-0.000 282 006 259 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 259(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
-0.000 282 006 259(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:

|-0.000 282 006 259| = 0.000 282 006 259


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: 0.000 282 006 259.

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) 0.000 282 006 259 × 2 = 0 + 0.000 564 012 518;
  • 2) 0.000 564 012 518 × 2 = 0 + 0.001 128 025 036;
  • 3) 0.001 128 025 036 × 2 = 0 + 0.002 256 050 072;
  • 4) 0.002 256 050 072 × 2 = 0 + 0.004 512 100 144;
  • 5) 0.004 512 100 144 × 2 = 0 + 0.009 024 200 288;
  • 6) 0.009 024 200 288 × 2 = 0 + 0.018 048 400 576;
  • 7) 0.018 048 400 576 × 2 = 0 + 0.036 096 801 152;
  • 8) 0.036 096 801 152 × 2 = 0 + 0.072 193 602 304;
  • 9) 0.072 193 602 304 × 2 = 0 + 0.144 387 204 608;
  • 10) 0.144 387 204 608 × 2 = 0 + 0.288 774 409 216;
  • 11) 0.288 774 409 216 × 2 = 0 + 0.577 548 818 432;
  • 12) 0.577 548 818 432 × 2 = 1 + 0.155 097 636 864;
  • 13) 0.155 097 636 864 × 2 = 0 + 0.310 195 273 728;
  • 14) 0.310 195 273 728 × 2 = 0 + 0.620 390 547 456;
  • 15) 0.620 390 547 456 × 2 = 1 + 0.240 781 094 912;
  • 16) 0.240 781 094 912 × 2 = 0 + 0.481 562 189 824;
  • 17) 0.481 562 189 824 × 2 = 0 + 0.963 124 379 648;
  • 18) 0.963 124 379 648 × 2 = 1 + 0.926 248 759 296;
  • 19) 0.926 248 759 296 × 2 = 1 + 0.852 497 518 592;
  • 20) 0.852 497 518 592 × 2 = 1 + 0.704 995 037 184;
  • 21) 0.704 995 037 184 × 2 = 1 + 0.409 990 074 368;
  • 22) 0.409 990 074 368 × 2 = 0 + 0.819 980 148 736;
  • 23) 0.819 980 148 736 × 2 = 1 + 0.639 960 297 472;
  • 24) 0.639 960 297 472 × 2 = 1 + 0.279 920 594 944;
  • 25) 0.279 920 594 944 × 2 = 0 + 0.559 841 189 888;
  • 26) 0.559 841 189 888 × 2 = 1 + 0.119 682 379 776;
  • 27) 0.119 682 379 776 × 2 = 0 + 0.239 364 759 552;
  • 28) 0.239 364 759 552 × 2 = 0 + 0.478 729 519 104;
  • 29) 0.478 729 519 104 × 2 = 0 + 0.957 459 038 208;
  • 30) 0.957 459 038 208 × 2 = 1 + 0.914 918 076 416;
  • 31) 0.914 918 076 416 × 2 = 1 + 0.829 836 152 832;
  • 32) 0.829 836 152 832 × 2 = 1 + 0.659 672 305 664;
  • 33) 0.659 672 305 664 × 2 = 1 + 0.319 344 611 328;
  • 34) 0.319 344 611 328 × 2 = 0 + 0.638 689 222 656;
  • 35) 0.638 689 222 656 × 2 = 1 + 0.277 378 445 312;
  • 36) 0.277 378 445 312 × 2 = 0 + 0.554 756 890 624;
  • 37) 0.554 756 890 624 × 2 = 1 + 0.109 513 781 248;
  • 38) 0.109 513 781 248 × 2 = 0 + 0.219 027 562 496;
  • 39) 0.219 027 562 496 × 2 = 0 + 0.438 055 124 992;
  • 40) 0.438 055 124 992 × 2 = 0 + 0.876 110 249 984;
  • 41) 0.876 110 249 984 × 2 = 1 + 0.752 220 499 968;
  • 42) 0.752 220 499 968 × 2 = 1 + 0.504 440 999 936;
  • 43) 0.504 440 999 936 × 2 = 1 + 0.008 881 999 872;
  • 44) 0.008 881 999 872 × 2 = 0 + 0.017 763 999 744;
  • 45) 0.017 763 999 744 × 2 = 0 + 0.035 527 999 488;
  • 46) 0.035 527 999 488 × 2 = 0 + 0.071 055 998 976;
  • 47) 0.071 055 998 976 × 2 = 0 + 0.142 111 997 952;
  • 48) 0.142 111 997 952 × 2 = 0 + 0.284 223 995 904;
  • 49) 0.284 223 995 904 × 2 = 0 + 0.568 447 991 808;
  • 50) 0.568 447 991 808 × 2 = 1 + 0.136 895 983 616;
  • 51) 0.136 895 983 616 × 2 = 0 + 0.273 791 967 232;
  • 52) 0.273 791 967 232 × 2 = 0 + 0.547 583 934 464;
  • 53) 0.547 583 934 464 × 2 = 1 + 0.095 167 868 928;
  • 54) 0.095 167 868 928 × 2 = 0 + 0.190 335 737 856;
  • 55) 0.190 335 737 856 × 2 = 0 + 0.380 671 475 712;
  • 56) 0.380 671 475 712 × 2 = 0 + 0.761 342 951 424;
  • 57) 0.761 342 951 424 × 2 = 1 + 0.522 685 902 848;
  • 58) 0.522 685 902 848 × 2 = 1 + 0.045 371 805 696;
  • 59) 0.045 371 805 696 × 2 = 0 + 0.090 743 611 392;
  • 60) 0.090 743 611 392 × 2 = 0 + 0.181 487 222 784;
  • 61) 0.181 487 222 784 × 2 = 0 + 0.362 974 445 568;
  • 62) 0.362 974 445 568 × 2 = 0 + 0.725 948 891 136;
  • 63) 0.725 948 891 136 × 2 = 1 + 0.451 897 782 272;
  • 64) 0.451 897 782 272 × 2 = 0 + 0.903 795 564 544;

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:


0.000 282 006 259(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 1000 1110 0000 0100 1000 1100 0010(2)

6. Positive number before normalization:

0.000 282 006 259(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 1000 1110 0000 0100 1000 1100 0010(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:


0.000 282 006 259(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 1000 1110 0000 0100 1000 1100 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 1000 1110 0000 0100 1000 1100 0010(2) × 20 =

1.0010 0111 1011 0100 0111 1010 1000 1110 0000 0100 1000 1100 0010(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 1010 1000 1110 0000 0100 1000 1100 0010


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 1010 1000 1110 0000 0100 1000 1100 0010 =

0010 0111 1011 0100 0111 1010 1000 1110 0000 0100 1000 1100 0010


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 1010 1000 1110 0000 0100 1000 1100 0010


Decimal number -0.000 282 006 259 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1010 1000 1110 0000 0100 1000 1100 0010

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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