0.648 419 777 325 504 832 966 877 056 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.648 419 777 325 504 832 966 877 056 7(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
0.648 419 777 325 504 832 966 877 056 7(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: 0.648 419 777 325 504 832 966 877 056 7.

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) 0.648 419 777 325 504 832 966 877 056 7 × 2 = 1 + 0.296 839 554 651 009 665 933 754 113 4;
  • 2) 0.296 839 554 651 009 665 933 754 113 4 × 2 = 0 + 0.593 679 109 302 019 331 867 508 226 8;
  • 3) 0.593 679 109 302 019 331 867 508 226 8 × 2 = 1 + 0.187 358 218 604 038 663 735 016 453 6;
  • 4) 0.187 358 218 604 038 663 735 016 453 6 × 2 = 0 + 0.374 716 437 208 077 327 470 032 907 2;
  • 5) 0.374 716 437 208 077 327 470 032 907 2 × 2 = 0 + 0.749 432 874 416 154 654 940 065 814 4;
  • 6) 0.749 432 874 416 154 654 940 065 814 4 × 2 = 1 + 0.498 865 748 832 309 309 880 131 628 8;
  • 7) 0.498 865 748 832 309 309 880 131 628 8 × 2 = 0 + 0.997 731 497 664 618 619 760 263 257 6;
  • 8) 0.997 731 497 664 618 619 760 263 257 6 × 2 = 1 + 0.995 462 995 329 237 239 520 526 515 2;
  • 9) 0.995 462 995 329 237 239 520 526 515 2 × 2 = 1 + 0.990 925 990 658 474 479 041 053 030 4;
  • 10) 0.990 925 990 658 474 479 041 053 030 4 × 2 = 1 + 0.981 851 981 316 948 958 082 106 060 8;
  • 11) 0.981 851 981 316 948 958 082 106 060 8 × 2 = 1 + 0.963 703 962 633 897 916 164 212 121 6;
  • 12) 0.963 703 962 633 897 916 164 212 121 6 × 2 = 1 + 0.927 407 925 267 795 832 328 424 243 2;
  • 13) 0.927 407 925 267 795 832 328 424 243 2 × 2 = 1 + 0.854 815 850 535 591 664 656 848 486 4;
  • 14) 0.854 815 850 535 591 664 656 848 486 4 × 2 = 1 + 0.709 631 701 071 183 329 313 696 972 8;
  • 15) 0.709 631 701 071 183 329 313 696 972 8 × 2 = 1 + 0.419 263 402 142 366 658 627 393 945 6;
  • 16) 0.419 263 402 142 366 658 627 393 945 6 × 2 = 0 + 0.838 526 804 284 733 317 254 787 891 2;
  • 17) 0.838 526 804 284 733 317 254 787 891 2 × 2 = 1 + 0.677 053 608 569 466 634 509 575 782 4;
  • 18) 0.677 053 608 569 466 634 509 575 782 4 × 2 = 1 + 0.354 107 217 138 933 269 019 151 564 8;
  • 19) 0.354 107 217 138 933 269 019 151 564 8 × 2 = 0 + 0.708 214 434 277 866 538 038 303 129 6;
  • 20) 0.708 214 434 277 866 538 038 303 129 6 × 2 = 1 + 0.416 428 868 555 733 076 076 606 259 2;
  • 21) 0.416 428 868 555 733 076 076 606 259 2 × 2 = 0 + 0.832 857 737 111 466 152 153 212 518 4;
  • 22) 0.832 857 737 111 466 152 153 212 518 4 × 2 = 1 + 0.665 715 474 222 932 304 306 425 036 8;
  • 23) 0.665 715 474 222 932 304 306 425 036 8 × 2 = 1 + 0.331 430 948 445 864 608 612 850 073 6;
  • 24) 0.331 430 948 445 864 608 612 850 073 6 × 2 = 0 + 0.662 861 896 891 729 217 225 700 147 2;
  • 25) 0.662 861 896 891 729 217 225 700 147 2 × 2 = 1 + 0.325 723 793 783 458 434 451 400 294 4;
  • 26) 0.325 723 793 783 458 434 451 400 294 4 × 2 = 0 + 0.651 447 587 566 916 868 902 800 588 8;
  • 27) 0.651 447 587 566 916 868 902 800 588 8 × 2 = 1 + 0.302 895 175 133 833 737 805 601 177 6;
  • 28) 0.302 895 175 133 833 737 805 601 177 6 × 2 = 0 + 0.605 790 350 267 667 475 611 202 355 2;
  • 29) 0.605 790 350 267 667 475 611 202 355 2 × 2 = 1 + 0.211 580 700 535 334 951 222 404 710 4;
  • 30) 0.211 580 700 535 334 951 222 404 710 4 × 2 = 0 + 0.423 161 401 070 669 902 444 809 420 8;
  • 31) 0.423 161 401 070 669 902 444 809 420 8 × 2 = 0 + 0.846 322 802 141 339 804 889 618 841 6;
  • 32) 0.846 322 802 141 339 804 889 618 841 6 × 2 = 1 + 0.692 645 604 282 679 609 779 237 683 2;
  • 33) 0.692 645 604 282 679 609 779 237 683 2 × 2 = 1 + 0.385 291 208 565 359 219 558 475 366 4;
  • 34) 0.385 291 208 565 359 219 558 475 366 4 × 2 = 0 + 0.770 582 417 130 718 439 116 950 732 8;
  • 35) 0.770 582 417 130 718 439 116 950 732 8 × 2 = 1 + 0.541 164 834 261 436 878 233 901 465 6;
  • 36) 0.541 164 834 261 436 878 233 901 465 6 × 2 = 1 + 0.082 329 668 522 873 756 467 802 931 2;
  • 37) 0.082 329 668 522 873 756 467 802 931 2 × 2 = 0 + 0.164 659 337 045 747 512 935 605 862 4;
  • 38) 0.164 659 337 045 747 512 935 605 862 4 × 2 = 0 + 0.329 318 674 091 495 025 871 211 724 8;
  • 39) 0.329 318 674 091 495 025 871 211 724 8 × 2 = 0 + 0.658 637 348 182 990 051 742 423 449 6;
  • 40) 0.658 637 348 182 990 051 742 423 449 6 × 2 = 1 + 0.317 274 696 365 980 103 484 846 899 2;
  • 41) 0.317 274 696 365 980 103 484 846 899 2 × 2 = 0 + 0.634 549 392 731 960 206 969 693 798 4;
  • 42) 0.634 549 392 731 960 206 969 693 798 4 × 2 = 1 + 0.269 098 785 463 920 413 939 387 596 8;
  • 43) 0.269 098 785 463 920 413 939 387 596 8 × 2 = 0 + 0.538 197 570 927 840 827 878 775 193 6;
  • 44) 0.538 197 570 927 840 827 878 775 193 6 × 2 = 1 + 0.076 395 141 855 681 655 757 550 387 2;
  • 45) 0.076 395 141 855 681 655 757 550 387 2 × 2 = 0 + 0.152 790 283 711 363 311 515 100 774 4;
  • 46) 0.152 790 283 711 363 311 515 100 774 4 × 2 = 0 + 0.305 580 567 422 726 623 030 201 548 8;
  • 47) 0.305 580 567 422 726 623 030 201 548 8 × 2 = 0 + 0.611 161 134 845 453 246 060 403 097 6;
  • 48) 0.611 161 134 845 453 246 060 403 097 6 × 2 = 1 + 0.222 322 269 690 906 492 120 806 195 2;
  • 49) 0.222 322 269 690 906 492 120 806 195 2 × 2 = 0 + 0.444 644 539 381 812 984 241 612 390 4;
  • 50) 0.444 644 539 381 812 984 241 612 390 4 × 2 = 0 + 0.889 289 078 763 625 968 483 224 780 8;
  • 51) 0.889 289 078 763 625 968 483 224 780 8 × 2 = 1 + 0.778 578 157 527 251 936 966 449 561 6;
  • 52) 0.778 578 157 527 251 936 966 449 561 6 × 2 = 1 + 0.557 156 315 054 503 873 932 899 123 2;
  • 53) 0.557 156 315 054 503 873 932 899 123 2 × 2 = 1 + 0.114 312 630 109 007 747 865 798 246 4;

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:


0.648 419 777 325 504 832 966 877 056 7(10) =

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

5. Positive number before normalization:

0.648 419 777 325 504 832 966 877 056 7(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:


0.648 419 777 325 504 832 966 877 056 7(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 0.648 419 777 325 504 832 966 877 056 7 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