-0.000 282 006 341 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 341(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
-0.000 282 006 341(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:

|-0.000 282 006 341| = 0.000 282 006 341


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: 0.000 282 006 341.

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) 0.000 282 006 341 × 2 = 0 + 0.000 564 012 682;
  • 2) 0.000 564 012 682 × 2 = 0 + 0.001 128 025 364;
  • 3) 0.001 128 025 364 × 2 = 0 + 0.002 256 050 728;
  • 4) 0.002 256 050 728 × 2 = 0 + 0.004 512 101 456;
  • 5) 0.004 512 101 456 × 2 = 0 + 0.009 024 202 912;
  • 6) 0.009 024 202 912 × 2 = 0 + 0.018 048 405 824;
  • 7) 0.018 048 405 824 × 2 = 0 + 0.036 096 811 648;
  • 8) 0.036 096 811 648 × 2 = 0 + 0.072 193 623 296;
  • 9) 0.072 193 623 296 × 2 = 0 + 0.144 387 246 592;
  • 10) 0.144 387 246 592 × 2 = 0 + 0.288 774 493 184;
  • 11) 0.288 774 493 184 × 2 = 0 + 0.577 548 986 368;
  • 12) 0.577 548 986 368 × 2 = 1 + 0.155 097 972 736;
  • 13) 0.155 097 972 736 × 2 = 0 + 0.310 195 945 472;
  • 14) 0.310 195 945 472 × 2 = 0 + 0.620 391 890 944;
  • 15) 0.620 391 890 944 × 2 = 1 + 0.240 783 781 888;
  • 16) 0.240 783 781 888 × 2 = 0 + 0.481 567 563 776;
  • 17) 0.481 567 563 776 × 2 = 0 + 0.963 135 127 552;
  • 18) 0.963 135 127 552 × 2 = 1 + 0.926 270 255 104;
  • 19) 0.926 270 255 104 × 2 = 1 + 0.852 540 510 208;
  • 20) 0.852 540 510 208 × 2 = 1 + 0.705 081 020 416;
  • 21) 0.705 081 020 416 × 2 = 1 + 0.410 162 040 832;
  • 22) 0.410 162 040 832 × 2 = 0 + 0.820 324 081 664;
  • 23) 0.820 324 081 664 × 2 = 1 + 0.640 648 163 328;
  • 24) 0.640 648 163 328 × 2 = 1 + 0.281 296 326 656;
  • 25) 0.281 296 326 656 × 2 = 0 + 0.562 592 653 312;
  • 26) 0.562 592 653 312 × 2 = 1 + 0.125 185 306 624;
  • 27) 0.125 185 306 624 × 2 = 0 + 0.250 370 613 248;
  • 28) 0.250 370 613 248 × 2 = 0 + 0.500 741 226 496;
  • 29) 0.500 741 226 496 × 2 = 1 + 0.001 482 452 992;
  • 30) 0.001 482 452 992 × 2 = 0 + 0.002 964 905 984;
  • 31) 0.002 964 905 984 × 2 = 0 + 0.005 929 811 968;
  • 32) 0.005 929 811 968 × 2 = 0 + 0.011 859 623 936;
  • 33) 0.011 859 623 936 × 2 = 0 + 0.023 719 247 872;
  • 34) 0.023 719 247 872 × 2 = 0 + 0.047 438 495 744;
  • 35) 0.047 438 495 744 × 2 = 0 + 0.094 876 991 488;
  • 36) 0.094 876 991 488 × 2 = 0 + 0.189 753 982 976;
  • 37) 0.189 753 982 976 × 2 = 0 + 0.379 507 965 952;
  • 38) 0.379 507 965 952 × 2 = 0 + 0.759 015 931 904;
  • 39) 0.759 015 931 904 × 2 = 1 + 0.518 031 863 808;
  • 40) 0.518 031 863 808 × 2 = 1 + 0.036 063 727 616;
  • 41) 0.036 063 727 616 × 2 = 0 + 0.072 127 455 232;
  • 42) 0.072 127 455 232 × 2 = 0 + 0.144 254 910 464;
  • 43) 0.144 254 910 464 × 2 = 0 + 0.288 509 820 928;
  • 44) 0.288 509 820 928 × 2 = 0 + 0.577 019 641 856;
  • 45) 0.577 019 641 856 × 2 = 1 + 0.154 039 283 712;
  • 46) 0.154 039 283 712 × 2 = 0 + 0.308 078 567 424;
  • 47) 0.308 078 567 424 × 2 = 0 + 0.616 157 134 848;
  • 48) 0.616 157 134 848 × 2 = 1 + 0.232 314 269 696;
  • 49) 0.232 314 269 696 × 2 = 0 + 0.464 628 539 392;
  • 50) 0.464 628 539 392 × 2 = 0 + 0.929 257 078 784;
  • 51) 0.929 257 078 784 × 2 = 1 + 0.858 514 157 568;
  • 52) 0.858 514 157 568 × 2 = 1 + 0.717 028 315 136;
  • 53) 0.717 028 315 136 × 2 = 1 + 0.434 056 630 272;
  • 54) 0.434 056 630 272 × 2 = 0 + 0.868 113 260 544;
  • 55) 0.868 113 260 544 × 2 = 1 + 0.736 226 521 088;
  • 56) 0.736 226 521 088 × 2 = 1 + 0.472 453 042 176;
  • 57) 0.472 453 042 176 × 2 = 0 + 0.944 906 084 352;
  • 58) 0.944 906 084 352 × 2 = 1 + 0.889 812 168 704;
  • 59) 0.889 812 168 704 × 2 = 1 + 0.779 624 337 408;
  • 60) 0.779 624 337 408 × 2 = 1 + 0.559 248 674 816;
  • 61) 0.559 248 674 816 × 2 = 1 + 0.118 497 349 632;
  • 62) 0.118 497 349 632 × 2 = 0 + 0.236 994 699 264;
  • 63) 0.236 994 699 264 × 2 = 0 + 0.473 989 398 528;
  • 64) 0.473 989 398 528 × 2 = 0 + 0.947 978 797 056;

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:


0.000 282 006 341(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 0011 0000 1001 0011 1011 0111 1000(2)

6. Positive number before normalization:

0.000 282 006 341(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 0011 0000 1001 0011 1011 0111 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:


0.000 282 006 341(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 0011 0000 1001 0011 1011 0111 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 0011 0000 1001 0011 1011 0111 1000(2) × 20 =

1.0010 0111 1011 0100 1000 0000 0011 0000 1001 0011 1011 0111 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 1000 0000 0011 0000 1001 0011 1011 0111 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 1000 0000 0011 0000 1001 0011 1011 0111 1000 =

0010 0111 1011 0100 1000 0000 0011 0000 1001 0011 1011 0111 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 1000 0000 0011 0000 1001 0011 1011 0111 1000


Decimal number -0.000 282 006 341 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0000 0011 0000 1001 0011 1011 0111 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