-0.000 282 006 291 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 291(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
-0.000 282 006 291(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:

|-0.000 282 006 291| = 0.000 282 006 291


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: 0.000 282 006 291.

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) 0.000 282 006 291 × 2 = 0 + 0.000 564 012 582;
  • 2) 0.000 564 012 582 × 2 = 0 + 0.001 128 025 164;
  • 3) 0.001 128 025 164 × 2 = 0 + 0.002 256 050 328;
  • 4) 0.002 256 050 328 × 2 = 0 + 0.004 512 100 656;
  • 5) 0.004 512 100 656 × 2 = 0 + 0.009 024 201 312;
  • 6) 0.009 024 201 312 × 2 = 0 + 0.018 048 402 624;
  • 7) 0.018 048 402 624 × 2 = 0 + 0.036 096 805 248;
  • 8) 0.036 096 805 248 × 2 = 0 + 0.072 193 610 496;
  • 9) 0.072 193 610 496 × 2 = 0 + 0.144 387 220 992;
  • 10) 0.144 387 220 992 × 2 = 0 + 0.288 774 441 984;
  • 11) 0.288 774 441 984 × 2 = 0 + 0.577 548 883 968;
  • 12) 0.577 548 883 968 × 2 = 1 + 0.155 097 767 936;
  • 13) 0.155 097 767 936 × 2 = 0 + 0.310 195 535 872;
  • 14) 0.310 195 535 872 × 2 = 0 + 0.620 391 071 744;
  • 15) 0.620 391 071 744 × 2 = 1 + 0.240 782 143 488;
  • 16) 0.240 782 143 488 × 2 = 0 + 0.481 564 286 976;
  • 17) 0.481 564 286 976 × 2 = 0 + 0.963 128 573 952;
  • 18) 0.963 128 573 952 × 2 = 1 + 0.926 257 147 904;
  • 19) 0.926 257 147 904 × 2 = 1 + 0.852 514 295 808;
  • 20) 0.852 514 295 808 × 2 = 1 + 0.705 028 591 616;
  • 21) 0.705 028 591 616 × 2 = 1 + 0.410 057 183 232;
  • 22) 0.410 057 183 232 × 2 = 0 + 0.820 114 366 464;
  • 23) 0.820 114 366 464 × 2 = 1 + 0.640 228 732 928;
  • 24) 0.640 228 732 928 × 2 = 1 + 0.280 457 465 856;
  • 25) 0.280 457 465 856 × 2 = 0 + 0.560 914 931 712;
  • 26) 0.560 914 931 712 × 2 = 1 + 0.121 829 863 424;
  • 27) 0.121 829 863 424 × 2 = 0 + 0.243 659 726 848;
  • 28) 0.243 659 726 848 × 2 = 0 + 0.487 319 453 696;
  • 29) 0.487 319 453 696 × 2 = 0 + 0.974 638 907 392;
  • 30) 0.974 638 907 392 × 2 = 1 + 0.949 277 814 784;
  • 31) 0.949 277 814 784 × 2 = 1 + 0.898 555 629 568;
  • 32) 0.898 555 629 568 × 2 = 1 + 0.797 111 259 136;
  • 33) 0.797 111 259 136 × 2 = 1 + 0.594 222 518 272;
  • 34) 0.594 222 518 272 × 2 = 1 + 0.188 445 036 544;
  • 35) 0.188 445 036 544 × 2 = 0 + 0.376 890 073 088;
  • 36) 0.376 890 073 088 × 2 = 0 + 0.753 780 146 176;
  • 37) 0.753 780 146 176 × 2 = 1 + 0.507 560 292 352;
  • 38) 0.507 560 292 352 × 2 = 1 + 0.015 120 584 704;
  • 39) 0.015 120 584 704 × 2 = 0 + 0.030 241 169 408;
  • 40) 0.030 241 169 408 × 2 = 0 + 0.060 482 338 816;
  • 41) 0.060 482 338 816 × 2 = 0 + 0.120 964 677 632;
  • 42) 0.120 964 677 632 × 2 = 0 + 0.241 929 355 264;
  • 43) 0.241 929 355 264 × 2 = 0 + 0.483 858 710 528;
  • 44) 0.483 858 710 528 × 2 = 0 + 0.967 717 421 056;
  • 45) 0.967 717 421 056 × 2 = 1 + 0.935 434 842 112;
  • 46) 0.935 434 842 112 × 2 = 1 + 0.870 869 684 224;
  • 47) 0.870 869 684 224 × 2 = 1 + 0.741 739 368 448;
  • 48) 0.741 739 368 448 × 2 = 1 + 0.483 478 736 896;
  • 49) 0.483 478 736 896 × 2 = 0 + 0.966 957 473 792;
  • 50) 0.966 957 473 792 × 2 = 1 + 0.933 914 947 584;
  • 51) 0.933 914 947 584 × 2 = 1 + 0.867 829 895 168;
  • 52) 0.867 829 895 168 × 2 = 1 + 0.735 659 790 336;
  • 53) 0.735 659 790 336 × 2 = 1 + 0.471 319 580 672;
  • 54) 0.471 319 580 672 × 2 = 0 + 0.942 639 161 344;
  • 55) 0.942 639 161 344 × 2 = 1 + 0.885 278 322 688;
  • 56) 0.885 278 322 688 × 2 = 1 + 0.770 556 645 376;
  • 57) 0.770 556 645 376 × 2 = 1 + 0.541 113 290 752;
  • 58) 0.541 113 290 752 × 2 = 1 + 0.082 226 581 504;
  • 59) 0.082 226 581 504 × 2 = 0 + 0.164 453 163 008;
  • 60) 0.164 453 163 008 × 2 = 0 + 0.328 906 326 016;
  • 61) 0.328 906 326 016 × 2 = 0 + 0.657 812 652 032;
  • 62) 0.657 812 652 032 × 2 = 1 + 0.315 625 304 064;
  • 63) 0.315 625 304 064 × 2 = 0 + 0.631 250 608 128;
  • 64) 0.631 250 608 128 × 2 = 1 + 0.262 501 216 256;

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:


0.000 282 006 291(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 1100 0000 1111 0111 1011 1100 0101(2)

6. Positive number before normalization:

0.000 282 006 291(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 1100 0000 1111 0111 1011 1100 0101(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:


0.000 282 006 291(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 1100 0000 1111 0111 1011 1100 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 1100 0000 1111 0111 1011 1100 0101(2) × 20 =

1.0010 0111 1011 0100 0111 1100 1100 0000 1111 0111 1011 1100 0101(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 1100 1100 0000 1111 0111 1011 1100 0101


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 1100 1100 0000 1111 0111 1011 1100 0101 =

0010 0111 1011 0100 0111 1100 1100 0000 1111 0111 1011 1100 0101


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 1100 1100 0000 1111 0111 1011 1100 0101


Decimal number -0.000 282 006 291 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1100 1100 0000 1111 0111 1011 1100 0101

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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