-0.000 282 181 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 181(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
-0.000 282 181(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:

|-0.000 282 181| = 0.000 282 181


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: 0.000 282 181.

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) 0.000 282 181 × 2 = 0 + 0.000 564 362;
  • 2) 0.000 564 362 × 2 = 0 + 0.001 128 724;
  • 3) 0.001 128 724 × 2 = 0 + 0.002 257 448;
  • 4) 0.002 257 448 × 2 = 0 + 0.004 514 896;
  • 5) 0.004 514 896 × 2 = 0 + 0.009 029 792;
  • 6) 0.009 029 792 × 2 = 0 + 0.018 059 584;
  • 7) 0.018 059 584 × 2 = 0 + 0.036 119 168;
  • 8) 0.036 119 168 × 2 = 0 + 0.072 238 336;
  • 9) 0.072 238 336 × 2 = 0 + 0.144 476 672;
  • 10) 0.144 476 672 × 2 = 0 + 0.288 953 344;
  • 11) 0.288 953 344 × 2 = 0 + 0.577 906 688;
  • 12) 0.577 906 688 × 2 = 1 + 0.155 813 376;
  • 13) 0.155 813 376 × 2 = 0 + 0.311 626 752;
  • 14) 0.311 626 752 × 2 = 0 + 0.623 253 504;
  • 15) 0.623 253 504 × 2 = 1 + 0.246 507 008;
  • 16) 0.246 507 008 × 2 = 0 + 0.493 014 016;
  • 17) 0.493 014 016 × 2 = 0 + 0.986 028 032;
  • 18) 0.986 028 032 × 2 = 1 + 0.972 056 064;
  • 19) 0.972 056 064 × 2 = 1 + 0.944 112 128;
  • 20) 0.944 112 128 × 2 = 1 + 0.888 224 256;
  • 21) 0.888 224 256 × 2 = 1 + 0.776 448 512;
  • 22) 0.776 448 512 × 2 = 1 + 0.552 897 024;
  • 23) 0.552 897 024 × 2 = 1 + 0.105 794 048;
  • 24) 0.105 794 048 × 2 = 0 + 0.211 588 096;
  • 25) 0.211 588 096 × 2 = 0 + 0.423 176 192;
  • 26) 0.423 176 192 × 2 = 0 + 0.846 352 384;
  • 27) 0.846 352 384 × 2 = 1 + 0.692 704 768;
  • 28) 0.692 704 768 × 2 = 1 + 0.385 409 536;
  • 29) 0.385 409 536 × 2 = 0 + 0.770 819 072;
  • 30) 0.770 819 072 × 2 = 1 + 0.541 638 144;
  • 31) 0.541 638 144 × 2 = 1 + 0.083 276 288;
  • 32) 0.083 276 288 × 2 = 0 + 0.166 552 576;
  • 33) 0.166 552 576 × 2 = 0 + 0.333 105 152;
  • 34) 0.333 105 152 × 2 = 0 + 0.666 210 304;
  • 35) 0.666 210 304 × 2 = 1 + 0.332 420 608;
  • 36) 0.332 420 608 × 2 = 0 + 0.664 841 216;
  • 37) 0.664 841 216 × 2 = 1 + 0.329 682 432;
  • 38) 0.329 682 432 × 2 = 0 + 0.659 364 864;
  • 39) 0.659 364 864 × 2 = 1 + 0.318 729 728;
  • 40) 0.318 729 728 × 2 = 0 + 0.637 459 456;
  • 41) 0.637 459 456 × 2 = 1 + 0.274 918 912;
  • 42) 0.274 918 912 × 2 = 0 + 0.549 837 824;
  • 43) 0.549 837 824 × 2 = 1 + 0.099 675 648;
  • 44) 0.099 675 648 × 2 = 0 + 0.199 351 296;
  • 45) 0.199 351 296 × 2 = 0 + 0.398 702 592;
  • 46) 0.398 702 592 × 2 = 0 + 0.797 405 184;
  • 47) 0.797 405 184 × 2 = 1 + 0.594 810 368;
  • 48) 0.594 810 368 × 2 = 1 + 0.189 620 736;
  • 49) 0.189 620 736 × 2 = 0 + 0.379 241 472;
  • 50) 0.379 241 472 × 2 = 0 + 0.758 482 944;
  • 51) 0.758 482 944 × 2 = 1 + 0.516 965 888;
  • 52) 0.516 965 888 × 2 = 1 + 0.033 931 776;
  • 53) 0.033 931 776 × 2 = 0 + 0.067 863 552;
  • 54) 0.067 863 552 × 2 = 0 + 0.135 727 104;
  • 55) 0.135 727 104 × 2 = 0 + 0.271 454 208;
  • 56) 0.271 454 208 × 2 = 0 + 0.542 908 416;
  • 57) 0.542 908 416 × 2 = 1 + 0.085 816 832;
  • 58) 0.085 816 832 × 2 = 0 + 0.171 633 664;
  • 59) 0.171 633 664 × 2 = 0 + 0.343 267 328;
  • 60) 0.343 267 328 × 2 = 0 + 0.686 534 656;
  • 61) 0.686 534 656 × 2 = 1 + 0.373 069 312;
  • 62) 0.373 069 312 × 2 = 0 + 0.746 138 624;
  • 63) 0.746 138 624 × 2 = 1 + 0.492 277 248;
  • 64) 0.492 277 248 × 2 = 0 + 0.984 554 496;

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:


0.000 282 181(10) =

0.0000 0000 0001 0010 0111 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010(2)

6. Positive number before normalization:

0.000 282 181(10) =

0.0000 0000 0001 0010 0111 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010(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:


0.000 282 181(10) =

0.0000 0000 0001 0010 0111 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010(2) =

0.0000 0000 0001 0010 0111 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010(2) × 20 =

1.0010 0111 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010(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 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010


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 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010 =

0010 0111 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010


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 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010


Decimal number -0.000 282 181 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1110 0011 0110 0010 1010 1010 0011 0011 0000 1000 1010

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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