-0.000 282 009 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 009 4(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
-0.000 282 009 4(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:

|-0.000 282 009 4| = 0.000 282 009 4


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: 0.000 282 009 4.

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) 0.000 282 009 4 × 2 = 0 + 0.000 564 018 8;
  • 2) 0.000 564 018 8 × 2 = 0 + 0.001 128 037 6;
  • 3) 0.001 128 037 6 × 2 = 0 + 0.002 256 075 2;
  • 4) 0.002 256 075 2 × 2 = 0 + 0.004 512 150 4;
  • 5) 0.004 512 150 4 × 2 = 0 + 0.009 024 300 8;
  • 6) 0.009 024 300 8 × 2 = 0 + 0.018 048 601 6;
  • 7) 0.018 048 601 6 × 2 = 0 + 0.036 097 203 2;
  • 8) 0.036 097 203 2 × 2 = 0 + 0.072 194 406 4;
  • 9) 0.072 194 406 4 × 2 = 0 + 0.144 388 812 8;
  • 10) 0.144 388 812 8 × 2 = 0 + 0.288 777 625 6;
  • 11) 0.288 777 625 6 × 2 = 0 + 0.577 555 251 2;
  • 12) 0.577 555 251 2 × 2 = 1 + 0.155 110 502 4;
  • 13) 0.155 110 502 4 × 2 = 0 + 0.310 221 004 8;
  • 14) 0.310 221 004 8 × 2 = 0 + 0.620 442 009 6;
  • 15) 0.620 442 009 6 × 2 = 1 + 0.240 884 019 2;
  • 16) 0.240 884 019 2 × 2 = 0 + 0.481 768 038 4;
  • 17) 0.481 768 038 4 × 2 = 0 + 0.963 536 076 8;
  • 18) 0.963 536 076 8 × 2 = 1 + 0.927 072 153 6;
  • 19) 0.927 072 153 6 × 2 = 1 + 0.854 144 307 2;
  • 20) 0.854 144 307 2 × 2 = 1 + 0.708 288 614 4;
  • 21) 0.708 288 614 4 × 2 = 1 + 0.416 577 228 8;
  • 22) 0.416 577 228 8 × 2 = 0 + 0.833 154 457 6;
  • 23) 0.833 154 457 6 × 2 = 1 + 0.666 308 915 2;
  • 24) 0.666 308 915 2 × 2 = 1 + 0.332 617 830 4;
  • 25) 0.332 617 830 4 × 2 = 0 + 0.665 235 660 8;
  • 26) 0.665 235 660 8 × 2 = 1 + 0.330 471 321 6;
  • 27) 0.330 471 321 6 × 2 = 0 + 0.660 942 643 2;
  • 28) 0.660 942 643 2 × 2 = 1 + 0.321 885 286 4;
  • 29) 0.321 885 286 4 × 2 = 0 + 0.643 770 572 8;
  • 30) 0.643 770 572 8 × 2 = 1 + 0.287 541 145 6;
  • 31) 0.287 541 145 6 × 2 = 0 + 0.575 082 291 2;
  • 32) 0.575 082 291 2 × 2 = 1 + 0.150 164 582 4;
  • 33) 0.150 164 582 4 × 2 = 0 + 0.300 329 164 8;
  • 34) 0.300 329 164 8 × 2 = 0 + 0.600 658 329 6;
  • 35) 0.600 658 329 6 × 2 = 1 + 0.201 316 659 2;
  • 36) 0.201 316 659 2 × 2 = 0 + 0.402 633 318 4;
  • 37) 0.402 633 318 4 × 2 = 0 + 0.805 266 636 8;
  • 38) 0.805 266 636 8 × 2 = 1 + 0.610 533 273 6;
  • 39) 0.610 533 273 6 × 2 = 1 + 0.221 066 547 2;
  • 40) 0.221 066 547 2 × 2 = 0 + 0.442 133 094 4;
  • 41) 0.442 133 094 4 × 2 = 0 + 0.884 266 188 8;
  • 42) 0.884 266 188 8 × 2 = 1 + 0.768 532 377 6;
  • 43) 0.768 532 377 6 × 2 = 1 + 0.537 064 755 2;
  • 44) 0.537 064 755 2 × 2 = 1 + 0.074 129 510 4;
  • 45) 0.074 129 510 4 × 2 = 0 + 0.148 259 020 8;
  • 46) 0.148 259 020 8 × 2 = 0 + 0.296 518 041 6;
  • 47) 0.296 518 041 6 × 2 = 0 + 0.593 036 083 2;
  • 48) 0.593 036 083 2 × 2 = 1 + 0.186 072 166 4;
  • 49) 0.186 072 166 4 × 2 = 0 + 0.372 144 332 8;
  • 50) 0.372 144 332 8 × 2 = 0 + 0.744 288 665 6;
  • 51) 0.744 288 665 6 × 2 = 1 + 0.488 577 331 2;
  • 52) 0.488 577 331 2 × 2 = 0 + 0.977 154 662 4;
  • 53) 0.977 154 662 4 × 2 = 1 + 0.954 309 324 8;
  • 54) 0.954 309 324 8 × 2 = 1 + 0.908 618 649 6;
  • 55) 0.908 618 649 6 × 2 = 1 + 0.817 237 299 2;
  • 56) 0.817 237 299 2 × 2 = 1 + 0.634 474 598 4;
  • 57) 0.634 474 598 4 × 2 = 1 + 0.268 949 196 8;
  • 58) 0.268 949 196 8 × 2 = 0 + 0.537 898 393 6;
  • 59) 0.537 898 393 6 × 2 = 1 + 0.075 796 787 2;
  • 60) 0.075 796 787 2 × 2 = 0 + 0.151 593 574 4;
  • 61) 0.151 593 574 4 × 2 = 0 + 0.303 187 148 8;
  • 62) 0.303 187 148 8 × 2 = 0 + 0.606 374 297 6;
  • 63) 0.606 374 297 6 × 2 = 1 + 0.212 748 595 2;
  • 64) 0.212 748 595 2 × 2 = 0 + 0.425 497 190 4;

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:


0.000 282 009 4(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 0010 0110 0111 0001 0010 1111 1010 0010(2)

6. Positive number before normalization:

0.000 282 009 4(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 0010 0110 0111 0001 0010 1111 1010 0010(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:


0.000 282 009 4(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 0010 0110 0111 0001 0010 1111 1010 0010(2) =

0.0000 0000 0001 0010 0111 1011 0101 0101 0010 0110 0111 0001 0010 1111 1010 0010(2) × 20 =

1.0010 0111 1011 0101 0101 0010 0110 0111 0001 0010 1111 1010 0010(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 0101 0010 0110 0111 0001 0010 1111 1010 0010


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 0101 0010 0110 0111 0001 0010 1111 1010 0010 =

0010 0111 1011 0101 0101 0010 0110 0111 0001 0010 1111 1010 0010


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 0101 0010 0110 0111 0001 0010 1111 1010 0010


Decimal number -0.000 282 009 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0101 0010 0110 0111 0001 0010 1111 1010 0010

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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