0.974 013 318 541 727 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 727 5(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
0.974 013 318 541 727 5(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: 0.974 013 318 541 727 5.

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) 0.974 013 318 541 727 5 × 2 = 1 + 0.948 026 637 083 455;
  • 2) 0.948 026 637 083 455 × 2 = 1 + 0.896 053 274 166 91;
  • 3) 0.896 053 274 166 91 × 2 = 1 + 0.792 106 548 333 82;
  • 4) 0.792 106 548 333 82 × 2 = 1 + 0.584 213 096 667 64;
  • 5) 0.584 213 096 667 64 × 2 = 1 + 0.168 426 193 335 28;
  • 6) 0.168 426 193 335 28 × 2 = 0 + 0.336 852 386 670 56;
  • 7) 0.336 852 386 670 56 × 2 = 0 + 0.673 704 773 341 12;
  • 8) 0.673 704 773 341 12 × 2 = 1 + 0.347 409 546 682 24;
  • 9) 0.347 409 546 682 24 × 2 = 0 + 0.694 819 093 364 48;
  • 10) 0.694 819 093 364 48 × 2 = 1 + 0.389 638 186 728 96;
  • 11) 0.389 638 186 728 96 × 2 = 0 + 0.779 276 373 457 92;
  • 12) 0.779 276 373 457 92 × 2 = 1 + 0.558 552 746 915 84;
  • 13) 0.558 552 746 915 84 × 2 = 1 + 0.117 105 493 831 68;
  • 14) 0.117 105 493 831 68 × 2 = 0 + 0.234 210 987 663 36;
  • 15) 0.234 210 987 663 36 × 2 = 0 + 0.468 421 975 326 72;
  • 16) 0.468 421 975 326 72 × 2 = 0 + 0.936 843 950 653 44;
  • 17) 0.936 843 950 653 44 × 2 = 1 + 0.873 687 901 306 88;
  • 18) 0.873 687 901 306 88 × 2 = 1 + 0.747 375 802 613 76;
  • 19) 0.747 375 802 613 76 × 2 = 1 + 0.494 751 605 227 52;
  • 20) 0.494 751 605 227 52 × 2 = 0 + 0.989 503 210 455 04;
  • 21) 0.989 503 210 455 04 × 2 = 1 + 0.979 006 420 910 08;
  • 22) 0.979 006 420 910 08 × 2 = 1 + 0.958 012 841 820 16;
  • 23) 0.958 012 841 820 16 × 2 = 1 + 0.916 025 683 640 32;
  • 24) 0.916 025 683 640 32 × 2 = 1 + 0.832 051 367 280 64;
  • 25) 0.832 051 367 280 64 × 2 = 1 + 0.664 102 734 561 28;
  • 26) 0.664 102 734 561 28 × 2 = 1 + 0.328 205 469 122 56;
  • 27) 0.328 205 469 122 56 × 2 = 0 + 0.656 410 938 245 12;
  • 28) 0.656 410 938 245 12 × 2 = 1 + 0.312 821 876 490 24;
  • 29) 0.312 821 876 490 24 × 2 = 0 + 0.625 643 752 980 48;
  • 30) 0.625 643 752 980 48 × 2 = 1 + 0.251 287 505 960 96;
  • 31) 0.251 287 505 960 96 × 2 = 0 + 0.502 575 011 921 92;
  • 32) 0.502 575 011 921 92 × 2 = 1 + 0.005 150 023 843 84;
  • 33) 0.005 150 023 843 84 × 2 = 0 + 0.010 300 047 687 68;
  • 34) 0.010 300 047 687 68 × 2 = 0 + 0.020 600 095 375 36;
  • 35) 0.020 600 095 375 36 × 2 = 0 + 0.041 200 190 750 72;
  • 36) 0.041 200 190 750 72 × 2 = 0 + 0.082 400 381 501 44;
  • 37) 0.082 400 381 501 44 × 2 = 0 + 0.164 800 763 002 88;
  • 38) 0.164 800 763 002 88 × 2 = 0 + 0.329 601 526 005 76;
  • 39) 0.329 601 526 005 76 × 2 = 0 + 0.659 203 052 011 52;
  • 40) 0.659 203 052 011 52 × 2 = 1 + 0.318 406 104 023 04;
  • 41) 0.318 406 104 023 04 × 2 = 0 + 0.636 812 208 046 08;
  • 42) 0.636 812 208 046 08 × 2 = 1 + 0.273 624 416 092 16;
  • 43) 0.273 624 416 092 16 × 2 = 0 + 0.547 248 832 184 32;
  • 44) 0.547 248 832 184 32 × 2 = 1 + 0.094 497 664 368 64;
  • 45) 0.094 497 664 368 64 × 2 = 0 + 0.188 995 328 737 28;
  • 46) 0.188 995 328 737 28 × 2 = 0 + 0.377 990 657 474 56;
  • 47) 0.377 990 657 474 56 × 2 = 0 + 0.755 981 314 949 12;
  • 48) 0.755 981 314 949 12 × 2 = 1 + 0.511 962 629 898 24;
  • 49) 0.511 962 629 898 24 × 2 = 1 + 0.023 925 259 796 48;
  • 50) 0.023 925 259 796 48 × 2 = 0 + 0.047 850 519 592 96;
  • 51) 0.047 850 519 592 96 × 2 = 0 + 0.095 701 039 185 92;
  • 52) 0.095 701 039 185 92 × 2 = 0 + 0.191 402 078 371 84;
  • 53) 0.191 402 078 371 84 × 2 = 0 + 0.382 804 156 743 68;

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:


0.974 013 318 541 727 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 1000 0(2)

5. Positive number before normalization:

0.974 013 318 541 727 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 1000 0(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:


0.974 013 318 541 727 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 1000 0(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 1000 0(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0011 0000(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):
1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0011 0000


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. 1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0011 0000 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0011 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 1110

Mantissa (52 bits) =
1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0011 0000


Decimal number 0.974 013 318 541 727 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0011 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