-0.000 282 010 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 010 62(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
-0.000 282 010 62(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:

|-0.000 282 010 62| = 0.000 282 010 62


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: 0.000 282 010 62.

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) 0.000 282 010 62 × 2 = 0 + 0.000 564 021 24;
  • 2) 0.000 564 021 24 × 2 = 0 + 0.001 128 042 48;
  • 3) 0.001 128 042 48 × 2 = 0 + 0.002 256 084 96;
  • 4) 0.002 256 084 96 × 2 = 0 + 0.004 512 169 92;
  • 5) 0.004 512 169 92 × 2 = 0 + 0.009 024 339 84;
  • 6) 0.009 024 339 84 × 2 = 0 + 0.018 048 679 68;
  • 7) 0.018 048 679 68 × 2 = 0 + 0.036 097 359 36;
  • 8) 0.036 097 359 36 × 2 = 0 + 0.072 194 718 72;
  • 9) 0.072 194 718 72 × 2 = 0 + 0.144 389 437 44;
  • 10) 0.144 389 437 44 × 2 = 0 + 0.288 778 874 88;
  • 11) 0.288 778 874 88 × 2 = 0 + 0.577 557 749 76;
  • 12) 0.577 557 749 76 × 2 = 1 + 0.155 115 499 52;
  • 13) 0.155 115 499 52 × 2 = 0 + 0.310 230 999 04;
  • 14) 0.310 230 999 04 × 2 = 0 + 0.620 461 998 08;
  • 15) 0.620 461 998 08 × 2 = 1 + 0.240 923 996 16;
  • 16) 0.240 923 996 16 × 2 = 0 + 0.481 847 992 32;
  • 17) 0.481 847 992 32 × 2 = 0 + 0.963 695 984 64;
  • 18) 0.963 695 984 64 × 2 = 1 + 0.927 391 969 28;
  • 19) 0.927 391 969 28 × 2 = 1 + 0.854 783 938 56;
  • 20) 0.854 783 938 56 × 2 = 1 + 0.709 567 877 12;
  • 21) 0.709 567 877 12 × 2 = 1 + 0.419 135 754 24;
  • 22) 0.419 135 754 24 × 2 = 0 + 0.838 271 508 48;
  • 23) 0.838 271 508 48 × 2 = 1 + 0.676 543 016 96;
  • 24) 0.676 543 016 96 × 2 = 1 + 0.353 086 033 92;
  • 25) 0.353 086 033 92 × 2 = 0 + 0.706 172 067 84;
  • 26) 0.706 172 067 84 × 2 = 1 + 0.412 344 135 68;
  • 27) 0.412 344 135 68 × 2 = 0 + 0.824 688 271 36;
  • 28) 0.824 688 271 36 × 2 = 1 + 0.649 376 542 72;
  • 29) 0.649 376 542 72 × 2 = 1 + 0.298 753 085 44;
  • 30) 0.298 753 085 44 × 2 = 0 + 0.597 506 170 88;
  • 31) 0.597 506 170 88 × 2 = 1 + 0.195 012 341 76;
  • 32) 0.195 012 341 76 × 2 = 0 + 0.390 024 683 52;
  • 33) 0.390 024 683 52 × 2 = 0 + 0.780 049 367 04;
  • 34) 0.780 049 367 04 × 2 = 1 + 0.560 098 734 08;
  • 35) 0.560 098 734 08 × 2 = 1 + 0.120 197 468 16;
  • 36) 0.120 197 468 16 × 2 = 0 + 0.240 394 936 32;
  • 37) 0.240 394 936 32 × 2 = 0 + 0.480 789 872 64;
  • 38) 0.480 789 872 64 × 2 = 0 + 0.961 579 745 28;
  • 39) 0.961 579 745 28 × 2 = 1 + 0.923 159 490 56;
  • 40) 0.923 159 490 56 × 2 = 1 + 0.846 318 981 12;
  • 41) 0.846 318 981 12 × 2 = 1 + 0.692 637 962 24;
  • 42) 0.692 637 962 24 × 2 = 1 + 0.385 275 924 48;
  • 43) 0.385 275 924 48 × 2 = 0 + 0.770 551 848 96;
  • 44) 0.770 551 848 96 × 2 = 1 + 0.541 103 697 92;
  • 45) 0.541 103 697 92 × 2 = 1 + 0.082 207 395 84;
  • 46) 0.082 207 395 84 × 2 = 0 + 0.164 414 791 68;
  • 47) 0.164 414 791 68 × 2 = 0 + 0.328 829 583 36;
  • 48) 0.328 829 583 36 × 2 = 0 + 0.657 659 166 72;
  • 49) 0.657 659 166 72 × 2 = 1 + 0.315 318 333 44;
  • 50) 0.315 318 333 44 × 2 = 0 + 0.630 636 666 88;
  • 51) 0.630 636 666 88 × 2 = 1 + 0.261 273 333 76;
  • 52) 0.261 273 333 76 × 2 = 0 + 0.522 546 667 52;
  • 53) 0.522 546 667 52 × 2 = 1 + 0.045 093 335 04;
  • 54) 0.045 093 335 04 × 2 = 0 + 0.090 186 670 08;
  • 55) 0.090 186 670 08 × 2 = 0 + 0.180 373 340 16;
  • 56) 0.180 373 340 16 × 2 = 0 + 0.360 746 680 32;
  • 57) 0.360 746 680 32 × 2 = 0 + 0.721 493 360 64;
  • 58) 0.721 493 360 64 × 2 = 1 + 0.442 986 721 28;
  • 59) 0.442 986 721 28 × 2 = 0 + 0.885 973 442 56;
  • 60) 0.885 973 442 56 × 2 = 1 + 0.771 946 885 12;
  • 61) 0.771 946 885 12 × 2 = 1 + 0.543 893 770 24;
  • 62) 0.543 893 770 24 × 2 = 1 + 0.087 787 540 48;
  • 63) 0.087 787 540 48 × 2 = 0 + 0.175 575 080 96;
  • 64) 0.175 575 080 96 × 2 = 0 + 0.351 150 161 92;

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:


0.000 282 010 62(10) =

0.0000 0000 0001 0010 0111 1011 0101 1010 0110 0011 1101 1000 1010 1000 0101 1100(2)

6. Positive number before normalization:

0.000 282 010 62(10) =

0.0000 0000 0001 0010 0111 1011 0101 1010 0110 0011 1101 1000 1010 1000 0101 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:


0.000 282 010 62(10) =

0.0000 0000 0001 0010 0111 1011 0101 1010 0110 0011 1101 1000 1010 1000 0101 1100(2) =

0.0000 0000 0001 0010 0111 1011 0101 1010 0110 0011 1101 1000 1010 1000 0101 1100(2) × 20 =

1.0010 0111 1011 0101 1010 0110 0011 1101 1000 1010 1000 0101 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 0101 1010 0110 0011 1101 1000 1010 1000 0101 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 0101 1010 0110 0011 1101 1000 1010 1000 0101 1100 =

0010 0111 1011 0101 1010 0110 0011 1101 1000 1010 1000 0101 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 0101 1010 0110 0011 1101 1000 1010 1000 0101 1100


Decimal number -0.000 282 010 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 1010 0110 0011 1101 1000 1010 1000 0101 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