0.957 603 280 698 574 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.957 603 280 698 574 47(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
0.957 603 280 698 574 47(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: 0.957 603 280 698 574 47.

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) 0.957 603 280 698 574 47 × 2 = 1 + 0.915 206 561 397 148 94;
  • 2) 0.915 206 561 397 148 94 × 2 = 1 + 0.830 413 122 794 297 88;
  • 3) 0.830 413 122 794 297 88 × 2 = 1 + 0.660 826 245 588 595 76;
  • 4) 0.660 826 245 588 595 76 × 2 = 1 + 0.321 652 491 177 191 52;
  • 5) 0.321 652 491 177 191 52 × 2 = 0 + 0.643 304 982 354 383 04;
  • 6) 0.643 304 982 354 383 04 × 2 = 1 + 0.286 609 964 708 766 08;
  • 7) 0.286 609 964 708 766 08 × 2 = 0 + 0.573 219 929 417 532 16;
  • 8) 0.573 219 929 417 532 16 × 2 = 1 + 0.146 439 858 835 064 32;
  • 9) 0.146 439 858 835 064 32 × 2 = 0 + 0.292 879 717 670 128 64;
  • 10) 0.292 879 717 670 128 64 × 2 = 0 + 0.585 759 435 340 257 28;
  • 11) 0.585 759 435 340 257 28 × 2 = 1 + 0.171 518 870 680 514 56;
  • 12) 0.171 518 870 680 514 56 × 2 = 0 + 0.343 037 741 361 029 12;
  • 13) 0.343 037 741 361 029 12 × 2 = 0 + 0.686 075 482 722 058 24;
  • 14) 0.686 075 482 722 058 24 × 2 = 1 + 0.372 150 965 444 116 48;
  • 15) 0.372 150 965 444 116 48 × 2 = 0 + 0.744 301 930 888 232 96;
  • 16) 0.744 301 930 888 232 96 × 2 = 1 + 0.488 603 861 776 465 92;
  • 17) 0.488 603 861 776 465 92 × 2 = 0 + 0.977 207 723 552 931 84;
  • 18) 0.977 207 723 552 931 84 × 2 = 1 + 0.954 415 447 105 863 68;
  • 19) 0.954 415 447 105 863 68 × 2 = 1 + 0.908 830 894 211 727 36;
  • 20) 0.908 830 894 211 727 36 × 2 = 1 + 0.817 661 788 423 454 72;
  • 21) 0.817 661 788 423 454 72 × 2 = 1 + 0.635 323 576 846 909 44;
  • 22) 0.635 323 576 846 909 44 × 2 = 1 + 0.270 647 153 693 818 88;
  • 23) 0.270 647 153 693 818 88 × 2 = 0 + 0.541 294 307 387 637 76;
  • 24) 0.541 294 307 387 637 76 × 2 = 1 + 0.082 588 614 775 275 52;
  • 25) 0.082 588 614 775 275 52 × 2 = 0 + 0.165 177 229 550 551 04;
  • 26) 0.165 177 229 550 551 04 × 2 = 0 + 0.330 354 459 101 102 08;
  • 27) 0.330 354 459 101 102 08 × 2 = 0 + 0.660 708 918 202 204 16;
  • 28) 0.660 708 918 202 204 16 × 2 = 1 + 0.321 417 836 404 408 32;
  • 29) 0.321 417 836 404 408 32 × 2 = 0 + 0.642 835 672 808 816 64;
  • 30) 0.642 835 672 808 816 64 × 2 = 1 + 0.285 671 345 617 633 28;
  • 31) 0.285 671 345 617 633 28 × 2 = 0 + 0.571 342 691 235 266 56;
  • 32) 0.571 342 691 235 266 56 × 2 = 1 + 0.142 685 382 470 533 12;
  • 33) 0.142 685 382 470 533 12 × 2 = 0 + 0.285 370 764 941 066 24;
  • 34) 0.285 370 764 941 066 24 × 2 = 0 + 0.570 741 529 882 132 48;
  • 35) 0.570 741 529 882 132 48 × 2 = 1 + 0.141 483 059 764 264 96;
  • 36) 0.141 483 059 764 264 96 × 2 = 0 + 0.282 966 119 528 529 92;
  • 37) 0.282 966 119 528 529 92 × 2 = 0 + 0.565 932 239 057 059 84;
  • 38) 0.565 932 239 057 059 84 × 2 = 1 + 0.131 864 478 114 119 68;
  • 39) 0.131 864 478 114 119 68 × 2 = 0 + 0.263 728 956 228 239 36;
  • 40) 0.263 728 956 228 239 36 × 2 = 0 + 0.527 457 912 456 478 72;
  • 41) 0.527 457 912 456 478 72 × 2 = 1 + 0.054 915 824 912 957 44;
  • 42) 0.054 915 824 912 957 44 × 2 = 0 + 0.109 831 649 825 914 88;
  • 43) 0.109 831 649 825 914 88 × 2 = 0 + 0.219 663 299 651 829 76;
  • 44) 0.219 663 299 651 829 76 × 2 = 0 + 0.439 326 599 303 659 52;
  • 45) 0.439 326 599 303 659 52 × 2 = 0 + 0.878 653 198 607 319 04;
  • 46) 0.878 653 198 607 319 04 × 2 = 1 + 0.757 306 397 214 638 08;
  • 47) 0.757 306 397 214 638 08 × 2 = 1 + 0.514 612 794 429 276 16;
  • 48) 0.514 612 794 429 276 16 × 2 = 1 + 0.029 225 588 858 552 32;
  • 49) 0.029 225 588 858 552 32 × 2 = 0 + 0.058 451 177 717 104 64;
  • 50) 0.058 451 177 717 104 64 × 2 = 0 + 0.116 902 355 434 209 28;
  • 51) 0.116 902 355 434 209 28 × 2 = 0 + 0.233 804 710 868 418 56;
  • 52) 0.233 804 710 868 418 56 × 2 = 0 + 0.467 609 421 736 837 12;
  • 53) 0.467 609 421 736 837 12 × 2 = 0 + 0.935 218 843 473 674 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:


0.957 603 280 698 574 47(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0111 0000 0(2)

5. Positive number before normalization:

0.957 603 280 698 574 47(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0111 0000 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:


0.957 603 280 698 574 47(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0111 0000 0(2) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0111 0000 0(2) × 20 =

1.1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1110 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):
1.1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1110 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. 1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1110 0000 =

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


Decimal number 0.957 603 280 698 574 47 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 1110 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