-0.000 282 006 29 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 29(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 29(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 29| = 0.000 282 006 29


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 29.

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 29 × 2 = 0 + 0.000 564 012 58;
  • 2) 0.000 564 012 58 × 2 = 0 + 0.001 128 025 16;
  • 3) 0.001 128 025 16 × 2 = 0 + 0.002 256 050 32;
  • 4) 0.002 256 050 32 × 2 = 0 + 0.004 512 100 64;
  • 5) 0.004 512 100 64 × 2 = 0 + 0.009 024 201 28;
  • 6) 0.009 024 201 28 × 2 = 0 + 0.018 048 402 56;
  • 7) 0.018 048 402 56 × 2 = 0 + 0.036 096 805 12;
  • 8) 0.036 096 805 12 × 2 = 0 + 0.072 193 610 24;
  • 9) 0.072 193 610 24 × 2 = 0 + 0.144 387 220 48;
  • 10) 0.144 387 220 48 × 2 = 0 + 0.288 774 440 96;
  • 11) 0.288 774 440 96 × 2 = 0 + 0.577 548 881 92;
  • 12) 0.577 548 881 92 × 2 = 1 + 0.155 097 763 84;
  • 13) 0.155 097 763 84 × 2 = 0 + 0.310 195 527 68;
  • 14) 0.310 195 527 68 × 2 = 0 + 0.620 391 055 36;
  • 15) 0.620 391 055 36 × 2 = 1 + 0.240 782 110 72;
  • 16) 0.240 782 110 72 × 2 = 0 + 0.481 564 221 44;
  • 17) 0.481 564 221 44 × 2 = 0 + 0.963 128 442 88;
  • 18) 0.963 128 442 88 × 2 = 1 + 0.926 256 885 76;
  • 19) 0.926 256 885 76 × 2 = 1 + 0.852 513 771 52;
  • 20) 0.852 513 771 52 × 2 = 1 + 0.705 027 543 04;
  • 21) 0.705 027 543 04 × 2 = 1 + 0.410 055 086 08;
  • 22) 0.410 055 086 08 × 2 = 0 + 0.820 110 172 16;
  • 23) 0.820 110 172 16 × 2 = 1 + 0.640 220 344 32;
  • 24) 0.640 220 344 32 × 2 = 1 + 0.280 440 688 64;
  • 25) 0.280 440 688 64 × 2 = 0 + 0.560 881 377 28;
  • 26) 0.560 881 377 28 × 2 = 1 + 0.121 762 754 56;
  • 27) 0.121 762 754 56 × 2 = 0 + 0.243 525 509 12;
  • 28) 0.243 525 509 12 × 2 = 0 + 0.487 051 018 24;
  • 29) 0.487 051 018 24 × 2 = 0 + 0.974 102 036 48;
  • 30) 0.974 102 036 48 × 2 = 1 + 0.948 204 072 96;
  • 31) 0.948 204 072 96 × 2 = 1 + 0.896 408 145 92;
  • 32) 0.896 408 145 92 × 2 = 1 + 0.792 816 291 84;
  • 33) 0.792 816 291 84 × 2 = 1 + 0.585 632 583 68;
  • 34) 0.585 632 583 68 × 2 = 1 + 0.171 265 167 36;
  • 35) 0.171 265 167 36 × 2 = 0 + 0.342 530 334 72;
  • 36) 0.342 530 334 72 × 2 = 0 + 0.685 060 669 44;
  • 37) 0.685 060 669 44 × 2 = 1 + 0.370 121 338 88;
  • 38) 0.370 121 338 88 × 2 = 0 + 0.740 242 677 76;
  • 39) 0.740 242 677 76 × 2 = 1 + 0.480 485 355 52;
  • 40) 0.480 485 355 52 × 2 = 0 + 0.960 970 711 04;
  • 41) 0.960 970 711 04 × 2 = 1 + 0.921 941 422 08;
  • 42) 0.921 941 422 08 × 2 = 1 + 0.843 882 844 16;
  • 43) 0.843 882 844 16 × 2 = 1 + 0.687 765 688 32;
  • 44) 0.687 765 688 32 × 2 = 1 + 0.375 531 376 64;
  • 45) 0.375 531 376 64 × 2 = 0 + 0.751 062 753 28;
  • 46) 0.751 062 753 28 × 2 = 1 + 0.502 125 506 56;
  • 47) 0.502 125 506 56 × 2 = 1 + 0.004 251 013 12;
  • 48) 0.004 251 013 12 × 2 = 0 + 0.008 502 026 24;
  • 49) 0.008 502 026 24 × 2 = 0 + 0.017 004 052 48;
  • 50) 0.017 004 052 48 × 2 = 0 + 0.034 008 104 96;
  • 51) 0.034 008 104 96 × 2 = 0 + 0.068 016 209 92;
  • 52) 0.068 016 209 92 × 2 = 0 + 0.136 032 419 84;
  • 53) 0.136 032 419 84 × 2 = 0 + 0.272 064 839 68;
  • 54) 0.272 064 839 68 × 2 = 0 + 0.544 129 679 36;
  • 55) 0.544 129 679 36 × 2 = 1 + 0.088 259 358 72;
  • 56) 0.088 259 358 72 × 2 = 0 + 0.176 518 717 44;
  • 57) 0.176 518 717 44 × 2 = 0 + 0.353 037 434 88;
  • 58) 0.353 037 434 88 × 2 = 0 + 0.706 074 869 76;
  • 59) 0.706 074 869 76 × 2 = 1 + 0.412 149 739 52;
  • 60) 0.412 149 739 52 × 2 = 0 + 0.824 299 479 04;
  • 61) 0.824 299 479 04 × 2 = 1 + 0.648 598 958 08;
  • 62) 0.648 598 958 08 × 2 = 1 + 0.297 197 916 16;
  • 63) 0.297 197 916 16 × 2 = 0 + 0.594 395 832 32;
  • 64) 0.594 395 832 32 × 2 = 1 + 0.188 791 664 64;

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 29(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 1010 1111 0110 0000 0010 0010 1101(2)

6. Positive number before normalization:

0.000 282 006 29(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 1010 1111 0110 0000 0010 0010 1101(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 29(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 1010 1111 0110 0000 0010 0010 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 1010 1111 0110 0000 0010 0010 1101(2) × 20 =

1.0010 0111 1011 0100 0111 1100 1010 1111 0110 0000 0010 0010 1101(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 0111 1100 1010 1111 0110 0000 0010 0010 1101


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 0111 1100 1010 1111 0110 0000 0010 0010 1101 =

0010 0111 1011 0100 0111 1100 1010 1111 0110 0000 0010 0010 1101


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 0111 1100 1010 1111 0110 0000 0010 0010 1101


Decimal number -0.000 282 006 29 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1100 1010 1111 0110 0000 0010 0010 1101

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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