0.974 013 318 541 737 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 737(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
0.974 013 318 541 737(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: 0.974 013 318 541 737.

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) 0.974 013 318 541 737 × 2 = 1 + 0.948 026 637 083 474;
  • 2) 0.948 026 637 083 474 × 2 = 1 + 0.896 053 274 166 948;
  • 3) 0.896 053 274 166 948 × 2 = 1 + 0.792 106 548 333 896;
  • 4) 0.792 106 548 333 896 × 2 = 1 + 0.584 213 096 667 792;
  • 5) 0.584 213 096 667 792 × 2 = 1 + 0.168 426 193 335 584;
  • 6) 0.168 426 193 335 584 × 2 = 0 + 0.336 852 386 671 168;
  • 7) 0.336 852 386 671 168 × 2 = 0 + 0.673 704 773 342 336;
  • 8) 0.673 704 773 342 336 × 2 = 1 + 0.347 409 546 684 672;
  • 9) 0.347 409 546 684 672 × 2 = 0 + 0.694 819 093 369 344;
  • 10) 0.694 819 093 369 344 × 2 = 1 + 0.389 638 186 738 688;
  • 11) 0.389 638 186 738 688 × 2 = 0 + 0.779 276 373 477 376;
  • 12) 0.779 276 373 477 376 × 2 = 1 + 0.558 552 746 954 752;
  • 13) 0.558 552 746 954 752 × 2 = 1 + 0.117 105 493 909 504;
  • 14) 0.117 105 493 909 504 × 2 = 0 + 0.234 210 987 819 008;
  • 15) 0.234 210 987 819 008 × 2 = 0 + 0.468 421 975 638 016;
  • 16) 0.468 421 975 638 016 × 2 = 0 + 0.936 843 951 276 032;
  • 17) 0.936 843 951 276 032 × 2 = 1 + 0.873 687 902 552 064;
  • 18) 0.873 687 902 552 064 × 2 = 1 + 0.747 375 805 104 128;
  • 19) 0.747 375 805 104 128 × 2 = 1 + 0.494 751 610 208 256;
  • 20) 0.494 751 610 208 256 × 2 = 0 + 0.989 503 220 416 512;
  • 21) 0.989 503 220 416 512 × 2 = 1 + 0.979 006 440 833 024;
  • 22) 0.979 006 440 833 024 × 2 = 1 + 0.958 012 881 666 048;
  • 23) 0.958 012 881 666 048 × 2 = 1 + 0.916 025 763 332 096;
  • 24) 0.916 025 763 332 096 × 2 = 1 + 0.832 051 526 664 192;
  • 25) 0.832 051 526 664 192 × 2 = 1 + 0.664 103 053 328 384;
  • 26) 0.664 103 053 328 384 × 2 = 1 + 0.328 206 106 656 768;
  • 27) 0.328 206 106 656 768 × 2 = 0 + 0.656 412 213 313 536;
  • 28) 0.656 412 213 313 536 × 2 = 1 + 0.312 824 426 627 072;
  • 29) 0.312 824 426 627 072 × 2 = 0 + 0.625 648 853 254 144;
  • 30) 0.625 648 853 254 144 × 2 = 1 + 0.251 297 706 508 288;
  • 31) 0.251 297 706 508 288 × 2 = 0 + 0.502 595 413 016 576;
  • 32) 0.502 595 413 016 576 × 2 = 1 + 0.005 190 826 033 152;
  • 33) 0.005 190 826 033 152 × 2 = 0 + 0.010 381 652 066 304;
  • 34) 0.010 381 652 066 304 × 2 = 0 + 0.020 763 304 132 608;
  • 35) 0.020 763 304 132 608 × 2 = 0 + 0.041 526 608 265 216;
  • 36) 0.041 526 608 265 216 × 2 = 0 + 0.083 053 216 530 432;
  • 37) 0.083 053 216 530 432 × 2 = 0 + 0.166 106 433 060 864;
  • 38) 0.166 106 433 060 864 × 2 = 0 + 0.332 212 866 121 728;
  • 39) 0.332 212 866 121 728 × 2 = 0 + 0.664 425 732 243 456;
  • 40) 0.664 425 732 243 456 × 2 = 1 + 0.328 851 464 486 912;
  • 41) 0.328 851 464 486 912 × 2 = 0 + 0.657 702 928 973 824;
  • 42) 0.657 702 928 973 824 × 2 = 1 + 0.315 405 857 947 648;
  • 43) 0.315 405 857 947 648 × 2 = 0 + 0.630 811 715 895 296;
  • 44) 0.630 811 715 895 296 × 2 = 1 + 0.261 623 431 790 592;
  • 45) 0.261 623 431 790 592 × 2 = 0 + 0.523 246 863 581 184;
  • 46) 0.523 246 863 581 184 × 2 = 1 + 0.046 493 727 162 368;
  • 47) 0.046 493 727 162 368 × 2 = 0 + 0.092 987 454 324 736;
  • 48) 0.092 987 454 324 736 × 2 = 0 + 0.185 974 908 649 472;
  • 49) 0.185 974 908 649 472 × 2 = 0 + 0.371 949 817 298 944;
  • 50) 0.371 949 817 298 944 × 2 = 0 + 0.743 899 634 597 888;
  • 51) 0.743 899 634 597 888 × 2 = 1 + 0.487 799 269 195 776;
  • 52) 0.487 799 269 195 776 × 2 = 0 + 0.975 598 538 391 552;
  • 53) 0.975 598 538 391 552 × 2 = 1 + 0.951 197 076 783 104;

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:


0.974 013 318 541 737(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0100 0010 1(2)

5. Positive number before normalization:

0.974 013 318 541 737(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0100 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:


0.974 013 318 541 737(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0100 0010 1(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0100 0010 1(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 1000 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):
1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 1000 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. 1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 1000 0101 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 1000 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) =
1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 1000 0101


Decimal number 0.974 013 318 541 737 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 1000 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