0.648 419 777 325 504 832 966 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.648 419 777 325 504 832 966 4(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
0.648 419 777 325 504 832 966 4(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: 0.648 419 777 325 504 832 966 4.

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) 0.648 419 777 325 504 832 966 4 × 2 = 1 + 0.296 839 554 651 009 665 932 8;
  • 2) 0.296 839 554 651 009 665 932 8 × 2 = 0 + 0.593 679 109 302 019 331 865 6;
  • 3) 0.593 679 109 302 019 331 865 6 × 2 = 1 + 0.187 358 218 604 038 663 731 2;
  • 4) 0.187 358 218 604 038 663 731 2 × 2 = 0 + 0.374 716 437 208 077 327 462 4;
  • 5) 0.374 716 437 208 077 327 462 4 × 2 = 0 + 0.749 432 874 416 154 654 924 8;
  • 6) 0.749 432 874 416 154 654 924 8 × 2 = 1 + 0.498 865 748 832 309 309 849 6;
  • 7) 0.498 865 748 832 309 309 849 6 × 2 = 0 + 0.997 731 497 664 618 619 699 2;
  • 8) 0.997 731 497 664 618 619 699 2 × 2 = 1 + 0.995 462 995 329 237 239 398 4;
  • 9) 0.995 462 995 329 237 239 398 4 × 2 = 1 + 0.990 925 990 658 474 478 796 8;
  • 10) 0.990 925 990 658 474 478 796 8 × 2 = 1 + 0.981 851 981 316 948 957 593 6;
  • 11) 0.981 851 981 316 948 957 593 6 × 2 = 1 + 0.963 703 962 633 897 915 187 2;
  • 12) 0.963 703 962 633 897 915 187 2 × 2 = 1 + 0.927 407 925 267 795 830 374 4;
  • 13) 0.927 407 925 267 795 830 374 4 × 2 = 1 + 0.854 815 850 535 591 660 748 8;
  • 14) 0.854 815 850 535 591 660 748 8 × 2 = 1 + 0.709 631 701 071 183 321 497 6;
  • 15) 0.709 631 701 071 183 321 497 6 × 2 = 1 + 0.419 263 402 142 366 642 995 2;
  • 16) 0.419 263 402 142 366 642 995 2 × 2 = 0 + 0.838 526 804 284 733 285 990 4;
  • 17) 0.838 526 804 284 733 285 990 4 × 2 = 1 + 0.677 053 608 569 466 571 980 8;
  • 18) 0.677 053 608 569 466 571 980 8 × 2 = 1 + 0.354 107 217 138 933 143 961 6;
  • 19) 0.354 107 217 138 933 143 961 6 × 2 = 0 + 0.708 214 434 277 866 287 923 2;
  • 20) 0.708 214 434 277 866 287 923 2 × 2 = 1 + 0.416 428 868 555 732 575 846 4;
  • 21) 0.416 428 868 555 732 575 846 4 × 2 = 0 + 0.832 857 737 111 465 151 692 8;
  • 22) 0.832 857 737 111 465 151 692 8 × 2 = 1 + 0.665 715 474 222 930 303 385 6;
  • 23) 0.665 715 474 222 930 303 385 6 × 2 = 1 + 0.331 430 948 445 860 606 771 2;
  • 24) 0.331 430 948 445 860 606 771 2 × 2 = 0 + 0.662 861 896 891 721 213 542 4;
  • 25) 0.662 861 896 891 721 213 542 4 × 2 = 1 + 0.325 723 793 783 442 427 084 8;
  • 26) 0.325 723 793 783 442 427 084 8 × 2 = 0 + 0.651 447 587 566 884 854 169 6;
  • 27) 0.651 447 587 566 884 854 169 6 × 2 = 1 + 0.302 895 175 133 769 708 339 2;
  • 28) 0.302 895 175 133 769 708 339 2 × 2 = 0 + 0.605 790 350 267 539 416 678 4;
  • 29) 0.605 790 350 267 539 416 678 4 × 2 = 1 + 0.211 580 700 535 078 833 356 8;
  • 30) 0.211 580 700 535 078 833 356 8 × 2 = 0 + 0.423 161 401 070 157 666 713 6;
  • 31) 0.423 161 401 070 157 666 713 6 × 2 = 0 + 0.846 322 802 140 315 333 427 2;
  • 32) 0.846 322 802 140 315 333 427 2 × 2 = 1 + 0.692 645 604 280 630 666 854 4;
  • 33) 0.692 645 604 280 630 666 854 4 × 2 = 1 + 0.385 291 208 561 261 333 708 8;
  • 34) 0.385 291 208 561 261 333 708 8 × 2 = 0 + 0.770 582 417 122 522 667 417 6;
  • 35) 0.770 582 417 122 522 667 417 6 × 2 = 1 + 0.541 164 834 245 045 334 835 2;
  • 36) 0.541 164 834 245 045 334 835 2 × 2 = 1 + 0.082 329 668 490 090 669 670 4;
  • 37) 0.082 329 668 490 090 669 670 4 × 2 = 0 + 0.164 659 336 980 181 339 340 8;
  • 38) 0.164 659 336 980 181 339 340 8 × 2 = 0 + 0.329 318 673 960 362 678 681 6;
  • 39) 0.329 318 673 960 362 678 681 6 × 2 = 0 + 0.658 637 347 920 725 357 363 2;
  • 40) 0.658 637 347 920 725 357 363 2 × 2 = 1 + 0.317 274 695 841 450 714 726 4;
  • 41) 0.317 274 695 841 450 714 726 4 × 2 = 0 + 0.634 549 391 682 901 429 452 8;
  • 42) 0.634 549 391 682 901 429 452 8 × 2 = 1 + 0.269 098 783 365 802 858 905 6;
  • 43) 0.269 098 783 365 802 858 905 6 × 2 = 0 + 0.538 197 566 731 605 717 811 2;
  • 44) 0.538 197 566 731 605 717 811 2 × 2 = 1 + 0.076 395 133 463 211 435 622 4;
  • 45) 0.076 395 133 463 211 435 622 4 × 2 = 0 + 0.152 790 266 926 422 871 244 8;
  • 46) 0.152 790 266 926 422 871 244 8 × 2 = 0 + 0.305 580 533 852 845 742 489 6;
  • 47) 0.305 580 533 852 845 742 489 6 × 2 = 0 + 0.611 161 067 705 691 484 979 2;
  • 48) 0.611 161 067 705 691 484 979 2 × 2 = 1 + 0.222 322 135 411 382 969 958 4;
  • 49) 0.222 322 135 411 382 969 958 4 × 2 = 0 + 0.444 644 270 822 765 939 916 8;
  • 50) 0.444 644 270 822 765 939 916 8 × 2 = 0 + 0.889 288 541 645 531 879 833 6;
  • 51) 0.889 288 541 645 531 879 833 6 × 2 = 1 + 0.778 577 083 291 063 759 667 2;
  • 52) 0.778 577 083 291 063 759 667 2 × 2 = 1 + 0.557 154 166 582 127 519 334 4;
  • 53) 0.557 154 166 582 127 519 334 4 × 2 = 1 + 0.114 308 333 164 255 038 668 8;

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:


0.648 419 777 325 504 832 966 4(10) =

0.1010 0101 1111 1110 1101 0110 1010 1001 1011 0001 0101 0001 0011 1(2)

5. Positive number before normalization:

0.648 419 777 325 504 832 966 4(10) =

0.1010 0101 1111 1110 1101 0110 1010 1001 1011 0001 0101 0001 0011 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:


0.648 419 777 325 504 832 966 4(10) =

0.1010 0101 1111 1110 1101 0110 1010 1001 1011 0001 0101 0001 0011 1(2) =

0.1010 0101 1111 1110 1101 0110 1010 1001 1011 0001 0101 0001 0011 1(2) × 20 =

1.0100 1011 1111 1101 1010 1101 0101 0011 0110 0010 1010 0010 0111(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):
1.0100 1011 1111 1101 1010 1101 0101 0011 0110 0010 1010 0010 0111


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. 0100 1011 1111 1101 1010 1101 0101 0011 0110 0010 1010 0010 0111 =

0100 1011 1111 1101 1010 1101 0101 0011 0110 0010 1010 0010 0111


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) =
0100 1011 1111 1101 1010 1101 0101 0011 0110 0010 1010 0010 0111


Decimal number 0.648 419 777 325 504 832 966 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0100 1011 1111 1101 1010 1101 0101 0011 0110 0010 1010 0010 0111

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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