-0.000 282 396 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 396(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
-0.000 282 396(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:

|-0.000 282 396| = 0.000 282 396


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: 0.000 282 396.

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) 0.000 282 396 × 2 = 0 + 0.000 564 792;
  • 2) 0.000 564 792 × 2 = 0 + 0.001 129 584;
  • 3) 0.001 129 584 × 2 = 0 + 0.002 259 168;
  • 4) 0.002 259 168 × 2 = 0 + 0.004 518 336;
  • 5) 0.004 518 336 × 2 = 0 + 0.009 036 672;
  • 6) 0.009 036 672 × 2 = 0 + 0.018 073 344;
  • 7) 0.018 073 344 × 2 = 0 + 0.036 146 688;
  • 8) 0.036 146 688 × 2 = 0 + 0.072 293 376;
  • 9) 0.072 293 376 × 2 = 0 + 0.144 586 752;
  • 10) 0.144 586 752 × 2 = 0 + 0.289 173 504;
  • 11) 0.289 173 504 × 2 = 0 + 0.578 347 008;
  • 12) 0.578 347 008 × 2 = 1 + 0.156 694 016;
  • 13) 0.156 694 016 × 2 = 0 + 0.313 388 032;
  • 14) 0.313 388 032 × 2 = 0 + 0.626 776 064;
  • 15) 0.626 776 064 × 2 = 1 + 0.253 552 128;
  • 16) 0.253 552 128 × 2 = 0 + 0.507 104 256;
  • 17) 0.507 104 256 × 2 = 1 + 0.014 208 512;
  • 18) 0.014 208 512 × 2 = 0 + 0.028 417 024;
  • 19) 0.028 417 024 × 2 = 0 + 0.056 834 048;
  • 20) 0.056 834 048 × 2 = 0 + 0.113 668 096;
  • 21) 0.113 668 096 × 2 = 0 + 0.227 336 192;
  • 22) 0.227 336 192 × 2 = 0 + 0.454 672 384;
  • 23) 0.454 672 384 × 2 = 0 + 0.909 344 768;
  • 24) 0.909 344 768 × 2 = 1 + 0.818 689 536;
  • 25) 0.818 689 536 × 2 = 1 + 0.637 379 072;
  • 26) 0.637 379 072 × 2 = 1 + 0.274 758 144;
  • 27) 0.274 758 144 × 2 = 0 + 0.549 516 288;
  • 28) 0.549 516 288 × 2 = 1 + 0.099 032 576;
  • 29) 0.099 032 576 × 2 = 0 + 0.198 065 152;
  • 30) 0.198 065 152 × 2 = 0 + 0.396 130 304;
  • 31) 0.396 130 304 × 2 = 0 + 0.792 260 608;
  • 32) 0.792 260 608 × 2 = 1 + 0.584 521 216;
  • 33) 0.584 521 216 × 2 = 1 + 0.169 042 432;
  • 34) 0.169 042 432 × 2 = 0 + 0.338 084 864;
  • 35) 0.338 084 864 × 2 = 0 + 0.676 169 728;
  • 36) 0.676 169 728 × 2 = 1 + 0.352 339 456;
  • 37) 0.352 339 456 × 2 = 0 + 0.704 678 912;
  • 38) 0.704 678 912 × 2 = 1 + 0.409 357 824;
  • 39) 0.409 357 824 × 2 = 0 + 0.818 715 648;
  • 40) 0.818 715 648 × 2 = 1 + 0.637 431 296;
  • 41) 0.637 431 296 × 2 = 1 + 0.274 862 592;
  • 42) 0.274 862 592 × 2 = 0 + 0.549 725 184;
  • 43) 0.549 725 184 × 2 = 1 + 0.099 450 368;
  • 44) 0.099 450 368 × 2 = 0 + 0.198 900 736;
  • 45) 0.198 900 736 × 2 = 0 + 0.397 801 472;
  • 46) 0.397 801 472 × 2 = 0 + 0.795 602 944;
  • 47) 0.795 602 944 × 2 = 1 + 0.591 205 888;
  • 48) 0.591 205 888 × 2 = 1 + 0.182 411 776;
  • 49) 0.182 411 776 × 2 = 0 + 0.364 823 552;
  • 50) 0.364 823 552 × 2 = 0 + 0.729 647 104;
  • 51) 0.729 647 104 × 2 = 1 + 0.459 294 208;
  • 52) 0.459 294 208 × 2 = 0 + 0.918 588 416;
  • 53) 0.918 588 416 × 2 = 1 + 0.837 176 832;
  • 54) 0.837 176 832 × 2 = 1 + 0.674 353 664;
  • 55) 0.674 353 664 × 2 = 1 + 0.348 707 328;
  • 56) 0.348 707 328 × 2 = 0 + 0.697 414 656;
  • 57) 0.697 414 656 × 2 = 1 + 0.394 829 312;
  • 58) 0.394 829 312 × 2 = 0 + 0.789 658 624;
  • 59) 0.789 658 624 × 2 = 1 + 0.579 317 248;
  • 60) 0.579 317 248 × 2 = 1 + 0.158 634 496;
  • 61) 0.158 634 496 × 2 = 0 + 0.317 268 992;
  • 62) 0.317 268 992 × 2 = 0 + 0.634 537 984;
  • 63) 0.634 537 984 × 2 = 1 + 0.269 075 968;
  • 64) 0.269 075 968 × 2 = 0 + 0.538 151 936;

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:


0.000 282 396(10) =

0.0000 0000 0001 0010 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 0010(2)

6. Positive number before normalization:

0.000 282 396(10) =

0.0000 0000 0001 0010 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 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:


0.000 282 396(10) =

0.0000 0000 0001 0010 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 0010(2) =

0.0000 0000 0001 0010 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 0010(2) × 20 =

1.0010 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 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 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 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 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 0010 =

0010 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 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 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 0010


Decimal number -0.000 282 396 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0001 1101 0001 1001 0101 1010 0011 0010 1110 1011 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