-0.000 282 403 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 403(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
-0.000 282 403(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:

|-0.000 282 403| = 0.000 282 403


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: 0.000 282 403.

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) 0.000 282 403 × 2 = 0 + 0.000 564 806;
  • 2) 0.000 564 806 × 2 = 0 + 0.001 129 612;
  • 3) 0.001 129 612 × 2 = 0 + 0.002 259 224;
  • 4) 0.002 259 224 × 2 = 0 + 0.004 518 448;
  • 5) 0.004 518 448 × 2 = 0 + 0.009 036 896;
  • 6) 0.009 036 896 × 2 = 0 + 0.018 073 792;
  • 7) 0.018 073 792 × 2 = 0 + 0.036 147 584;
  • 8) 0.036 147 584 × 2 = 0 + 0.072 295 168;
  • 9) 0.072 295 168 × 2 = 0 + 0.144 590 336;
  • 10) 0.144 590 336 × 2 = 0 + 0.289 180 672;
  • 11) 0.289 180 672 × 2 = 0 + 0.578 361 344;
  • 12) 0.578 361 344 × 2 = 1 + 0.156 722 688;
  • 13) 0.156 722 688 × 2 = 0 + 0.313 445 376;
  • 14) 0.313 445 376 × 2 = 0 + 0.626 890 752;
  • 15) 0.626 890 752 × 2 = 1 + 0.253 781 504;
  • 16) 0.253 781 504 × 2 = 0 + 0.507 563 008;
  • 17) 0.507 563 008 × 2 = 1 + 0.015 126 016;
  • 18) 0.015 126 016 × 2 = 0 + 0.030 252 032;
  • 19) 0.030 252 032 × 2 = 0 + 0.060 504 064;
  • 20) 0.060 504 064 × 2 = 0 + 0.121 008 128;
  • 21) 0.121 008 128 × 2 = 0 + 0.242 016 256;
  • 22) 0.242 016 256 × 2 = 0 + 0.484 032 512;
  • 23) 0.484 032 512 × 2 = 0 + 0.968 065 024;
  • 24) 0.968 065 024 × 2 = 1 + 0.936 130 048;
  • 25) 0.936 130 048 × 2 = 1 + 0.872 260 096;
  • 26) 0.872 260 096 × 2 = 1 + 0.744 520 192;
  • 27) 0.744 520 192 × 2 = 1 + 0.489 040 384;
  • 28) 0.489 040 384 × 2 = 0 + 0.978 080 768;
  • 29) 0.978 080 768 × 2 = 1 + 0.956 161 536;
  • 30) 0.956 161 536 × 2 = 1 + 0.912 323 072;
  • 31) 0.912 323 072 × 2 = 1 + 0.824 646 144;
  • 32) 0.824 646 144 × 2 = 1 + 0.649 292 288;
  • 33) 0.649 292 288 × 2 = 1 + 0.298 584 576;
  • 34) 0.298 584 576 × 2 = 0 + 0.597 169 152;
  • 35) 0.597 169 152 × 2 = 1 + 0.194 338 304;
  • 36) 0.194 338 304 × 2 = 0 + 0.388 676 608;
  • 37) 0.388 676 608 × 2 = 0 + 0.777 353 216;
  • 38) 0.777 353 216 × 2 = 1 + 0.554 706 432;
  • 39) 0.554 706 432 × 2 = 1 + 0.109 412 864;
  • 40) 0.109 412 864 × 2 = 0 + 0.218 825 728;
  • 41) 0.218 825 728 × 2 = 0 + 0.437 651 456;
  • 42) 0.437 651 456 × 2 = 0 + 0.875 302 912;
  • 43) 0.875 302 912 × 2 = 1 + 0.750 605 824;
  • 44) 0.750 605 824 × 2 = 1 + 0.501 211 648;
  • 45) 0.501 211 648 × 2 = 1 + 0.002 423 296;
  • 46) 0.002 423 296 × 2 = 0 + 0.004 846 592;
  • 47) 0.004 846 592 × 2 = 0 + 0.009 693 184;
  • 48) 0.009 693 184 × 2 = 0 + 0.019 386 368;
  • 49) 0.019 386 368 × 2 = 0 + 0.038 772 736;
  • 50) 0.038 772 736 × 2 = 0 + 0.077 545 472;
  • 51) 0.077 545 472 × 2 = 0 + 0.155 090 944;
  • 52) 0.155 090 944 × 2 = 0 + 0.310 181 888;
  • 53) 0.310 181 888 × 2 = 0 + 0.620 363 776;
  • 54) 0.620 363 776 × 2 = 1 + 0.240 727 552;
  • 55) 0.240 727 552 × 2 = 0 + 0.481 455 104;
  • 56) 0.481 455 104 × 2 = 0 + 0.962 910 208;
  • 57) 0.962 910 208 × 2 = 1 + 0.925 820 416;
  • 58) 0.925 820 416 × 2 = 1 + 0.851 640 832;
  • 59) 0.851 640 832 × 2 = 1 + 0.703 281 664;
  • 60) 0.703 281 664 × 2 = 1 + 0.406 563 328;
  • 61) 0.406 563 328 × 2 = 0 + 0.813 126 656;
  • 62) 0.813 126 656 × 2 = 1 + 0.626 253 312;
  • 63) 0.626 253 312 × 2 = 1 + 0.252 506 624;
  • 64) 0.252 506 624 × 2 = 0 + 0.505 013 248;

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:


0.000 282 403(10) =

0.0000 0000 0001 0010 1000 0001 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110(2)

6. Positive number before normalization:

0.000 282 403(10) =

0.0000 0000 0001 0010 1000 0001 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110(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:


0.000 282 403(10) =

0.0000 0000 0001 0010 1000 0001 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110(2) =

0.0000 0000 0001 0010 1000 0001 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110(2) × 20 =

1.0010 1000 0001 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110(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 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110


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 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110 =

0010 1000 0001 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110


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 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110


Decimal number -0.000 282 403 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0001 1110 1111 1010 0110 0011 1000 0000 0100 1111 0110

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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