-0.000 282 005 925 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 925 1(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
-0.000 282 005 925 1(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:

|-0.000 282 005 925 1| = 0.000 282 005 925 1


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: 0.000 282 005 925 1.

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) 0.000 282 005 925 1 × 2 = 0 + 0.000 564 011 850 2;
  • 2) 0.000 564 011 850 2 × 2 = 0 + 0.001 128 023 700 4;
  • 3) 0.001 128 023 700 4 × 2 = 0 + 0.002 256 047 400 8;
  • 4) 0.002 256 047 400 8 × 2 = 0 + 0.004 512 094 801 6;
  • 5) 0.004 512 094 801 6 × 2 = 0 + 0.009 024 189 603 2;
  • 6) 0.009 024 189 603 2 × 2 = 0 + 0.018 048 379 206 4;
  • 7) 0.018 048 379 206 4 × 2 = 0 + 0.036 096 758 412 8;
  • 8) 0.036 096 758 412 8 × 2 = 0 + 0.072 193 516 825 6;
  • 9) 0.072 193 516 825 6 × 2 = 0 + 0.144 387 033 651 2;
  • 10) 0.144 387 033 651 2 × 2 = 0 + 0.288 774 067 302 4;
  • 11) 0.288 774 067 302 4 × 2 = 0 + 0.577 548 134 604 8;
  • 12) 0.577 548 134 604 8 × 2 = 1 + 0.155 096 269 209 6;
  • 13) 0.155 096 269 209 6 × 2 = 0 + 0.310 192 538 419 2;
  • 14) 0.310 192 538 419 2 × 2 = 0 + 0.620 385 076 838 4;
  • 15) 0.620 385 076 838 4 × 2 = 1 + 0.240 770 153 676 8;
  • 16) 0.240 770 153 676 8 × 2 = 0 + 0.481 540 307 353 6;
  • 17) 0.481 540 307 353 6 × 2 = 0 + 0.963 080 614 707 2;
  • 18) 0.963 080 614 707 2 × 2 = 1 + 0.926 161 229 414 4;
  • 19) 0.926 161 229 414 4 × 2 = 1 + 0.852 322 458 828 8;
  • 20) 0.852 322 458 828 8 × 2 = 1 + 0.704 644 917 657 6;
  • 21) 0.704 644 917 657 6 × 2 = 1 + 0.409 289 835 315 2;
  • 22) 0.409 289 835 315 2 × 2 = 0 + 0.818 579 670 630 4;
  • 23) 0.818 579 670 630 4 × 2 = 1 + 0.637 159 341 260 8;
  • 24) 0.637 159 341 260 8 × 2 = 1 + 0.274 318 682 521 6;
  • 25) 0.274 318 682 521 6 × 2 = 0 + 0.548 637 365 043 2;
  • 26) 0.548 637 365 043 2 × 2 = 1 + 0.097 274 730 086 4;
  • 27) 0.097 274 730 086 4 × 2 = 0 + 0.194 549 460 172 8;
  • 28) 0.194 549 460 172 8 × 2 = 0 + 0.389 098 920 345 6;
  • 29) 0.389 098 920 345 6 × 2 = 0 + 0.778 197 840 691 2;
  • 30) 0.778 197 840 691 2 × 2 = 1 + 0.556 395 681 382 4;
  • 31) 0.556 395 681 382 4 × 2 = 1 + 0.112 791 362 764 8;
  • 32) 0.112 791 362 764 8 × 2 = 0 + 0.225 582 725 529 6;
  • 33) 0.225 582 725 529 6 × 2 = 0 + 0.451 165 451 059 2;
  • 34) 0.451 165 451 059 2 × 2 = 0 + 0.902 330 902 118 4;
  • 35) 0.902 330 902 118 4 × 2 = 1 + 0.804 661 804 236 8;
  • 36) 0.804 661 804 236 8 × 2 = 1 + 0.609 323 608 473 6;
  • 37) 0.609 323 608 473 6 × 2 = 1 + 0.218 647 216 947 2;
  • 38) 0.218 647 216 947 2 × 2 = 0 + 0.437 294 433 894 4;
  • 39) 0.437 294 433 894 4 × 2 = 0 + 0.874 588 867 788 8;
  • 40) 0.874 588 867 788 8 × 2 = 1 + 0.749 177 735 577 6;
  • 41) 0.749 177 735 577 6 × 2 = 1 + 0.498 355 471 155 2;
  • 42) 0.498 355 471 155 2 × 2 = 0 + 0.996 710 942 310 4;
  • 43) 0.996 710 942 310 4 × 2 = 1 + 0.993 421 884 620 8;
  • 44) 0.993 421 884 620 8 × 2 = 1 + 0.986 843 769 241 6;
  • 45) 0.986 843 769 241 6 × 2 = 1 + 0.973 687 538 483 2;
  • 46) 0.973 687 538 483 2 × 2 = 1 + 0.947 375 076 966 4;
  • 47) 0.947 375 076 966 4 × 2 = 1 + 0.894 750 153 932 8;
  • 48) 0.894 750 153 932 8 × 2 = 1 + 0.789 500 307 865 6;
  • 49) 0.789 500 307 865 6 × 2 = 1 + 0.579 000 615 731 2;
  • 50) 0.579 000 615 731 2 × 2 = 1 + 0.158 001 231 462 4;
  • 51) 0.158 001 231 462 4 × 2 = 0 + 0.316 002 462 924 8;
  • 52) 0.316 002 462 924 8 × 2 = 0 + 0.632 004 925 849 6;
  • 53) 0.632 004 925 849 6 × 2 = 1 + 0.264 009 851 699 2;
  • 54) 0.264 009 851 699 2 × 2 = 0 + 0.528 019 703 398 4;
  • 55) 0.528 019 703 398 4 × 2 = 1 + 0.056 039 406 796 8;
  • 56) 0.056 039 406 796 8 × 2 = 0 + 0.112 078 813 593 6;
  • 57) 0.112 078 813 593 6 × 2 = 0 + 0.224 157 627 187 2;
  • 58) 0.224 157 627 187 2 × 2 = 0 + 0.448 315 254 374 4;
  • 59) 0.448 315 254 374 4 × 2 = 0 + 0.896 630 508 748 8;
  • 60) 0.896 630 508 748 8 × 2 = 1 + 0.793 261 017 497 6;
  • 61) 0.793 261 017 497 6 × 2 = 1 + 0.586 522 034 995 2;
  • 62) 0.586 522 034 995 2 × 2 = 1 + 0.173 044 069 990 4;
  • 63) 0.173 044 069 990 4 × 2 = 0 + 0.346 088 139 980 8;
  • 64) 0.346 088 139 980 8 × 2 = 0 + 0.692 176 279 961 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:


0.000 282 005 925 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1001 1011 1111 1100 1010 0001 1100(2)

6. Positive number before normalization:

0.000 282 005 925 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1001 1011 1111 1100 1010 0001 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:


0.000 282 005 925 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1001 1011 1111 1100 1010 0001 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1001 1011 1111 1100 1010 0001 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0011 1001 1011 1111 1100 1010 0001 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 0110 0011 1001 1011 1111 1100 1010 0001 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 0110 0011 1001 1011 1111 1100 1010 0001 1100 =

0010 0111 1011 0100 0110 0011 1001 1011 1111 1100 1010 0001 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 0110 0011 1001 1011 1111 1100 1010 0001 1100


Decimal number -0.000 282 005 925 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 1001 1011 1111 1100 1010 0001 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