0.974 013 318 541 726 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 726 1(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
0.974 013 318 541 726 1(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: 0.974 013 318 541 726 1.

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) 0.974 013 318 541 726 1 × 2 = 1 + 0.948 026 637 083 452 2;
  • 2) 0.948 026 637 083 452 2 × 2 = 1 + 0.896 053 274 166 904 4;
  • 3) 0.896 053 274 166 904 4 × 2 = 1 + 0.792 106 548 333 808 8;
  • 4) 0.792 106 548 333 808 8 × 2 = 1 + 0.584 213 096 667 617 6;
  • 5) 0.584 213 096 667 617 6 × 2 = 1 + 0.168 426 193 335 235 2;
  • 6) 0.168 426 193 335 235 2 × 2 = 0 + 0.336 852 386 670 470 4;
  • 7) 0.336 852 386 670 470 4 × 2 = 0 + 0.673 704 773 340 940 8;
  • 8) 0.673 704 773 340 940 8 × 2 = 1 + 0.347 409 546 681 881 6;
  • 9) 0.347 409 546 681 881 6 × 2 = 0 + 0.694 819 093 363 763 2;
  • 10) 0.694 819 093 363 763 2 × 2 = 1 + 0.389 638 186 727 526 4;
  • 11) 0.389 638 186 727 526 4 × 2 = 0 + 0.779 276 373 455 052 8;
  • 12) 0.779 276 373 455 052 8 × 2 = 1 + 0.558 552 746 910 105 6;
  • 13) 0.558 552 746 910 105 6 × 2 = 1 + 0.117 105 493 820 211 2;
  • 14) 0.117 105 493 820 211 2 × 2 = 0 + 0.234 210 987 640 422 4;
  • 15) 0.234 210 987 640 422 4 × 2 = 0 + 0.468 421 975 280 844 8;
  • 16) 0.468 421 975 280 844 8 × 2 = 0 + 0.936 843 950 561 689 6;
  • 17) 0.936 843 950 561 689 6 × 2 = 1 + 0.873 687 901 123 379 2;
  • 18) 0.873 687 901 123 379 2 × 2 = 1 + 0.747 375 802 246 758 4;
  • 19) 0.747 375 802 246 758 4 × 2 = 1 + 0.494 751 604 493 516 8;
  • 20) 0.494 751 604 493 516 8 × 2 = 0 + 0.989 503 208 987 033 6;
  • 21) 0.989 503 208 987 033 6 × 2 = 1 + 0.979 006 417 974 067 2;
  • 22) 0.979 006 417 974 067 2 × 2 = 1 + 0.958 012 835 948 134 4;
  • 23) 0.958 012 835 948 134 4 × 2 = 1 + 0.916 025 671 896 268 8;
  • 24) 0.916 025 671 896 268 8 × 2 = 1 + 0.832 051 343 792 537 6;
  • 25) 0.832 051 343 792 537 6 × 2 = 1 + 0.664 102 687 585 075 2;
  • 26) 0.664 102 687 585 075 2 × 2 = 1 + 0.328 205 375 170 150 4;
  • 27) 0.328 205 375 170 150 4 × 2 = 0 + 0.656 410 750 340 300 8;
  • 28) 0.656 410 750 340 300 8 × 2 = 1 + 0.312 821 500 680 601 6;
  • 29) 0.312 821 500 680 601 6 × 2 = 0 + 0.625 643 001 361 203 2;
  • 30) 0.625 643 001 361 203 2 × 2 = 1 + 0.251 286 002 722 406 4;
  • 31) 0.251 286 002 722 406 4 × 2 = 0 + 0.502 572 005 444 812 8;
  • 32) 0.502 572 005 444 812 8 × 2 = 1 + 0.005 144 010 889 625 6;
  • 33) 0.005 144 010 889 625 6 × 2 = 0 + 0.010 288 021 779 251 2;
  • 34) 0.010 288 021 779 251 2 × 2 = 0 + 0.020 576 043 558 502 4;
  • 35) 0.020 576 043 558 502 4 × 2 = 0 + 0.041 152 087 117 004 8;
  • 36) 0.041 152 087 117 004 8 × 2 = 0 + 0.082 304 174 234 009 6;
  • 37) 0.082 304 174 234 009 6 × 2 = 0 + 0.164 608 348 468 019 2;
  • 38) 0.164 608 348 468 019 2 × 2 = 0 + 0.329 216 696 936 038 4;
  • 39) 0.329 216 696 936 038 4 × 2 = 0 + 0.658 433 393 872 076 8;
  • 40) 0.658 433 393 872 076 8 × 2 = 1 + 0.316 866 787 744 153 6;
  • 41) 0.316 866 787 744 153 6 × 2 = 0 + 0.633 733 575 488 307 2;
  • 42) 0.633 733 575 488 307 2 × 2 = 1 + 0.267 467 150 976 614 4;
  • 43) 0.267 467 150 976 614 4 × 2 = 0 + 0.534 934 301 953 228 8;
  • 44) 0.534 934 301 953 228 8 × 2 = 1 + 0.069 868 603 906 457 6;
  • 45) 0.069 868 603 906 457 6 × 2 = 0 + 0.139 737 207 812 915 2;
  • 46) 0.139 737 207 812 915 2 × 2 = 0 + 0.279 474 415 625 830 4;
  • 47) 0.279 474 415 625 830 4 × 2 = 0 + 0.558 948 831 251 660 8;
  • 48) 0.558 948 831 251 660 8 × 2 = 1 + 0.117 897 662 503 321 6;
  • 49) 0.117 897 662 503 321 6 × 2 = 0 + 0.235 795 325 006 643 2;
  • 50) 0.235 795 325 006 643 2 × 2 = 0 + 0.471 590 650 013 286 4;
  • 51) 0.471 590 650 013 286 4 × 2 = 0 + 0.943 181 300 026 572 8;
  • 52) 0.943 181 300 026 572 8 × 2 = 1 + 0.886 362 600 053 145 6;
  • 53) 0.886 362 600 053 145 6 × 2 = 1 + 0.772 725 200 106 291 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:


0.974 013 318 541 726 1(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 0001 1(2)

5. Positive number before normalization:

0.974 013 318 541 726 1(10) =

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


0.974 013 318 541 726 1(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 0001 1(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 0001 1(2) × 20 =

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


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 0010 0011 =

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


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 0010 0011


Decimal number 0.974 013 318 541 726 1 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 0010 0011

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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