-0.000 282 006 311 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 311(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
-0.000 282 006 311(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:

|-0.000 282 006 311| = 0.000 282 006 311


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: 0.000 282 006 311.

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) 0.000 282 006 311 × 2 = 0 + 0.000 564 012 622;
  • 2) 0.000 564 012 622 × 2 = 0 + 0.001 128 025 244;
  • 3) 0.001 128 025 244 × 2 = 0 + 0.002 256 050 488;
  • 4) 0.002 256 050 488 × 2 = 0 + 0.004 512 100 976;
  • 5) 0.004 512 100 976 × 2 = 0 + 0.009 024 201 952;
  • 6) 0.009 024 201 952 × 2 = 0 + 0.018 048 403 904;
  • 7) 0.018 048 403 904 × 2 = 0 + 0.036 096 807 808;
  • 8) 0.036 096 807 808 × 2 = 0 + 0.072 193 615 616;
  • 9) 0.072 193 615 616 × 2 = 0 + 0.144 387 231 232;
  • 10) 0.144 387 231 232 × 2 = 0 + 0.288 774 462 464;
  • 11) 0.288 774 462 464 × 2 = 0 + 0.577 548 924 928;
  • 12) 0.577 548 924 928 × 2 = 1 + 0.155 097 849 856;
  • 13) 0.155 097 849 856 × 2 = 0 + 0.310 195 699 712;
  • 14) 0.310 195 699 712 × 2 = 0 + 0.620 391 399 424;
  • 15) 0.620 391 399 424 × 2 = 1 + 0.240 782 798 848;
  • 16) 0.240 782 798 848 × 2 = 0 + 0.481 565 597 696;
  • 17) 0.481 565 597 696 × 2 = 0 + 0.963 131 195 392;
  • 18) 0.963 131 195 392 × 2 = 1 + 0.926 262 390 784;
  • 19) 0.926 262 390 784 × 2 = 1 + 0.852 524 781 568;
  • 20) 0.852 524 781 568 × 2 = 1 + 0.705 049 563 136;
  • 21) 0.705 049 563 136 × 2 = 1 + 0.410 099 126 272;
  • 22) 0.410 099 126 272 × 2 = 0 + 0.820 198 252 544;
  • 23) 0.820 198 252 544 × 2 = 1 + 0.640 396 505 088;
  • 24) 0.640 396 505 088 × 2 = 1 + 0.280 793 010 176;
  • 25) 0.280 793 010 176 × 2 = 0 + 0.561 586 020 352;
  • 26) 0.561 586 020 352 × 2 = 1 + 0.123 172 040 704;
  • 27) 0.123 172 040 704 × 2 = 0 + 0.246 344 081 408;
  • 28) 0.246 344 081 408 × 2 = 0 + 0.492 688 162 816;
  • 29) 0.492 688 162 816 × 2 = 0 + 0.985 376 325 632;
  • 30) 0.985 376 325 632 × 2 = 1 + 0.970 752 651 264;
  • 31) 0.970 752 651 264 × 2 = 1 + 0.941 505 302 528;
  • 32) 0.941 505 302 528 × 2 = 1 + 0.883 010 605 056;
  • 33) 0.883 010 605 056 × 2 = 1 + 0.766 021 210 112;
  • 34) 0.766 021 210 112 × 2 = 1 + 0.532 042 420 224;
  • 35) 0.532 042 420 224 × 2 = 1 + 0.064 084 840 448;
  • 36) 0.064 084 840 448 × 2 = 0 + 0.128 169 680 896;
  • 37) 0.128 169 680 896 × 2 = 0 + 0.256 339 361 792;
  • 38) 0.256 339 361 792 × 2 = 0 + 0.512 678 723 584;
  • 39) 0.512 678 723 584 × 2 = 1 + 0.025 357 447 168;
  • 40) 0.025 357 447 168 × 2 = 0 + 0.050 714 894 336;
  • 41) 0.050 714 894 336 × 2 = 0 + 0.101 429 788 672;
  • 42) 0.101 429 788 672 × 2 = 0 + 0.202 859 577 344;
  • 43) 0.202 859 577 344 × 2 = 0 + 0.405 719 154 688;
  • 44) 0.405 719 154 688 × 2 = 0 + 0.811 438 309 376;
  • 45) 0.811 438 309 376 × 2 = 1 + 0.622 876 618 752;
  • 46) 0.622 876 618 752 × 2 = 1 + 0.245 753 237 504;
  • 47) 0.245 753 237 504 × 2 = 0 + 0.491 506 475 008;
  • 48) 0.491 506 475 008 × 2 = 0 + 0.983 012 950 016;
  • 49) 0.983 012 950 016 × 2 = 1 + 0.966 025 900 032;
  • 50) 0.966 025 900 032 × 2 = 1 + 0.932 051 800 064;
  • 51) 0.932 051 800 064 × 2 = 1 + 0.864 103 600 128;
  • 52) 0.864 103 600 128 × 2 = 1 + 0.728 207 200 256;
  • 53) 0.728 207 200 256 × 2 = 1 + 0.456 414 400 512;
  • 54) 0.456 414 400 512 × 2 = 0 + 0.912 828 801 024;
  • 55) 0.912 828 801 024 × 2 = 1 + 0.825 657 602 048;
  • 56) 0.825 657 602 048 × 2 = 1 + 0.651 315 204 096;
  • 57) 0.651 315 204 096 × 2 = 1 + 0.302 630 408 192;
  • 58) 0.302 630 408 192 × 2 = 0 + 0.605 260 816 384;
  • 59) 0.605 260 816 384 × 2 = 1 + 0.210 521 632 768;
  • 60) 0.210 521 632 768 × 2 = 0 + 0.421 043 265 536;
  • 61) 0.421 043 265 536 × 2 = 0 + 0.842 086 531 072;
  • 62) 0.842 086 531 072 × 2 = 1 + 0.684 173 062 144;
  • 63) 0.684 173 062 144 × 2 = 1 + 0.368 346 124 288;
  • 64) 0.368 346 124 288 × 2 = 0 + 0.736 692 248 576;

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:


0.000 282 006 311(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1110 0010 0000 1100 1111 1011 1010 0110(2)

6. Positive number before normalization:

0.000 282 006 311(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1110 0010 0000 1100 1111 1011 1010 0110(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:


0.000 282 006 311(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1110 0010 0000 1100 1111 1011 1010 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1110 0010 0000 1100 1111 1011 1010 0110(2) × 20 =

1.0010 0111 1011 0100 0111 1110 0010 0000 1100 1111 1011 1010 0110(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 0111 1110 0010 0000 1100 1111 1011 1010 0110


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 0111 1110 0010 0000 1100 1111 1011 1010 0110 =

0010 0111 1011 0100 0111 1110 0010 0000 1100 1111 1011 1010 0110


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 0111 1110 0010 0000 1100 1111 1011 1010 0110


Decimal number -0.000 282 006 311 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1110 0010 0000 1100 1111 1011 1010 0110

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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