-0.000 282 004 98 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 98(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
-0.000 282 004 98(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:

|-0.000 282 004 98| = 0.000 282 004 98


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: 0.000 282 004 98.

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) 0.000 282 004 98 × 2 = 0 + 0.000 564 009 96;
  • 2) 0.000 564 009 96 × 2 = 0 + 0.001 128 019 92;
  • 3) 0.001 128 019 92 × 2 = 0 + 0.002 256 039 84;
  • 4) 0.002 256 039 84 × 2 = 0 + 0.004 512 079 68;
  • 5) 0.004 512 079 68 × 2 = 0 + 0.009 024 159 36;
  • 6) 0.009 024 159 36 × 2 = 0 + 0.018 048 318 72;
  • 7) 0.018 048 318 72 × 2 = 0 + 0.036 096 637 44;
  • 8) 0.036 096 637 44 × 2 = 0 + 0.072 193 274 88;
  • 9) 0.072 193 274 88 × 2 = 0 + 0.144 386 549 76;
  • 10) 0.144 386 549 76 × 2 = 0 + 0.288 773 099 52;
  • 11) 0.288 773 099 52 × 2 = 0 + 0.577 546 199 04;
  • 12) 0.577 546 199 04 × 2 = 1 + 0.155 092 398 08;
  • 13) 0.155 092 398 08 × 2 = 0 + 0.310 184 796 16;
  • 14) 0.310 184 796 16 × 2 = 0 + 0.620 369 592 32;
  • 15) 0.620 369 592 32 × 2 = 1 + 0.240 739 184 64;
  • 16) 0.240 739 184 64 × 2 = 0 + 0.481 478 369 28;
  • 17) 0.481 478 369 28 × 2 = 0 + 0.962 956 738 56;
  • 18) 0.962 956 738 56 × 2 = 1 + 0.925 913 477 12;
  • 19) 0.925 913 477 12 × 2 = 1 + 0.851 826 954 24;
  • 20) 0.851 826 954 24 × 2 = 1 + 0.703 653 908 48;
  • 21) 0.703 653 908 48 × 2 = 1 + 0.407 307 816 96;
  • 22) 0.407 307 816 96 × 2 = 0 + 0.814 615 633 92;
  • 23) 0.814 615 633 92 × 2 = 1 + 0.629 231 267 84;
  • 24) 0.629 231 267 84 × 2 = 1 + 0.258 462 535 68;
  • 25) 0.258 462 535 68 × 2 = 0 + 0.516 925 071 36;
  • 26) 0.516 925 071 36 × 2 = 1 + 0.033 850 142 72;
  • 27) 0.033 850 142 72 × 2 = 0 + 0.067 700 285 44;
  • 28) 0.067 700 285 44 × 2 = 0 + 0.135 400 570 88;
  • 29) 0.135 400 570 88 × 2 = 0 + 0.270 801 141 76;
  • 30) 0.270 801 141 76 × 2 = 0 + 0.541 602 283 52;
  • 31) 0.541 602 283 52 × 2 = 1 + 0.083 204 567 04;
  • 32) 0.083 204 567 04 × 2 = 0 + 0.166 409 134 08;
  • 33) 0.166 409 134 08 × 2 = 0 + 0.332 818 268 16;
  • 34) 0.332 818 268 16 × 2 = 0 + 0.665 636 536 32;
  • 35) 0.665 636 536 32 × 2 = 1 + 0.331 273 072 64;
  • 36) 0.331 273 072 64 × 2 = 0 + 0.662 546 145 28;
  • 37) 0.662 546 145 28 × 2 = 1 + 0.325 092 290 56;
  • 38) 0.325 092 290 56 × 2 = 0 + 0.650 184 581 12;
  • 39) 0.650 184 581 12 × 2 = 1 + 0.300 369 162 24;
  • 40) 0.300 369 162 24 × 2 = 0 + 0.600 738 324 48;
  • 41) 0.600 738 324 48 × 2 = 1 + 0.201 476 648 96;
  • 42) 0.201 476 648 96 × 2 = 0 + 0.402 953 297 92;
  • 43) 0.402 953 297 92 × 2 = 0 + 0.805 906 595 84;
  • 44) 0.805 906 595 84 × 2 = 1 + 0.611 813 191 68;
  • 45) 0.611 813 191 68 × 2 = 1 + 0.223 626 383 36;
  • 46) 0.223 626 383 36 × 2 = 0 + 0.447 252 766 72;
  • 47) 0.447 252 766 72 × 2 = 0 + 0.894 505 533 44;
  • 48) 0.894 505 533 44 × 2 = 1 + 0.789 011 066 88;
  • 49) 0.789 011 066 88 × 2 = 1 + 0.578 022 133 76;
  • 50) 0.578 022 133 76 × 2 = 1 + 0.156 044 267 52;
  • 51) 0.156 044 267 52 × 2 = 0 + 0.312 088 535 04;
  • 52) 0.312 088 535 04 × 2 = 0 + 0.624 177 070 08;
  • 53) 0.624 177 070 08 × 2 = 1 + 0.248 354 140 16;
  • 54) 0.248 354 140 16 × 2 = 0 + 0.496 708 280 32;
  • 55) 0.496 708 280 32 × 2 = 0 + 0.993 416 560 64;
  • 56) 0.993 416 560 64 × 2 = 1 + 0.986 833 121 28;
  • 57) 0.986 833 121 28 × 2 = 1 + 0.973 666 242 56;
  • 58) 0.973 666 242 56 × 2 = 1 + 0.947 332 485 12;
  • 59) 0.947 332 485 12 × 2 = 1 + 0.894 664 970 24;
  • 60) 0.894 664 970 24 × 2 = 1 + 0.789 329 940 48;
  • 61) 0.789 329 940 48 × 2 = 1 + 0.578 659 880 96;
  • 62) 0.578 659 880 96 × 2 = 1 + 0.157 319 761 92;
  • 63) 0.157 319 761 92 × 2 = 0 + 0.314 639 523 84;
  • 64) 0.314 639 523 84 × 2 = 0 + 0.629 279 047 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:


0.000 282 004 98(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 0010 1010 1001 1001 1100 1001 1111 1100(2)

6. Positive number before normalization:

0.000 282 004 98(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 0010 1010 1001 1001 1100 1001 1111 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:


0.000 282 004 98(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 0010 1010 1001 1001 1100 1001 1111 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0010 0010 1010 1001 1001 1100 1001 1111 1100(2) × 20 =

1.0010 0111 1011 0100 0010 0010 1010 1001 1001 1100 1001 1111 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 0100 0010 0010 1010 1001 1001 1100 1001 1111 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 0100 0010 0010 1010 1001 1001 1100 1001 1111 1100 =

0010 0111 1011 0100 0010 0010 1010 1001 1001 1100 1001 1111 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 0100 0010 0010 1010 1001 1001 1100 1001 1111 1100


Decimal number -0.000 282 004 98 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0010 0010 1010 1001 1001 1100 1001 1111 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