0.957 603 280 698 573 646 935 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.957 603 280 698 573 646 935 72(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
0.957 603 280 698 573 646 935 72(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: 0.957 603 280 698 573 646 935 72.

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) 0.957 603 280 698 573 646 935 72 × 2 = 1 + 0.915 206 561 397 147 293 871 44;
  • 2) 0.915 206 561 397 147 293 871 44 × 2 = 1 + 0.830 413 122 794 294 587 742 88;
  • 3) 0.830 413 122 794 294 587 742 88 × 2 = 1 + 0.660 826 245 588 589 175 485 76;
  • 4) 0.660 826 245 588 589 175 485 76 × 2 = 1 + 0.321 652 491 177 178 350 971 52;
  • 5) 0.321 652 491 177 178 350 971 52 × 2 = 0 + 0.643 304 982 354 356 701 943 04;
  • 6) 0.643 304 982 354 356 701 943 04 × 2 = 1 + 0.286 609 964 708 713 403 886 08;
  • 7) 0.286 609 964 708 713 403 886 08 × 2 = 0 + 0.573 219 929 417 426 807 772 16;
  • 8) 0.573 219 929 417 426 807 772 16 × 2 = 1 + 0.146 439 858 834 853 615 544 32;
  • 9) 0.146 439 858 834 853 615 544 32 × 2 = 0 + 0.292 879 717 669 707 231 088 64;
  • 10) 0.292 879 717 669 707 231 088 64 × 2 = 0 + 0.585 759 435 339 414 462 177 28;
  • 11) 0.585 759 435 339 414 462 177 28 × 2 = 1 + 0.171 518 870 678 828 924 354 56;
  • 12) 0.171 518 870 678 828 924 354 56 × 2 = 0 + 0.343 037 741 357 657 848 709 12;
  • 13) 0.343 037 741 357 657 848 709 12 × 2 = 0 + 0.686 075 482 715 315 697 418 24;
  • 14) 0.686 075 482 715 315 697 418 24 × 2 = 1 + 0.372 150 965 430 631 394 836 48;
  • 15) 0.372 150 965 430 631 394 836 48 × 2 = 0 + 0.744 301 930 861 262 789 672 96;
  • 16) 0.744 301 930 861 262 789 672 96 × 2 = 1 + 0.488 603 861 722 525 579 345 92;
  • 17) 0.488 603 861 722 525 579 345 92 × 2 = 0 + 0.977 207 723 445 051 158 691 84;
  • 18) 0.977 207 723 445 051 158 691 84 × 2 = 1 + 0.954 415 446 890 102 317 383 68;
  • 19) 0.954 415 446 890 102 317 383 68 × 2 = 1 + 0.908 830 893 780 204 634 767 36;
  • 20) 0.908 830 893 780 204 634 767 36 × 2 = 1 + 0.817 661 787 560 409 269 534 72;
  • 21) 0.817 661 787 560 409 269 534 72 × 2 = 1 + 0.635 323 575 120 818 539 069 44;
  • 22) 0.635 323 575 120 818 539 069 44 × 2 = 1 + 0.270 647 150 241 637 078 138 88;
  • 23) 0.270 647 150 241 637 078 138 88 × 2 = 0 + 0.541 294 300 483 274 156 277 76;
  • 24) 0.541 294 300 483 274 156 277 76 × 2 = 1 + 0.082 588 600 966 548 312 555 52;
  • 25) 0.082 588 600 966 548 312 555 52 × 2 = 0 + 0.165 177 201 933 096 625 111 04;
  • 26) 0.165 177 201 933 096 625 111 04 × 2 = 0 + 0.330 354 403 866 193 250 222 08;
  • 27) 0.330 354 403 866 193 250 222 08 × 2 = 0 + 0.660 708 807 732 386 500 444 16;
  • 28) 0.660 708 807 732 386 500 444 16 × 2 = 1 + 0.321 417 615 464 773 000 888 32;
  • 29) 0.321 417 615 464 773 000 888 32 × 2 = 0 + 0.642 835 230 929 546 001 776 64;
  • 30) 0.642 835 230 929 546 001 776 64 × 2 = 1 + 0.285 670 461 859 092 003 553 28;
  • 31) 0.285 670 461 859 092 003 553 28 × 2 = 0 + 0.571 340 923 718 184 007 106 56;
  • 32) 0.571 340 923 718 184 007 106 56 × 2 = 1 + 0.142 681 847 436 368 014 213 12;
  • 33) 0.142 681 847 436 368 014 213 12 × 2 = 0 + 0.285 363 694 872 736 028 426 24;
  • 34) 0.285 363 694 872 736 028 426 24 × 2 = 0 + 0.570 727 389 745 472 056 852 48;
  • 35) 0.570 727 389 745 472 056 852 48 × 2 = 1 + 0.141 454 779 490 944 113 704 96;
  • 36) 0.141 454 779 490 944 113 704 96 × 2 = 0 + 0.282 909 558 981 888 227 409 92;
  • 37) 0.282 909 558 981 888 227 409 92 × 2 = 0 + 0.565 819 117 963 776 454 819 84;
  • 38) 0.565 819 117 963 776 454 819 84 × 2 = 1 + 0.131 638 235 927 552 909 639 68;
  • 39) 0.131 638 235 927 552 909 639 68 × 2 = 0 + 0.263 276 471 855 105 819 279 36;
  • 40) 0.263 276 471 855 105 819 279 36 × 2 = 0 + 0.526 552 943 710 211 638 558 72;
  • 41) 0.526 552 943 710 211 638 558 72 × 2 = 1 + 0.053 105 887 420 423 277 117 44;
  • 42) 0.053 105 887 420 423 277 117 44 × 2 = 0 + 0.106 211 774 840 846 554 234 88;
  • 43) 0.106 211 774 840 846 554 234 88 × 2 = 0 + 0.212 423 549 681 693 108 469 76;
  • 44) 0.212 423 549 681 693 108 469 76 × 2 = 0 + 0.424 847 099 363 386 216 939 52;
  • 45) 0.424 847 099 363 386 216 939 52 × 2 = 0 + 0.849 694 198 726 772 433 879 04;
  • 46) 0.849 694 198 726 772 433 879 04 × 2 = 1 + 0.699 388 397 453 544 867 758 08;
  • 47) 0.699 388 397 453 544 867 758 08 × 2 = 1 + 0.398 776 794 907 089 735 516 16;
  • 48) 0.398 776 794 907 089 735 516 16 × 2 = 0 + 0.797 553 589 814 179 471 032 32;
  • 49) 0.797 553 589 814 179 471 032 32 × 2 = 1 + 0.595 107 179 628 358 942 064 64;
  • 50) 0.595 107 179 628 358 942 064 64 × 2 = 1 + 0.190 214 359 256 717 884 129 28;
  • 51) 0.190 214 359 256 717 884 129 28 × 2 = 0 + 0.380 428 718 513 435 768 258 56;
  • 52) 0.380 428 718 513 435 768 258 56 × 2 = 0 + 0.760 857 437 026 871 536 517 12;
  • 53) 0.760 857 437 026 871 536 517 12 × 2 = 1 + 0.521 714 874 053 743 073 034 24;

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:


0.957 603 280 698 573 646 935 72(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2)

5. Positive number before normalization:

0.957 603 280 698 573 646 935 72(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 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:


0.957 603 280 698 573 646 935 72(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2) × 20 =

1.1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001(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):
1.1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


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. 1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001 =

1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


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) =
1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


Decimal number 0.957 603 280 698 573 646 935 72 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001

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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