0.522 136 891 326 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.522 136 891 326(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
0.522 136 891 326(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: 0.522 136 891 326.

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) 0.522 136 891 326 × 2 = 1 + 0.044 273 782 652;
  • 2) 0.044 273 782 652 × 2 = 0 + 0.088 547 565 304;
  • 3) 0.088 547 565 304 × 2 = 0 + 0.177 095 130 608;
  • 4) 0.177 095 130 608 × 2 = 0 + 0.354 190 261 216;
  • 5) 0.354 190 261 216 × 2 = 0 + 0.708 380 522 432;
  • 6) 0.708 380 522 432 × 2 = 1 + 0.416 761 044 864;
  • 7) 0.416 761 044 864 × 2 = 0 + 0.833 522 089 728;
  • 8) 0.833 522 089 728 × 2 = 1 + 0.667 044 179 456;
  • 9) 0.667 044 179 456 × 2 = 1 + 0.334 088 358 912;
  • 10) 0.334 088 358 912 × 2 = 0 + 0.668 176 717 824;
  • 11) 0.668 176 717 824 × 2 = 1 + 0.336 353 435 648;
  • 12) 0.336 353 435 648 × 2 = 0 + 0.672 706 871 296;
  • 13) 0.672 706 871 296 × 2 = 1 + 0.345 413 742 592;
  • 14) 0.345 413 742 592 × 2 = 0 + 0.690 827 485 184;
  • 15) 0.690 827 485 184 × 2 = 1 + 0.381 654 970 368;
  • 16) 0.381 654 970 368 × 2 = 0 + 0.763 309 940 736;
  • 17) 0.763 309 940 736 × 2 = 1 + 0.526 619 881 472;
  • 18) 0.526 619 881 472 × 2 = 1 + 0.053 239 762 944;
  • 19) 0.053 239 762 944 × 2 = 0 + 0.106 479 525 888;
  • 20) 0.106 479 525 888 × 2 = 0 + 0.212 959 051 776;
  • 21) 0.212 959 051 776 × 2 = 0 + 0.425 918 103 552;
  • 22) 0.425 918 103 552 × 2 = 0 + 0.851 836 207 104;
  • 23) 0.851 836 207 104 × 2 = 1 + 0.703 672 414 208;
  • 24) 0.703 672 414 208 × 2 = 1 + 0.407 344 828 416;
  • 25) 0.407 344 828 416 × 2 = 0 + 0.814 689 656 832;
  • 26) 0.814 689 656 832 × 2 = 1 + 0.629 379 313 664;
  • 27) 0.629 379 313 664 × 2 = 1 + 0.258 758 627 328;
  • 28) 0.258 758 627 328 × 2 = 0 + 0.517 517 254 656;
  • 29) 0.517 517 254 656 × 2 = 1 + 0.035 034 509 312;
  • 30) 0.035 034 509 312 × 2 = 0 + 0.070 069 018 624;
  • 31) 0.070 069 018 624 × 2 = 0 + 0.140 138 037 248;
  • 32) 0.140 138 037 248 × 2 = 0 + 0.280 276 074 496;
  • 33) 0.280 276 074 496 × 2 = 0 + 0.560 552 148 992;
  • 34) 0.560 552 148 992 × 2 = 1 + 0.121 104 297 984;
  • 35) 0.121 104 297 984 × 2 = 0 + 0.242 208 595 968;
  • 36) 0.242 208 595 968 × 2 = 0 + 0.484 417 191 936;
  • 37) 0.484 417 191 936 × 2 = 0 + 0.968 834 383 872;
  • 38) 0.968 834 383 872 × 2 = 1 + 0.937 668 767 744;
  • 39) 0.937 668 767 744 × 2 = 1 + 0.875 337 535 488;
  • 40) 0.875 337 535 488 × 2 = 1 + 0.750 675 070 976;
  • 41) 0.750 675 070 976 × 2 = 1 + 0.501 350 141 952;
  • 42) 0.501 350 141 952 × 2 = 1 + 0.002 700 283 904;
  • 43) 0.002 700 283 904 × 2 = 0 + 0.005 400 567 808;
  • 44) 0.005 400 567 808 × 2 = 0 + 0.010 801 135 616;
  • 45) 0.010 801 135 616 × 2 = 0 + 0.021 602 271 232;
  • 46) 0.021 602 271 232 × 2 = 0 + 0.043 204 542 464;
  • 47) 0.043 204 542 464 × 2 = 0 + 0.086 409 084 928;
  • 48) 0.086 409 084 928 × 2 = 0 + 0.172 818 169 856;
  • 49) 0.172 818 169 856 × 2 = 0 + 0.345 636 339 712;
  • 50) 0.345 636 339 712 × 2 = 0 + 0.691 272 679 424;
  • 51) 0.691 272 679 424 × 2 = 1 + 0.382 545 358 848;
  • 52) 0.382 545 358 848 × 2 = 0 + 0.765 090 717 696;
  • 53) 0.765 090 717 696 × 2 = 1 + 0.530 181 435 392;

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:


0.522 136 891 326(10) =

0.1000 0101 1010 1010 1100 0011 0110 1000 0100 0111 1100 0000 0010 1(2)

5. Positive number before normalization:

0.522 136 891 326(10) =

0.1000 0101 1010 1010 1100 0011 0110 1000 0100 0111 1100 0000 0010 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.522 136 891 326(10) =

0.1000 0101 1010 1010 1100 0011 0110 1000 0100 0111 1100 0000 0010 1(2) =

0.1000 0101 1010 1010 1100 0011 0110 1000 0100 0111 1100 0000 0010 1(2) × 20 =

1.0000 1011 0101 0101 1000 0110 1101 0000 1000 1111 1000 0000 0101(2) × 2-1


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): -1

Mantissa (not normalized):
1.0000 1011 0101 0101 1000 0110 1101 0000 1000 1111 1000 0000 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 0000 1011 0101 0101 1000 0110 1101 0000 1000 1111 1000 0000 0101 =

0000 1011 0101 0101 1000 0110 1101 0000 1000 1111 1000 0000 0101


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 1110

Mantissa (52 bits) =
0000 1011 0101 0101 1000 0110 1101 0000 1000 1111 1000 0000 0101


Decimal number 0.522 136 891 326 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0000 1011 0101 0101 1000 0110 1101 0000 1000 1111 1000 0000 0101

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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