-0.000 282 266 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 266(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
-0.000 282 266(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:

|-0.000 282 266| = 0.000 282 266


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: 0.000 282 266.

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) 0.000 282 266 × 2 = 0 + 0.000 564 532;
  • 2) 0.000 564 532 × 2 = 0 + 0.001 129 064;
  • 3) 0.001 129 064 × 2 = 0 + 0.002 258 128;
  • 4) 0.002 258 128 × 2 = 0 + 0.004 516 256;
  • 5) 0.004 516 256 × 2 = 0 + 0.009 032 512;
  • 6) 0.009 032 512 × 2 = 0 + 0.018 065 024;
  • 7) 0.018 065 024 × 2 = 0 + 0.036 130 048;
  • 8) 0.036 130 048 × 2 = 0 + 0.072 260 096;
  • 9) 0.072 260 096 × 2 = 0 + 0.144 520 192;
  • 10) 0.144 520 192 × 2 = 0 + 0.289 040 384;
  • 11) 0.289 040 384 × 2 = 0 + 0.578 080 768;
  • 12) 0.578 080 768 × 2 = 1 + 0.156 161 536;
  • 13) 0.156 161 536 × 2 = 0 + 0.312 323 072;
  • 14) 0.312 323 072 × 2 = 0 + 0.624 646 144;
  • 15) 0.624 646 144 × 2 = 1 + 0.249 292 288;
  • 16) 0.249 292 288 × 2 = 0 + 0.498 584 576;
  • 17) 0.498 584 576 × 2 = 0 + 0.997 169 152;
  • 18) 0.997 169 152 × 2 = 1 + 0.994 338 304;
  • 19) 0.994 338 304 × 2 = 1 + 0.988 676 608;
  • 20) 0.988 676 608 × 2 = 1 + 0.977 353 216;
  • 21) 0.977 353 216 × 2 = 1 + 0.954 706 432;
  • 22) 0.954 706 432 × 2 = 1 + 0.909 412 864;
  • 23) 0.909 412 864 × 2 = 1 + 0.818 825 728;
  • 24) 0.818 825 728 × 2 = 1 + 0.637 651 456;
  • 25) 0.637 651 456 × 2 = 1 + 0.275 302 912;
  • 26) 0.275 302 912 × 2 = 0 + 0.550 605 824;
  • 27) 0.550 605 824 × 2 = 1 + 0.101 211 648;
  • 28) 0.101 211 648 × 2 = 0 + 0.202 423 296;
  • 29) 0.202 423 296 × 2 = 0 + 0.404 846 592;
  • 30) 0.404 846 592 × 2 = 0 + 0.809 693 184;
  • 31) 0.809 693 184 × 2 = 1 + 0.619 386 368;
  • 32) 0.619 386 368 × 2 = 1 + 0.238 772 736;
  • 33) 0.238 772 736 × 2 = 0 + 0.477 545 472;
  • 34) 0.477 545 472 × 2 = 0 + 0.955 090 944;
  • 35) 0.955 090 944 × 2 = 1 + 0.910 181 888;
  • 36) 0.910 181 888 × 2 = 1 + 0.820 363 776;
  • 37) 0.820 363 776 × 2 = 1 + 0.640 727 552;
  • 38) 0.640 727 552 × 2 = 1 + 0.281 455 104;
  • 39) 0.281 455 104 × 2 = 0 + 0.562 910 208;
  • 40) 0.562 910 208 × 2 = 1 + 0.125 820 416;
  • 41) 0.125 820 416 × 2 = 0 + 0.251 640 832;
  • 42) 0.251 640 832 × 2 = 0 + 0.503 281 664;
  • 43) 0.503 281 664 × 2 = 1 + 0.006 563 328;
  • 44) 0.006 563 328 × 2 = 0 + 0.013 126 656;
  • 45) 0.013 126 656 × 2 = 0 + 0.026 253 312;
  • 46) 0.026 253 312 × 2 = 0 + 0.052 506 624;
  • 47) 0.052 506 624 × 2 = 0 + 0.105 013 248;
  • 48) 0.105 013 248 × 2 = 0 + 0.210 026 496;
  • 49) 0.210 026 496 × 2 = 0 + 0.420 052 992;
  • 50) 0.420 052 992 × 2 = 0 + 0.840 105 984;
  • 51) 0.840 105 984 × 2 = 1 + 0.680 211 968;
  • 52) 0.680 211 968 × 2 = 1 + 0.360 423 936;
  • 53) 0.360 423 936 × 2 = 0 + 0.720 847 872;
  • 54) 0.720 847 872 × 2 = 1 + 0.441 695 744;
  • 55) 0.441 695 744 × 2 = 0 + 0.883 391 488;
  • 56) 0.883 391 488 × 2 = 1 + 0.766 782 976;
  • 57) 0.766 782 976 × 2 = 1 + 0.533 565 952;
  • 58) 0.533 565 952 × 2 = 1 + 0.067 131 904;
  • 59) 0.067 131 904 × 2 = 0 + 0.134 263 808;
  • 60) 0.134 263 808 × 2 = 0 + 0.268 527 616;
  • 61) 0.268 527 616 × 2 = 0 + 0.537 055 232;
  • 62) 0.537 055 232 × 2 = 1 + 0.074 110 464;
  • 63) 0.074 110 464 × 2 = 0 + 0.148 220 928;
  • 64) 0.148 220 928 × 2 = 0 + 0.296 441 856;

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:


0.000 282 266(10) =

0.0000 0000 0001 0010 0111 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100(2)

6. Positive number before normalization:

0.000 282 266(10) =

0.0000 0000 0001 0010 0111 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100(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:


0.000 282 266(10) =

0.0000 0000 0001 0010 0111 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100(2) =

0.0000 0000 0001 0010 0111 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100(2) × 20 =

1.0010 0111 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100(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 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100


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 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100 =

0010 0111 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100


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 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100


Decimal number -0.000 282 266 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1111 1010 0011 0011 1101 0010 0000 0011 0101 1100 0100

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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