-0.000 282 005 913 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 6(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
-0.000 282 005 913 6(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:

|-0.000 282 005 913 6| = 0.000 282 005 913 6


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: 0.000 282 005 913 6.

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) 0.000 282 005 913 6 × 2 = 0 + 0.000 564 011 827 2;
  • 2) 0.000 564 011 827 2 × 2 = 0 + 0.001 128 023 654 4;
  • 3) 0.001 128 023 654 4 × 2 = 0 + 0.002 256 047 308 8;
  • 4) 0.002 256 047 308 8 × 2 = 0 + 0.004 512 094 617 6;
  • 5) 0.004 512 094 617 6 × 2 = 0 + 0.009 024 189 235 2;
  • 6) 0.009 024 189 235 2 × 2 = 0 + 0.018 048 378 470 4;
  • 7) 0.018 048 378 470 4 × 2 = 0 + 0.036 096 756 940 8;
  • 8) 0.036 096 756 940 8 × 2 = 0 + 0.072 193 513 881 6;
  • 9) 0.072 193 513 881 6 × 2 = 0 + 0.144 387 027 763 2;
  • 10) 0.144 387 027 763 2 × 2 = 0 + 0.288 774 055 526 4;
  • 11) 0.288 774 055 526 4 × 2 = 0 + 0.577 548 111 052 8;
  • 12) 0.577 548 111 052 8 × 2 = 1 + 0.155 096 222 105 6;
  • 13) 0.155 096 222 105 6 × 2 = 0 + 0.310 192 444 211 2;
  • 14) 0.310 192 444 211 2 × 2 = 0 + 0.620 384 888 422 4;
  • 15) 0.620 384 888 422 4 × 2 = 1 + 0.240 769 776 844 8;
  • 16) 0.240 769 776 844 8 × 2 = 0 + 0.481 539 553 689 6;
  • 17) 0.481 539 553 689 6 × 2 = 0 + 0.963 079 107 379 2;
  • 18) 0.963 079 107 379 2 × 2 = 1 + 0.926 158 214 758 4;
  • 19) 0.926 158 214 758 4 × 2 = 1 + 0.852 316 429 516 8;
  • 20) 0.852 316 429 516 8 × 2 = 1 + 0.704 632 859 033 6;
  • 21) 0.704 632 859 033 6 × 2 = 1 + 0.409 265 718 067 2;
  • 22) 0.409 265 718 067 2 × 2 = 0 + 0.818 531 436 134 4;
  • 23) 0.818 531 436 134 4 × 2 = 1 + 0.637 062 872 268 8;
  • 24) 0.637 062 872 268 8 × 2 = 1 + 0.274 125 744 537 6;
  • 25) 0.274 125 744 537 6 × 2 = 0 + 0.548 251 489 075 2;
  • 26) 0.548 251 489 075 2 × 2 = 1 + 0.096 502 978 150 4;
  • 27) 0.096 502 978 150 4 × 2 = 0 + 0.193 005 956 300 8;
  • 28) 0.193 005 956 300 8 × 2 = 0 + 0.386 011 912 601 6;
  • 29) 0.386 011 912 601 6 × 2 = 0 + 0.772 023 825 203 2;
  • 30) 0.772 023 825 203 2 × 2 = 1 + 0.544 047 650 406 4;
  • 31) 0.544 047 650 406 4 × 2 = 1 + 0.088 095 300 812 8;
  • 32) 0.088 095 300 812 8 × 2 = 0 + 0.176 190 601 625 6;
  • 33) 0.176 190 601 625 6 × 2 = 0 + 0.352 381 203 251 2;
  • 34) 0.352 381 203 251 2 × 2 = 0 + 0.704 762 406 502 4;
  • 35) 0.704 762 406 502 4 × 2 = 1 + 0.409 524 813 004 8;
  • 36) 0.409 524 813 004 8 × 2 = 0 + 0.819 049 626 009 6;
  • 37) 0.819 049 626 009 6 × 2 = 1 + 0.638 099 252 019 2;
  • 38) 0.638 099 252 019 2 × 2 = 1 + 0.276 198 504 038 4;
  • 39) 0.276 198 504 038 4 × 2 = 0 + 0.552 397 008 076 8;
  • 40) 0.552 397 008 076 8 × 2 = 1 + 0.104 794 016 153 6;
  • 41) 0.104 794 016 153 6 × 2 = 0 + 0.209 588 032 307 2;
  • 42) 0.209 588 032 307 2 × 2 = 0 + 0.419 176 064 614 4;
  • 43) 0.419 176 064 614 4 × 2 = 0 + 0.838 352 129 228 8;
  • 44) 0.838 352 129 228 8 × 2 = 1 + 0.676 704 258 457 6;
  • 45) 0.676 704 258 457 6 × 2 = 1 + 0.353 408 516 915 2;
  • 46) 0.353 408 516 915 2 × 2 = 0 + 0.706 817 033 830 4;
  • 47) 0.706 817 033 830 4 × 2 = 1 + 0.413 634 067 660 8;
  • 48) 0.413 634 067 660 8 × 2 = 0 + 0.827 268 135 321 6;
  • 49) 0.827 268 135 321 6 × 2 = 1 + 0.654 536 270 643 2;
  • 50) 0.654 536 270 643 2 × 2 = 1 + 0.309 072 541 286 4;
  • 51) 0.309 072 541 286 4 × 2 = 0 + 0.618 145 082 572 8;
  • 52) 0.618 145 082 572 8 × 2 = 1 + 0.236 290 165 145 6;
  • 53) 0.236 290 165 145 6 × 2 = 0 + 0.472 580 330 291 2;
  • 54) 0.472 580 330 291 2 × 2 = 0 + 0.945 160 660 582 4;
  • 55) 0.945 160 660 582 4 × 2 = 1 + 0.890 321 321 164 8;
  • 56) 0.890 321 321 164 8 × 2 = 1 + 0.780 642 642 329 6;
  • 57) 0.780 642 642 329 6 × 2 = 1 + 0.561 285 284 659 2;
  • 58) 0.561 285 284 659 2 × 2 = 1 + 0.122 570 569 318 4;
  • 59) 0.122 570 569 318 4 × 2 = 0 + 0.245 141 138 636 8;
  • 60) 0.245 141 138 636 8 × 2 = 0 + 0.490 282 277 273 6;
  • 61) 0.490 282 277 273 6 × 2 = 0 + 0.980 564 554 547 2;
  • 62) 0.980 564 554 547 2 × 2 = 1 + 0.961 129 109 094 4;
  • 63) 0.961 129 109 094 4 × 2 = 1 + 0.922 258 218 188 8;
  • 64) 0.922 258 218 188 8 × 2 = 1 + 0.844 516 436 377 6;

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:


0.000 282 005 913 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0001 1010 1101 0011 1100 0111(2)

6. Positive number before normalization:

0.000 282 005 913 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0001 1010 1101 0011 1100 0111(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:


0.000 282 005 913 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0001 1010 1101 0011 1100 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0001 1010 1101 0011 1100 0111(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 0001 1010 1101 0011 1100 0111(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 0110 0010 1101 0001 1010 1101 0011 1100 0111


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 0110 0010 1101 0001 1010 1101 0011 1100 0111 =

0010 0111 1011 0100 0110 0010 1101 0001 1010 1101 0011 1100 0111


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 0110 0010 1101 0001 1010 1101 0011 1100 0111


Decimal number -0.000 282 005 913 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 0001 1010 1101 0011 1100 0111

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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