2.718 281 828 459 045 235 360 287 471 352 662 497 947 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 459 045 235 360 287 471 352 662 497 947(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
2.718 281 828 459 045 235 360 287 471 352 662 497 947(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: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.718 281 828 459 045 235 360 287 471 352 662 497 947.

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) 0.718 281 828 459 045 235 360 287 471 352 662 497 947 × 2 = 1 + 0.436 563 656 918 090 470 720 574 942 705 324 995 894;
  • 2) 0.436 563 656 918 090 470 720 574 942 705 324 995 894 × 2 = 0 + 0.873 127 313 836 180 941 441 149 885 410 649 991 788;
  • 3) 0.873 127 313 836 180 941 441 149 885 410 649 991 788 × 2 = 1 + 0.746 254 627 672 361 882 882 299 770 821 299 983 576;
  • 4) 0.746 254 627 672 361 882 882 299 770 821 299 983 576 × 2 = 1 + 0.492 509 255 344 723 765 764 599 541 642 599 967 152;
  • 5) 0.492 509 255 344 723 765 764 599 541 642 599 967 152 × 2 = 0 + 0.985 018 510 689 447 531 529 199 083 285 199 934 304;
  • 6) 0.985 018 510 689 447 531 529 199 083 285 199 934 304 × 2 = 1 + 0.970 037 021 378 895 063 058 398 166 570 399 868 608;
  • 7) 0.970 037 021 378 895 063 058 398 166 570 399 868 608 × 2 = 1 + 0.940 074 042 757 790 126 116 796 333 140 799 737 216;
  • 8) 0.940 074 042 757 790 126 116 796 333 140 799 737 216 × 2 = 1 + 0.880 148 085 515 580 252 233 592 666 281 599 474 432;
  • 9) 0.880 148 085 515 580 252 233 592 666 281 599 474 432 × 2 = 1 + 0.760 296 171 031 160 504 467 185 332 563 198 948 864;
  • 10) 0.760 296 171 031 160 504 467 185 332 563 198 948 864 × 2 = 1 + 0.520 592 342 062 321 008 934 370 665 126 397 897 728;
  • 11) 0.520 592 342 062 321 008 934 370 665 126 397 897 728 × 2 = 1 + 0.041 184 684 124 642 017 868 741 330 252 795 795 456;
  • 12) 0.041 184 684 124 642 017 868 741 330 252 795 795 456 × 2 = 0 + 0.082 369 368 249 284 035 737 482 660 505 591 590 912;
  • 13) 0.082 369 368 249 284 035 737 482 660 505 591 590 912 × 2 = 0 + 0.164 738 736 498 568 071 474 965 321 011 183 181 824;
  • 14) 0.164 738 736 498 568 071 474 965 321 011 183 181 824 × 2 = 0 + 0.329 477 472 997 136 142 949 930 642 022 366 363 648;
  • 15) 0.329 477 472 997 136 142 949 930 642 022 366 363 648 × 2 = 0 + 0.658 954 945 994 272 285 899 861 284 044 732 727 296;
  • 16) 0.658 954 945 994 272 285 899 861 284 044 732 727 296 × 2 = 1 + 0.317 909 891 988 544 571 799 722 568 089 465 454 592;
  • 17) 0.317 909 891 988 544 571 799 722 568 089 465 454 592 × 2 = 0 + 0.635 819 783 977 089 143 599 445 136 178 930 909 184;
  • 18) 0.635 819 783 977 089 143 599 445 136 178 930 909 184 × 2 = 1 + 0.271 639 567 954 178 287 198 890 272 357 861 818 368;
  • 19) 0.271 639 567 954 178 287 198 890 272 357 861 818 368 × 2 = 0 + 0.543 279 135 908 356 574 397 780 544 715 723 636 736;
  • 20) 0.543 279 135 908 356 574 397 780 544 715 723 636 736 × 2 = 1 + 0.086 558 271 816 713 148 795 561 089 431 447 273 472;
  • 21) 0.086 558 271 816 713 148 795 561 089 431 447 273 472 × 2 = 0 + 0.173 116 543 633 426 297 591 122 178 862 894 546 944;
  • 22) 0.173 116 543 633 426 297 591 122 178 862 894 546 944 × 2 = 0 + 0.346 233 087 266 852 595 182 244 357 725 789 093 888;
  • 23) 0.346 233 087 266 852 595 182 244 357 725 789 093 888 × 2 = 0 + 0.692 466 174 533 705 190 364 488 715 451 578 187 776;
  • 24) 0.692 466 174 533 705 190 364 488 715 451 578 187 776 × 2 = 1 + 0.384 932 349 067 410 380 728 977 430 903 156 375 552;
  • 25) 0.384 932 349 067 410 380 728 977 430 903 156 375 552 × 2 = 0 + 0.769 864 698 134 820 761 457 954 861 806 312 751 104;
  • 26) 0.769 864 698 134 820 761 457 954 861 806 312 751 104 × 2 = 1 + 0.539 729 396 269 641 522 915 909 723 612 625 502 208;
  • 27) 0.539 729 396 269 641 522 915 909 723 612 625 502 208 × 2 = 1 + 0.079 458 792 539 283 045 831 819 447 225 251 004 416;
  • 28) 0.079 458 792 539 283 045 831 819 447 225 251 004 416 × 2 = 0 + 0.158 917 585 078 566 091 663 638 894 450 502 008 832;
  • 29) 0.158 917 585 078 566 091 663 638 894 450 502 008 832 × 2 = 0 + 0.317 835 170 157 132 183 327 277 788 901 004 017 664;
  • 30) 0.317 835 170 157 132 183 327 277 788 901 004 017 664 × 2 = 0 + 0.635 670 340 314 264 366 654 555 577 802 008 035 328;
  • 31) 0.635 670 340 314 264 366 654 555 577 802 008 035 328 × 2 = 1 + 0.271 340 680 628 528 733 309 111 155 604 016 070 656;
  • 32) 0.271 340 680 628 528 733 309 111 155 604 016 070 656 × 2 = 0 + 0.542 681 361 257 057 466 618 222 311 208 032 141 312;
  • 33) 0.542 681 361 257 057 466 618 222 311 208 032 141 312 × 2 = 1 + 0.085 362 722 514 114 933 236 444 622 416 064 282 624;
  • 34) 0.085 362 722 514 114 933 236 444 622 416 064 282 624 × 2 = 0 + 0.170 725 445 028 229 866 472 889 244 832 128 565 248;
  • 35) 0.170 725 445 028 229 866 472 889 244 832 128 565 248 × 2 = 0 + 0.341 450 890 056 459 732 945 778 489 664 257 130 496;
  • 36) 0.341 450 890 056 459 732 945 778 489 664 257 130 496 × 2 = 0 + 0.682 901 780 112 919 465 891 556 979 328 514 260 992;
  • 37) 0.682 901 780 112 919 465 891 556 979 328 514 260 992 × 2 = 1 + 0.365 803 560 225 838 931 783 113 958 657 028 521 984;
  • 38) 0.365 803 560 225 838 931 783 113 958 657 028 521 984 × 2 = 0 + 0.731 607 120 451 677 863 566 227 917 314 057 043 968;
  • 39) 0.731 607 120 451 677 863 566 227 917 314 057 043 968 × 2 = 1 + 0.463 214 240 903 355 727 132 455 834 628 114 087 936;
  • 40) 0.463 214 240 903 355 727 132 455 834 628 114 087 936 × 2 = 0 + 0.926 428 481 806 711 454 264 911 669 256 228 175 872;
  • 41) 0.926 428 481 806 711 454 264 911 669 256 228 175 872 × 2 = 1 + 0.852 856 963 613 422 908 529 823 338 512 456 351 744;
  • 42) 0.852 856 963 613 422 908 529 823 338 512 456 351 744 × 2 = 1 + 0.705 713 927 226 845 817 059 646 677 024 912 703 488;
  • 43) 0.705 713 927 226 845 817 059 646 677 024 912 703 488 × 2 = 1 + 0.411 427 854 453 691 634 119 293 354 049 825 406 976;
  • 44) 0.411 427 854 453 691 634 119 293 354 049 825 406 976 × 2 = 0 + 0.822 855 708 907 383 268 238 586 708 099 650 813 952;
  • 45) 0.822 855 708 907 383 268 238 586 708 099 650 813 952 × 2 = 1 + 0.645 711 417 814 766 536 477 173 416 199 301 627 904;
  • 46) 0.645 711 417 814 766 536 477 173 416 199 301 627 904 × 2 = 1 + 0.291 422 835 629 533 072 954 346 832 398 603 255 808;
  • 47) 0.291 422 835 629 533 072 954 346 832 398 603 255 808 × 2 = 0 + 0.582 845 671 259 066 145 908 693 664 797 206 511 616;
  • 48) 0.582 845 671 259 066 145 908 693 664 797 206 511 616 × 2 = 1 + 0.165 691 342 518 132 291 817 387 329 594 413 023 232;
  • 49) 0.165 691 342 518 132 291 817 387 329 594 413 023 232 × 2 = 0 + 0.331 382 685 036 264 583 634 774 659 188 826 046 464;
  • 50) 0.331 382 685 036 264 583 634 774 659 188 826 046 464 × 2 = 0 + 0.662 765 370 072 529 167 269 549 318 377 652 092 928;
  • 51) 0.662 765 370 072 529 167 269 549 318 377 652 092 928 × 2 = 1 + 0.325 530 740 145 058 334 539 098 636 755 304 185 856;
  • 52) 0.325 530 740 145 058 334 539 098 636 755 304 185 856 × 2 = 0 + 0.651 061 480 290 116 669 078 197 273 510 608 371 712;
  • 53) 0.651 061 480 290 116 669 078 197 273 510 608 371 712 × 2 = 1 + 0.302 122 960 580 233 338 156 394 547 021 216 743 424;

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:


0.718 281 828 459 045 235 360 287 471 352 662 497 947(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1110 1101 0010 1(2)

5. Positive number before normalization:

2.718 281 828 459 045 235 360 287 471 352 662 497 947(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1110 1101 0010 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.718 281 828 459 045 235 360 287 471 352 662 497 947(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1110 1101 0010 1(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1110 1101 0010 1(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0111 0110 1001 01(2) × 21


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):
1.0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0111 0110 1001 01


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0111 0110 1001 01 =

0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0111 0110 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) =
100 0000 0000

Mantissa (52 bits) =
0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0111 0110 1001


Decimal number 2.718 281 828 459 045 235 360 287 471 352 662 497 947 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0111 0110 1001

» Convert decimal 2.718 281 828 459 045 235 360 287 471 352 662 497 88 to 64 bit double precision IEEE 754 binary floating point representation standard

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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