-0.000 282 005 73 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 73(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 73(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 73| = 0.000 282 005 73


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 73.

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 73 × 2 = 0 + 0.000 564 011 46;
  • 2) 0.000 564 011 46 × 2 = 0 + 0.001 128 022 92;
  • 3) 0.001 128 022 92 × 2 = 0 + 0.002 256 045 84;
  • 4) 0.002 256 045 84 × 2 = 0 + 0.004 512 091 68;
  • 5) 0.004 512 091 68 × 2 = 0 + 0.009 024 183 36;
  • 6) 0.009 024 183 36 × 2 = 0 + 0.018 048 366 72;
  • 7) 0.018 048 366 72 × 2 = 0 + 0.036 096 733 44;
  • 8) 0.036 096 733 44 × 2 = 0 + 0.072 193 466 88;
  • 9) 0.072 193 466 88 × 2 = 0 + 0.144 386 933 76;
  • 10) 0.144 386 933 76 × 2 = 0 + 0.288 773 867 52;
  • 11) 0.288 773 867 52 × 2 = 0 + 0.577 547 735 04;
  • 12) 0.577 547 735 04 × 2 = 1 + 0.155 095 470 08;
  • 13) 0.155 095 470 08 × 2 = 0 + 0.310 190 940 16;
  • 14) 0.310 190 940 16 × 2 = 0 + 0.620 381 880 32;
  • 15) 0.620 381 880 32 × 2 = 1 + 0.240 763 760 64;
  • 16) 0.240 763 760 64 × 2 = 0 + 0.481 527 521 28;
  • 17) 0.481 527 521 28 × 2 = 0 + 0.963 055 042 56;
  • 18) 0.963 055 042 56 × 2 = 1 + 0.926 110 085 12;
  • 19) 0.926 110 085 12 × 2 = 1 + 0.852 220 170 24;
  • 20) 0.852 220 170 24 × 2 = 1 + 0.704 440 340 48;
  • 21) 0.704 440 340 48 × 2 = 1 + 0.408 880 680 96;
  • 22) 0.408 880 680 96 × 2 = 0 + 0.817 761 361 92;
  • 23) 0.817 761 361 92 × 2 = 1 + 0.635 522 723 84;
  • 24) 0.635 522 723 84 × 2 = 1 + 0.271 045 447 68;
  • 25) 0.271 045 447 68 × 2 = 0 + 0.542 090 895 36;
  • 26) 0.542 090 895 36 × 2 = 1 + 0.084 181 790 72;
  • 27) 0.084 181 790 72 × 2 = 0 + 0.168 363 581 44;
  • 28) 0.168 363 581 44 × 2 = 0 + 0.336 727 162 88;
  • 29) 0.336 727 162 88 × 2 = 0 + 0.673 454 325 76;
  • 30) 0.673 454 325 76 × 2 = 1 + 0.346 908 651 52;
  • 31) 0.346 908 651 52 × 2 = 0 + 0.693 817 303 04;
  • 32) 0.693 817 303 04 × 2 = 1 + 0.387 634 606 08;
  • 33) 0.387 634 606 08 × 2 = 0 + 0.775 269 212 16;
  • 34) 0.775 269 212 16 × 2 = 1 + 0.550 538 424 32;
  • 35) 0.550 538 424 32 × 2 = 1 + 0.101 076 848 64;
  • 36) 0.101 076 848 64 × 2 = 0 + 0.202 153 697 28;
  • 37) 0.202 153 697 28 × 2 = 0 + 0.404 307 394 56;
  • 38) 0.404 307 394 56 × 2 = 0 + 0.808 614 789 12;
  • 39) 0.808 614 789 12 × 2 = 1 + 0.617 229 578 24;
  • 40) 0.617 229 578 24 × 2 = 1 + 0.234 459 156 48;
  • 41) 0.234 459 156 48 × 2 = 0 + 0.468 918 312 96;
  • 42) 0.468 918 312 96 × 2 = 0 + 0.937 836 625 92;
  • 43) 0.937 836 625 92 × 2 = 1 + 0.875 673 251 84;
  • 44) 0.875 673 251 84 × 2 = 1 + 0.751 346 503 68;
  • 45) 0.751 346 503 68 × 2 = 1 + 0.502 693 007 36;
  • 46) 0.502 693 007 36 × 2 = 1 + 0.005 386 014 72;
  • 47) 0.005 386 014 72 × 2 = 0 + 0.010 772 029 44;
  • 48) 0.010 772 029 44 × 2 = 0 + 0.021 544 058 88;
  • 49) 0.021 544 058 88 × 2 = 0 + 0.043 088 117 76;
  • 50) 0.043 088 117 76 × 2 = 0 + 0.086 176 235 52;
  • 51) 0.086 176 235 52 × 2 = 0 + 0.172 352 471 04;
  • 52) 0.172 352 471 04 × 2 = 0 + 0.344 704 942 08;
  • 53) 0.344 704 942 08 × 2 = 0 + 0.689 409 884 16;
  • 54) 0.689 409 884 16 × 2 = 1 + 0.378 819 768 32;
  • 55) 0.378 819 768 32 × 2 = 0 + 0.757 639 536 64;
  • 56) 0.757 639 536 64 × 2 = 1 + 0.515 279 073 28;
  • 57) 0.515 279 073 28 × 2 = 1 + 0.030 558 146 56;
  • 58) 0.030 558 146 56 × 2 = 0 + 0.061 116 293 12;
  • 59) 0.061 116 293 12 × 2 = 0 + 0.122 232 586 24;
  • 60) 0.122 232 586 24 × 2 = 0 + 0.244 465 172 48;
  • 61) 0.244 465 172 48 × 2 = 0 + 0.488 930 344 96;
  • 62) 0.488 930 344 96 × 2 = 0 + 0.977 860 689 92;
  • 63) 0.977 860 689 92 × 2 = 1 + 0.955 721 379 84;
  • 64) 0.955 721 379 84 × 2 = 1 + 0.911 442 759 68;

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 73(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 0011 0011 1100 0000 0101 1000 0011(2)

6. Positive number before normalization:

0.000 282 005 73(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 0011 0011 1100 0000 0101 1000 0011(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 73(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 0011 0011 1100 0000 0101 1000 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 0011 0011 1100 0000 0101 1000 0011(2) × 20 =

1.0010 0111 1011 0100 0101 0110 0011 0011 1100 0000 0101 1000 0011(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 0110 0011 0011 1100 0000 0101 1000 0011


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 0110 0011 0011 1100 0000 0101 1000 0011 =

0010 0111 1011 0100 0101 0110 0011 0011 1100 0000 0101 1000 0011


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 0110 0011 0011 1100 0000 0101 1000 0011


Decimal number -0.000 282 005 73 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0110 0011 0011 1100 0000 0101 1000 0011

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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