0.026 916 749 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.026 916 749(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
0.026 916 749(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.026 916 749.

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) 0.026 916 749 × 2 = 0 + 0.053 833 498;
  • 2) 0.053 833 498 × 2 = 0 + 0.107 666 996;
  • 3) 0.107 666 996 × 2 = 0 + 0.215 333 992;
  • 4) 0.215 333 992 × 2 = 0 + 0.430 667 984;
  • 5) 0.430 667 984 × 2 = 0 + 0.861 335 968;
  • 6) 0.861 335 968 × 2 = 1 + 0.722 671 936;
  • 7) 0.722 671 936 × 2 = 1 + 0.445 343 872;
  • 8) 0.445 343 872 × 2 = 0 + 0.890 687 744;
  • 9) 0.890 687 744 × 2 = 1 + 0.781 375 488;
  • 10) 0.781 375 488 × 2 = 1 + 0.562 750 976;
  • 11) 0.562 750 976 × 2 = 1 + 0.125 501 952;
  • 12) 0.125 501 952 × 2 = 0 + 0.251 003 904;
  • 13) 0.251 003 904 × 2 = 0 + 0.502 007 808;
  • 14) 0.502 007 808 × 2 = 1 + 0.004 015 616;
  • 15) 0.004 015 616 × 2 = 0 + 0.008 031 232;
  • 16) 0.008 031 232 × 2 = 0 + 0.016 062 464;
  • 17) 0.016 062 464 × 2 = 0 + 0.032 124 928;
  • 18) 0.032 124 928 × 2 = 0 + 0.064 249 856;
  • 19) 0.064 249 856 × 2 = 0 + 0.128 499 712;
  • 20) 0.128 499 712 × 2 = 0 + 0.256 999 424;
  • 21) 0.256 999 424 × 2 = 0 + 0.513 998 848;
  • 22) 0.513 998 848 × 2 = 1 + 0.027 997 696;
  • 23) 0.027 997 696 × 2 = 0 + 0.055 995 392;
  • 24) 0.055 995 392 × 2 = 0 + 0.111 990 784;
  • 25) 0.111 990 784 × 2 = 0 + 0.223 981 568;
  • 26) 0.223 981 568 × 2 = 0 + 0.447 963 136;
  • 27) 0.447 963 136 × 2 = 0 + 0.895 926 272;
  • 28) 0.895 926 272 × 2 = 1 + 0.791 852 544;
  • 29) 0.791 852 544 × 2 = 1 + 0.583 705 088;
  • 30) 0.583 705 088 × 2 = 1 + 0.167 410 176;
  • 31) 0.167 410 176 × 2 = 0 + 0.334 820 352;
  • 32) 0.334 820 352 × 2 = 0 + 0.669 640 704;
  • 33) 0.669 640 704 × 2 = 1 + 0.339 281 408;
  • 34) 0.339 281 408 × 2 = 0 + 0.678 562 816;
  • 35) 0.678 562 816 × 2 = 1 + 0.357 125 632;
  • 36) 0.357 125 632 × 2 = 0 + 0.714 251 264;
  • 37) 0.714 251 264 × 2 = 1 + 0.428 502 528;
  • 38) 0.428 502 528 × 2 = 0 + 0.857 005 056;
  • 39) 0.857 005 056 × 2 = 1 + 0.714 010 112;
  • 40) 0.714 010 112 × 2 = 1 + 0.428 020 224;
  • 41) 0.428 020 224 × 2 = 0 + 0.856 040 448;
  • 42) 0.856 040 448 × 2 = 1 + 0.712 080 896;
  • 43) 0.712 080 896 × 2 = 1 + 0.424 161 792;
  • 44) 0.424 161 792 × 2 = 0 + 0.848 323 584;
  • 45) 0.848 323 584 × 2 = 1 + 0.696 647 168;
  • 46) 0.696 647 168 × 2 = 1 + 0.393 294 336;
  • 47) 0.393 294 336 × 2 = 0 + 0.786 588 672;
  • 48) 0.786 588 672 × 2 = 1 + 0.573 177 344;
  • 49) 0.573 177 344 × 2 = 1 + 0.146 354 688;
  • 50) 0.146 354 688 × 2 = 0 + 0.292 709 376;
  • 51) 0.292 709 376 × 2 = 0 + 0.585 418 752;
  • 52) 0.585 418 752 × 2 = 1 + 0.170 837 504;
  • 53) 0.170 837 504 × 2 = 0 + 0.341 675 008;
  • 54) 0.341 675 008 × 2 = 0 + 0.683 350 016;
  • 55) 0.683 350 016 × 2 = 1 + 0.366 700 032;
  • 56) 0.366 700 032 × 2 = 0 + 0.733 400 064;
  • 57) 0.733 400 064 × 2 = 1 + 0.466 800 128;
  • 58) 0.466 800 128 × 2 = 0 + 0.933 600 256;

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).


4. 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:


0.026 916 749(10) =

0.0000 0110 1110 0100 0000 0100 0001 1100 1010 1011 0110 1101 1001 0010 10(2)

5. Positive number before normalization:

0.026 916 749(10) =

0.0000 0110 1110 0100 0000 0100 0001 1100 1010 1011 0110 1101 1001 0010 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.026 916 749(10) =

0.0000 0110 1110 0100 0000 0100 0001 1100 1010 1011 0110 1101 1001 0010 10(2) =

0.0000 0110 1110 0100 0000 0100 0001 1100 1010 1011 0110 1101 1001 0010 10(2) × 20 =

1.1011 1001 0000 0001 0000 0111 0010 1010 1101 1011 0110 0100 1010(2) × 2-6


7. 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 0 (a positive number)

Exponent (unadjusted): -6

Mantissa (not normalized):
1.1011 1001 0000 0001 0000 0111 0010 1010 1101 1011 0110 0100 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


9. 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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


11. 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. 1011 1001 0000 0001 0000 0111 0010 1010 1101 1011 0110 0100 1010 =

1011 1001 0000 0001 0000 0111 0010 1010 1101 1011 0110 0100 1010


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
1011 1001 0000 0001 0000 0111 0010 1010 1101 1011 0110 0100 1010


Decimal number 0.026 916 749 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 1011 1001 0000 0001 0000 0111 0010 1010 1101 1011 0110 0100 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