-0.000 282 008 25 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 008 25(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
-0.000 282 008 25(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:

|-0.000 282 008 25| = 0.000 282 008 25


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: 0.000 282 008 25.

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) 0.000 282 008 25 × 2 = 0 + 0.000 564 016 5;
  • 2) 0.000 564 016 5 × 2 = 0 + 0.001 128 033;
  • 3) 0.001 128 033 × 2 = 0 + 0.002 256 066;
  • 4) 0.002 256 066 × 2 = 0 + 0.004 512 132;
  • 5) 0.004 512 132 × 2 = 0 + 0.009 024 264;
  • 6) 0.009 024 264 × 2 = 0 + 0.018 048 528;
  • 7) 0.018 048 528 × 2 = 0 + 0.036 097 056;
  • 8) 0.036 097 056 × 2 = 0 + 0.072 194 112;
  • 9) 0.072 194 112 × 2 = 0 + 0.144 388 224;
  • 10) 0.144 388 224 × 2 = 0 + 0.288 776 448;
  • 11) 0.288 776 448 × 2 = 0 + 0.577 552 896;
  • 12) 0.577 552 896 × 2 = 1 + 0.155 105 792;
  • 13) 0.155 105 792 × 2 = 0 + 0.310 211 584;
  • 14) 0.310 211 584 × 2 = 0 + 0.620 423 168;
  • 15) 0.620 423 168 × 2 = 1 + 0.240 846 336;
  • 16) 0.240 846 336 × 2 = 0 + 0.481 692 672;
  • 17) 0.481 692 672 × 2 = 0 + 0.963 385 344;
  • 18) 0.963 385 344 × 2 = 1 + 0.926 770 688;
  • 19) 0.926 770 688 × 2 = 1 + 0.853 541 376;
  • 20) 0.853 541 376 × 2 = 1 + 0.707 082 752;
  • 21) 0.707 082 752 × 2 = 1 + 0.414 165 504;
  • 22) 0.414 165 504 × 2 = 0 + 0.828 331 008;
  • 23) 0.828 331 008 × 2 = 1 + 0.656 662 016;
  • 24) 0.656 662 016 × 2 = 1 + 0.313 324 032;
  • 25) 0.313 324 032 × 2 = 0 + 0.626 648 064;
  • 26) 0.626 648 064 × 2 = 1 + 0.253 296 128;
  • 27) 0.253 296 128 × 2 = 0 + 0.506 592 256;
  • 28) 0.506 592 256 × 2 = 1 + 0.013 184 512;
  • 29) 0.013 184 512 × 2 = 0 + 0.026 369 024;
  • 30) 0.026 369 024 × 2 = 0 + 0.052 738 048;
  • 31) 0.052 738 048 × 2 = 0 + 0.105 476 096;
  • 32) 0.105 476 096 × 2 = 0 + 0.210 952 192;
  • 33) 0.210 952 192 × 2 = 0 + 0.421 904 384;
  • 34) 0.421 904 384 × 2 = 0 + 0.843 808 768;
  • 35) 0.843 808 768 × 2 = 1 + 0.687 617 536;
  • 36) 0.687 617 536 × 2 = 1 + 0.375 235 072;
  • 37) 0.375 235 072 × 2 = 0 + 0.750 470 144;
  • 38) 0.750 470 144 × 2 = 1 + 0.500 940 288;
  • 39) 0.500 940 288 × 2 = 1 + 0.001 880 576;
  • 40) 0.001 880 576 × 2 = 0 + 0.003 761 152;
  • 41) 0.003 761 152 × 2 = 0 + 0.007 522 304;
  • 42) 0.007 522 304 × 2 = 0 + 0.015 044 608;
  • 43) 0.015 044 608 × 2 = 0 + 0.030 089 216;
  • 44) 0.030 089 216 × 2 = 0 + 0.060 178 432;
  • 45) 0.060 178 432 × 2 = 0 + 0.120 356 864;
  • 46) 0.120 356 864 × 2 = 0 + 0.240 713 728;
  • 47) 0.240 713 728 × 2 = 0 + 0.481 427 456;
  • 48) 0.481 427 456 × 2 = 0 + 0.962 854 912;
  • 49) 0.962 854 912 × 2 = 1 + 0.925 709 824;
  • 50) 0.925 709 824 × 2 = 1 + 0.851 419 648;
  • 51) 0.851 419 648 × 2 = 1 + 0.702 839 296;
  • 52) 0.702 839 296 × 2 = 1 + 0.405 678 592;
  • 53) 0.405 678 592 × 2 = 0 + 0.811 357 184;
  • 54) 0.811 357 184 × 2 = 1 + 0.622 714 368;
  • 55) 0.622 714 368 × 2 = 1 + 0.245 428 736;
  • 56) 0.245 428 736 × 2 = 0 + 0.490 857 472;
  • 57) 0.490 857 472 × 2 = 0 + 0.981 714 944;
  • 58) 0.981 714 944 × 2 = 1 + 0.963 429 888;
  • 59) 0.963 429 888 × 2 = 1 + 0.926 859 776;
  • 60) 0.926 859 776 × 2 = 1 + 0.853 719 552;
  • 61) 0.853 719 552 × 2 = 1 + 0.707 439 104;
  • 62) 0.707 439 104 × 2 = 1 + 0.414 878 208;
  • 63) 0.414 878 208 × 2 = 0 + 0.829 756 416;
  • 64) 0.829 756 416 × 2 = 1 + 0.659 512 832;

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:


0.000 282 008 25(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101(2)

6. Positive number before normalization:

0.000 282 008 25(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101(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:


0.000 282 008 25(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101(2) =

0.0000 0000 0001 0010 0111 1011 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101(2) × 20 =

1.0010 0111 1011 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101(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 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101


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 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101 =

0010 0111 1011 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101


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 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101


Decimal number -0.000 282 008 25 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0000 0011 0110 0000 0000 1111 0110 0111 1101

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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