8.535 179 390 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 8.535 179 390 6(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
8.535 179 390 6(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: 8.
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;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

8(10) =

1000(2)


3. Convert to binary (base 2) the fractional part: 0.535 179 390 6.

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) 0.535 179 390 6 × 2 = 1 + 0.070 358 781 2;
  • 2) 0.070 358 781 2 × 2 = 0 + 0.140 717 562 4;
  • 3) 0.140 717 562 4 × 2 = 0 + 0.281 435 124 8;
  • 4) 0.281 435 124 8 × 2 = 0 + 0.562 870 249 6;
  • 5) 0.562 870 249 6 × 2 = 1 + 0.125 740 499 2;
  • 6) 0.125 740 499 2 × 2 = 0 + 0.251 480 998 4;
  • 7) 0.251 480 998 4 × 2 = 0 + 0.502 961 996 8;
  • 8) 0.502 961 996 8 × 2 = 1 + 0.005 923 993 6;
  • 9) 0.005 923 993 6 × 2 = 0 + 0.011 847 987 2;
  • 10) 0.011 847 987 2 × 2 = 0 + 0.023 695 974 4;
  • 11) 0.023 695 974 4 × 2 = 0 + 0.047 391 948 8;
  • 12) 0.047 391 948 8 × 2 = 0 + 0.094 783 897 6;
  • 13) 0.094 783 897 6 × 2 = 0 + 0.189 567 795 2;
  • 14) 0.189 567 795 2 × 2 = 0 + 0.379 135 590 4;
  • 15) 0.379 135 590 4 × 2 = 0 + 0.758 271 180 8;
  • 16) 0.758 271 180 8 × 2 = 1 + 0.516 542 361 6;
  • 17) 0.516 542 361 6 × 2 = 1 + 0.033 084 723 2;
  • 18) 0.033 084 723 2 × 2 = 0 + 0.066 169 446 4;
  • 19) 0.066 169 446 4 × 2 = 0 + 0.132 338 892 8;
  • 20) 0.132 338 892 8 × 2 = 0 + 0.264 677 785 6;
  • 21) 0.264 677 785 6 × 2 = 0 + 0.529 355 571 2;
  • 22) 0.529 355 571 2 × 2 = 1 + 0.058 711 142 4;
  • 23) 0.058 711 142 4 × 2 = 0 + 0.117 422 284 8;
  • 24) 0.117 422 284 8 × 2 = 0 + 0.234 844 569 6;
  • 25) 0.234 844 569 6 × 2 = 0 + 0.469 689 139 2;
  • 26) 0.469 689 139 2 × 2 = 0 + 0.939 378 278 4;
  • 27) 0.939 378 278 4 × 2 = 1 + 0.878 756 556 8;
  • 28) 0.878 756 556 8 × 2 = 1 + 0.757 513 113 6;
  • 29) 0.757 513 113 6 × 2 = 1 + 0.515 026 227 2;
  • 30) 0.515 026 227 2 × 2 = 1 + 0.030 052 454 4;
  • 31) 0.030 052 454 4 × 2 = 0 + 0.060 104 908 8;
  • 32) 0.060 104 908 8 × 2 = 0 + 0.120 209 817 6;
  • 33) 0.120 209 817 6 × 2 = 0 + 0.240 419 635 2;
  • 34) 0.240 419 635 2 × 2 = 0 + 0.480 839 270 4;
  • 35) 0.480 839 270 4 × 2 = 0 + 0.961 678 540 8;
  • 36) 0.961 678 540 8 × 2 = 1 + 0.923 357 081 6;
  • 37) 0.923 357 081 6 × 2 = 1 + 0.846 714 163 2;
  • 38) 0.846 714 163 2 × 2 = 1 + 0.693 428 326 4;
  • 39) 0.693 428 326 4 × 2 = 1 + 0.386 856 652 8;
  • 40) 0.386 856 652 8 × 2 = 0 + 0.773 713 305 6;
  • 41) 0.773 713 305 6 × 2 = 1 + 0.547 426 611 2;
  • 42) 0.547 426 611 2 × 2 = 1 + 0.094 853 222 4;
  • 43) 0.094 853 222 4 × 2 = 0 + 0.189 706 444 8;
  • 44) 0.189 706 444 8 × 2 = 0 + 0.379 412 889 6;
  • 45) 0.379 412 889 6 × 2 = 0 + 0.758 825 779 2;
  • 46) 0.758 825 779 2 × 2 = 1 + 0.517 651 558 4;
  • 47) 0.517 651 558 4 × 2 = 1 + 0.035 303 116 8;
  • 48) 0.035 303 116 8 × 2 = 0 + 0.070 606 233 6;
  • 49) 0.070 606 233 6 × 2 = 0 + 0.141 212 467 2;
  • 50) 0.141 212 467 2 × 2 = 0 + 0.282 424 934 4;
  • 51) 0.282 424 934 4 × 2 = 0 + 0.564 849 868 8;
  • 52) 0.564 849 868 8 × 2 = 1 + 0.129 699 737 6;
  • 53) 0.129 699 737 6 × 2 = 0 + 0.259 399 475 2;

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:


0.535 179 390 6(10) =

0.1000 1001 0000 0001 1000 0100 0011 1100 0001 1110 1100 0110 0001 0(2)

5. Positive number before normalization:

8.535 179 390 6(10) =

1000.1000 1001 0000 0001 1000 0100 0011 1100 0001 1110 1100 0110 0001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


8.535 179 390 6(10) =

1000.1000 1001 0000 0001 1000 0100 0011 1100 0001 1110 1100 0110 0001 0(2) =

1000.1000 1001 0000 0001 1000 0100 0011 1100 0001 1110 1100 0110 0001 0(2) × 20 =

1.0001 0001 0010 0000 0011 0000 1000 0111 1000 0011 1101 1000 1100 0010(2) × 23


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): 3

Mantissa (not normalized):
1.0001 0001 0010 0000 0011 0000 1000 0111 1000 0011 1101 1000 1100 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

3 + 2(11-1) - 1 =

(3 + 1 023)(10) =

1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1026(10) =

100 0000 0010(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0001 0001 0010 0000 0011 0000 1000 0111 1000 0011 1101 1000 1100 0010 =

0001 0001 0010 0000 0011 0000 1000 0111 1000 0011 1101 1000 1100


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) =
100 0000 0010

Mantissa (52 bits) =
0001 0001 0010 0000 0011 0000 1000 0111 1000 0011 1101 1000 1100


Decimal number 8.535 179 390 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0010 - 0001 0001 0010 0000 0011 0000 1000 0111 1000 0011 1101 1000 1100

» Convert decimal 8.535 179 386 9 to 64 bit double precision IEEE 754 binary floating point representation standard

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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