-0.000 282 005 887 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 887 7(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
-0.000 282 005 887 7(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:

|-0.000 282 005 887 7| = 0.000 282 005 887 7


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: 0.000 282 005 887 7.

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) 0.000 282 005 887 7 × 2 = 0 + 0.000 564 011 775 4;
  • 2) 0.000 564 011 775 4 × 2 = 0 + 0.001 128 023 550 8;
  • 3) 0.001 128 023 550 8 × 2 = 0 + 0.002 256 047 101 6;
  • 4) 0.002 256 047 101 6 × 2 = 0 + 0.004 512 094 203 2;
  • 5) 0.004 512 094 203 2 × 2 = 0 + 0.009 024 188 406 4;
  • 6) 0.009 024 188 406 4 × 2 = 0 + 0.018 048 376 812 8;
  • 7) 0.018 048 376 812 8 × 2 = 0 + 0.036 096 753 625 6;
  • 8) 0.036 096 753 625 6 × 2 = 0 + 0.072 193 507 251 2;
  • 9) 0.072 193 507 251 2 × 2 = 0 + 0.144 387 014 502 4;
  • 10) 0.144 387 014 502 4 × 2 = 0 + 0.288 774 029 004 8;
  • 11) 0.288 774 029 004 8 × 2 = 0 + 0.577 548 058 009 6;
  • 12) 0.577 548 058 009 6 × 2 = 1 + 0.155 096 116 019 2;
  • 13) 0.155 096 116 019 2 × 2 = 0 + 0.310 192 232 038 4;
  • 14) 0.310 192 232 038 4 × 2 = 0 + 0.620 384 464 076 8;
  • 15) 0.620 384 464 076 8 × 2 = 1 + 0.240 768 928 153 6;
  • 16) 0.240 768 928 153 6 × 2 = 0 + 0.481 537 856 307 2;
  • 17) 0.481 537 856 307 2 × 2 = 0 + 0.963 075 712 614 4;
  • 18) 0.963 075 712 614 4 × 2 = 1 + 0.926 151 425 228 8;
  • 19) 0.926 151 425 228 8 × 2 = 1 + 0.852 302 850 457 6;
  • 20) 0.852 302 850 457 6 × 2 = 1 + 0.704 605 700 915 2;
  • 21) 0.704 605 700 915 2 × 2 = 1 + 0.409 211 401 830 4;
  • 22) 0.409 211 401 830 4 × 2 = 0 + 0.818 422 803 660 8;
  • 23) 0.818 422 803 660 8 × 2 = 1 + 0.636 845 607 321 6;
  • 24) 0.636 845 607 321 6 × 2 = 1 + 0.273 691 214 643 2;
  • 25) 0.273 691 214 643 2 × 2 = 0 + 0.547 382 429 286 4;
  • 26) 0.547 382 429 286 4 × 2 = 1 + 0.094 764 858 572 8;
  • 27) 0.094 764 858 572 8 × 2 = 0 + 0.189 529 717 145 6;
  • 28) 0.189 529 717 145 6 × 2 = 0 + 0.379 059 434 291 2;
  • 29) 0.379 059 434 291 2 × 2 = 0 + 0.758 118 868 582 4;
  • 30) 0.758 118 868 582 4 × 2 = 1 + 0.516 237 737 164 8;
  • 31) 0.516 237 737 164 8 × 2 = 1 + 0.032 475 474 329 6;
  • 32) 0.032 475 474 329 6 × 2 = 0 + 0.064 950 948 659 2;
  • 33) 0.064 950 948 659 2 × 2 = 0 + 0.129 901 897 318 4;
  • 34) 0.129 901 897 318 4 × 2 = 0 + 0.259 803 794 636 8;
  • 35) 0.259 803 794 636 8 × 2 = 0 + 0.519 607 589 273 6;
  • 36) 0.519 607 589 273 6 × 2 = 1 + 0.039 215 178 547 2;
  • 37) 0.039 215 178 547 2 × 2 = 0 + 0.078 430 357 094 4;
  • 38) 0.078 430 357 094 4 × 2 = 0 + 0.156 860 714 188 8;
  • 39) 0.156 860 714 188 8 × 2 = 0 + 0.313 721 428 377 6;
  • 40) 0.313 721 428 377 6 × 2 = 0 + 0.627 442 856 755 2;
  • 41) 0.627 442 856 755 2 × 2 = 1 + 0.254 885 713 510 4;
  • 42) 0.254 885 713 510 4 × 2 = 0 + 0.509 771 427 020 8;
  • 43) 0.509 771 427 020 8 × 2 = 1 + 0.019 542 854 041 6;
  • 44) 0.019 542 854 041 6 × 2 = 0 + 0.039 085 708 083 2;
  • 45) 0.039 085 708 083 2 × 2 = 0 + 0.078 171 416 166 4;
  • 46) 0.078 171 416 166 4 × 2 = 0 + 0.156 342 832 332 8;
  • 47) 0.156 342 832 332 8 × 2 = 0 + 0.312 685 664 665 6;
  • 48) 0.312 685 664 665 6 × 2 = 0 + 0.625 371 329 331 2;
  • 49) 0.625 371 329 331 2 × 2 = 1 + 0.250 742 658 662 4;
  • 50) 0.250 742 658 662 4 × 2 = 0 + 0.501 485 317 324 8;
  • 51) 0.501 485 317 324 8 × 2 = 1 + 0.002 970 634 649 6;
  • 52) 0.002 970 634 649 6 × 2 = 0 + 0.005 941 269 299 2;
  • 53) 0.005 941 269 299 2 × 2 = 0 + 0.011 882 538 598 4;
  • 54) 0.011 882 538 598 4 × 2 = 0 + 0.023 765 077 196 8;
  • 55) 0.023 765 077 196 8 × 2 = 0 + 0.047 530 154 393 6;
  • 56) 0.047 530 154 393 6 × 2 = 0 + 0.095 060 308 787 2;
  • 57) 0.095 060 308 787 2 × 2 = 0 + 0.190 120 617 574 4;
  • 58) 0.190 120 617 574 4 × 2 = 0 + 0.380 241 235 148 8;
  • 59) 0.380 241 235 148 8 × 2 = 0 + 0.760 482 470 297 6;
  • 60) 0.760 482 470 297 6 × 2 = 1 + 0.520 964 940 595 2;
  • 61) 0.520 964 940 595 2 × 2 = 1 + 0.041 929 881 190 4;
  • 62) 0.041 929 881 190 4 × 2 = 0 + 0.083 859 762 380 8;
  • 63) 0.083 859 762 380 8 × 2 = 0 + 0.167 719 524 761 6;
  • 64) 0.167 719 524 761 6 × 2 = 0 + 0.335 439 049 523 2;

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:


0.000 282 005 887 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 1010 0000 1010 0000 0001 1000(2)

6. Positive number before normalization:

0.000 282 005 887 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 1010 0000 1010 0000 0001 1000(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:


0.000 282 005 887 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 1010 0000 1010 0000 0001 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 1010 0000 1010 0000 0001 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0000 1010 0000 1010 0000 0001 1000(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 0001 0000 1010 0000 1010 0000 0001 1000


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 0001 0000 1010 0000 1010 0000 0001 1000 =

0010 0111 1011 0100 0110 0001 0000 1010 0000 1010 0000 0001 1000


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 0001 0000 1010 0000 1010 0000 0001 1000


Decimal number -0.000 282 005 887 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0000 1010 0000 1010 0000 0001 1000

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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