0.026 916 681 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.026 916 681(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
0.026 916 681(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: 0.026 916 681.

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) 0.026 916 681 × 2 = 0 + 0.053 833 362;
  • 2) 0.053 833 362 × 2 = 0 + 0.107 666 724;
  • 3) 0.107 666 724 × 2 = 0 + 0.215 333 448;
  • 4) 0.215 333 448 × 2 = 0 + 0.430 666 896;
  • 5) 0.430 666 896 × 2 = 0 + 0.861 333 792;
  • 6) 0.861 333 792 × 2 = 1 + 0.722 667 584;
  • 7) 0.722 667 584 × 2 = 1 + 0.445 335 168;
  • 8) 0.445 335 168 × 2 = 0 + 0.890 670 336;
  • 9) 0.890 670 336 × 2 = 1 + 0.781 340 672;
  • 10) 0.781 340 672 × 2 = 1 + 0.562 681 344;
  • 11) 0.562 681 344 × 2 = 1 + 0.125 362 688;
  • 12) 0.125 362 688 × 2 = 0 + 0.250 725 376;
  • 13) 0.250 725 376 × 2 = 0 + 0.501 450 752;
  • 14) 0.501 450 752 × 2 = 1 + 0.002 901 504;
  • 15) 0.002 901 504 × 2 = 0 + 0.005 803 008;
  • 16) 0.005 803 008 × 2 = 0 + 0.011 606 016;
  • 17) 0.011 606 016 × 2 = 0 + 0.023 212 032;
  • 18) 0.023 212 032 × 2 = 0 + 0.046 424 064;
  • 19) 0.046 424 064 × 2 = 0 + 0.092 848 128;
  • 20) 0.092 848 128 × 2 = 0 + 0.185 696 256;
  • 21) 0.185 696 256 × 2 = 0 + 0.371 392 512;
  • 22) 0.371 392 512 × 2 = 0 + 0.742 785 024;
  • 23) 0.742 785 024 × 2 = 1 + 0.485 570 048;
  • 24) 0.485 570 048 × 2 = 0 + 0.971 140 096;
  • 25) 0.971 140 096 × 2 = 1 + 0.942 280 192;
  • 26) 0.942 280 192 × 2 = 1 + 0.884 560 384;
  • 27) 0.884 560 384 × 2 = 1 + 0.769 120 768;
  • 28) 0.769 120 768 × 2 = 1 + 0.538 241 536;
  • 29) 0.538 241 536 × 2 = 1 + 0.076 483 072;
  • 30) 0.076 483 072 × 2 = 0 + 0.152 966 144;
  • 31) 0.152 966 144 × 2 = 0 + 0.305 932 288;
  • 32) 0.305 932 288 × 2 = 0 + 0.611 864 576;
  • 33) 0.611 864 576 × 2 = 1 + 0.223 729 152;
  • 34) 0.223 729 152 × 2 = 0 + 0.447 458 304;
  • 35) 0.447 458 304 × 2 = 0 + 0.894 916 608;
  • 36) 0.894 916 608 × 2 = 1 + 0.789 833 216;
  • 37) 0.789 833 216 × 2 = 1 + 0.579 666 432;
  • 38) 0.579 666 432 × 2 = 1 + 0.159 332 864;
  • 39) 0.159 332 864 × 2 = 0 + 0.318 665 728;
  • 40) 0.318 665 728 × 2 = 0 + 0.637 331 456;
  • 41) 0.637 331 456 × 2 = 1 + 0.274 662 912;
  • 42) 0.274 662 912 × 2 = 0 + 0.549 325 824;
  • 43) 0.549 325 824 × 2 = 1 + 0.098 651 648;
  • 44) 0.098 651 648 × 2 = 0 + 0.197 303 296;
  • 45) 0.197 303 296 × 2 = 0 + 0.394 606 592;
  • 46) 0.394 606 592 × 2 = 0 + 0.789 213 184;
  • 47) 0.789 213 184 × 2 = 1 + 0.578 426 368;
  • 48) 0.578 426 368 × 2 = 1 + 0.156 852 736;
  • 49) 0.156 852 736 × 2 = 0 + 0.313 705 472;
  • 50) 0.313 705 472 × 2 = 0 + 0.627 410 944;
  • 51) 0.627 410 944 × 2 = 1 + 0.254 821 888;
  • 52) 0.254 821 888 × 2 = 0 + 0.509 643 776;
  • 53) 0.509 643 776 × 2 = 1 + 0.019 287 552;
  • 54) 0.019 287 552 × 2 = 0 + 0.038 575 104;
  • 55) 0.038 575 104 × 2 = 0 + 0.077 150 208;
  • 56) 0.077 150 208 × 2 = 0 + 0.154 300 416;
  • 57) 0.154 300 416 × 2 = 0 + 0.308 600 832;
  • 58) 0.308 600 832 × 2 = 0 + 0.617 201 664;

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:


0.026 916 681(10) =

0.0000 0110 1110 0100 0000 0010 1111 1000 1001 1100 1010 0011 0010 1000 00(2)

5. Positive number before normalization:

0.026 916 681(10) =

0.0000 0110 1110 0100 0000 0010 1111 1000 1001 1100 1010 0011 0010 1000 00(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:


0.026 916 681(10) =

0.0000 0110 1110 0100 0000 0010 1111 1000 1001 1100 1010 0011 0010 1000 00(2) =

0.0000 0110 1110 0100 0000 0010 1111 1000 1001 1100 1010 0011 0010 1000 00(2) × 20 =

1.1011 1001 0000 0000 1011 1110 0010 0111 0010 1000 1100 1010 0000(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 0000 1011 1110 0010 0111 0010 1000 1100 1010 0000


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 0000 1011 1110 0010 0111 0010 1000 1100 1010 0000 =

1011 1001 0000 0000 1011 1110 0010 0111 0010 1000 1100 1010 0000


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 0000 1011 1110 0010 0111 0010 1000 1100 1010 0000


Decimal number 0.026 916 681 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 1011 1001 0000 0000 1011 1110 0010 0111 0010 1000 1100 1010 0000

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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