0010 0111 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111


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 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111


Decimal number -0.000 282 058 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 0010 0110 1011 1110 1001 1100 0000 1100 1010 1111

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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