2.356 194 490 192 344 928 846 982 537 459 627 163 148 313 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.356 194 490 192 344 928 846 982 537 459 627 163 148 313(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
2.356 194 490 192 344 928 846 982 537 459 627 163 148 313(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: 0.356 194 490 192 344 928 846 982 537 459 627 163 148 313.

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) 0.356 194 490 192 344 928 846 982 537 459 627 163 148 313 × 2 = 0 + 0.712 388 980 384 689 857 693 965 074 919 254 326 296 626;
  • 2) 0.712 388 980 384 689 857 693 965 074 919 254 326 296 626 × 2 = 1 + 0.424 777 960 769 379 715 387 930 149 838 508 652 593 252;
  • 3) 0.424 777 960 769 379 715 387 930 149 838 508 652 593 252 × 2 = 0 + 0.849 555 921 538 759 430 775 860 299 677 017 305 186 504;
  • 4) 0.849 555 921 538 759 430 775 860 299 677 017 305 186 504 × 2 = 1 + 0.699 111 843 077 518 861 551 720 599 354 034 610 373 008;
  • 5) 0.699 111 843 077 518 861 551 720 599 354 034 610 373 008 × 2 = 1 + 0.398 223 686 155 037 723 103 441 198 708 069 220 746 016;
  • 6) 0.398 223 686 155 037 723 103 441 198 708 069 220 746 016 × 2 = 0 + 0.796 447 372 310 075 446 206 882 397 416 138 441 492 032;
  • 7) 0.796 447 372 310 075 446 206 882 397 416 138 441 492 032 × 2 = 1 + 0.592 894 744 620 150 892 413 764 794 832 276 882 984 064;
  • 8) 0.592 894 744 620 150 892 413 764 794 832 276 882 984 064 × 2 = 1 + 0.185 789 489 240 301 784 827 529 589 664 553 765 968 128;
  • 9) 0.185 789 489 240 301 784 827 529 589 664 553 765 968 128 × 2 = 0 + 0.371 578 978 480 603 569 655 059 179 329 107 531 936 256;
  • 10) 0.371 578 978 480 603 569 655 059 179 329 107 531 936 256 × 2 = 0 + 0.743 157 956 961 207 139 310 118 358 658 215 063 872 512;
  • 11) 0.743 157 956 961 207 139 310 118 358 658 215 063 872 512 × 2 = 1 + 0.486 315 913 922 414 278 620 236 717 316 430 127 745 024;
  • 12) 0.486 315 913 922 414 278 620 236 717 316 430 127 745 024 × 2 = 0 + 0.972 631 827 844 828 557 240 473 434 632 860 255 490 048;
  • 13) 0.972 631 827 844 828 557 240 473 434 632 860 255 490 048 × 2 = 1 + 0.945 263 655 689 657 114 480 946 869 265 720 510 980 096;
  • 14) 0.945 263 655 689 657 114 480 946 869 265 720 510 980 096 × 2 = 1 + 0.890 527 311 379 314 228 961 893 738 531 441 021 960 192;
  • 15) 0.890 527 311 379 314 228 961 893 738 531 441 021 960 192 × 2 = 1 + 0.781 054 622 758 628 457 923 787 477 062 882 043 920 384;
  • 16) 0.781 054 622 758 628 457 923 787 477 062 882 043 920 384 × 2 = 1 + 0.562 109 245 517 256 915 847 574 954 125 764 087 840 768;
  • 17) 0.562 109 245 517 256 915 847 574 954 125 764 087 840 768 × 2 = 1 + 0.124 218 491 034 513 831 695 149 908 251 528 175 681 536;
  • 18) 0.124 218 491 034 513 831 695 149 908 251 528 175 681 536 × 2 = 0 + 0.248 436 982 069 027 663 390 299 816 503 056 351 363 072;
  • 19) 0.248 436 982 069 027 663 390 299 816 503 056 351 363 072 × 2 = 0 + 0.496 873 964 138 055 326 780 599 633 006 112 702 726 144;
  • 20) 0.496 873 964 138 055 326 780 599 633 006 112 702 726 144 × 2 = 0 + 0.993 747 928 276 110 653 561 199 266 012 225 405 452 288;
  • 21) 0.993 747 928 276 110 653 561 199 266 012 225 405 452 288 × 2 = 1 + 0.987 495 856 552 221 307 122 398 532 024 450 810 904 576;
  • 22) 0.987 495 856 552 221 307 122 398 532 024 450 810 904 576 × 2 = 1 + 0.974 991 713 104 442 614 244 797 064 048 901 621 809 152;
  • 23) 0.974 991 713 104 442 614 244 797 064 048 901 621 809 152 × 2 = 1 + 0.949 983 426 208 885 228 489 594 128 097 803 243 618 304;
  • 24) 0.949 983 426 208 885 228 489 594 128 097 803 243 618 304 × 2 = 1 + 0.899 966 852 417 770 456 979 188 256 195 606 487 236 608;
  • 25) 0.899 966 852 417 770 456 979 188 256 195 606 487 236 608 × 2 = 1 + 0.799 933 704 835 540 913 958 376 512 391 212 974 473 216;
  • 26) 0.799 933 704 835 540 913 958 376 512 391 212 974 473 216 × 2 = 1 + 0.599 867 409 671 081 827 916 753 024 782 425 948 946 432;
  • 27) 0.599 867 409 671 081 827 916 753 024 782 425 948 946 432 × 2 = 1 + 0.199 734 819 342 163 655 833 506 049 564 851 897 892 864;
  • 28) 0.199 734 819 342 163 655 833 506 049 564 851 897 892 864 × 2 = 0 + 0.399 469 638 684 327 311 667 012 099 129 703 795 785 728;
  • 29) 0.399 469 638 684 327 311 667 012 099 129 703 795 785 728 × 2 = 0 + 0.798 939 277 368 654 623 334 024 198 259 407 591 571 456;
  • 30) 0.798 939 277 368 654 623 334 024 198 259 407 591 571 456 × 2 = 1 + 0.597 878 554 737 309 246 668 048 396 518 815 183 142 912;
  • 31) 0.597 878 554 737 309 246 668 048 396 518 815 183 142 912 × 2 = 1 + 0.195 757 109 474 618 493 336 096 793 037 630 366 285 824;
  • 32) 0.195 757 109 474 618 493 336 096 793 037 630 366 285 824 × 2 = 0 + 0.391 514 218 949 236 986 672 193 586 075 260 732 571 648;
  • 33) 0.391 514 218 949 236 986 672 193 586 075 260 732 571 648 × 2 = 0 + 0.783 028 437 898 473 973 344 387 172 150 521 465 143 296;
  • 34) 0.783 028 437 898 473 973 344 387 172 150 521 465 143 296 × 2 = 1 + 0.566 056 875 796 947 946 688 774 344 301 042 930 286 592;
  • 35) 0.566 056 875 796 947 946 688 774 344 301 042 930 286 592 × 2 = 1 + 0.132 113 751 593 895 893 377 548 688 602 085 860 573 184;
  • 36) 0.132 113 751 593 895 893 377 548 688 602 085 860 573 184 × 2 = 0 + 0.264 227 503 187 791 786 755 097 377 204 171 721 146 368;
  • 37) 0.264 227 503 187 791 786 755 097 377 204 171 721 146 368 × 2 = 0 + 0.528 455 006 375 583 573 510 194 754 408 343 442 292 736;
  • 38) 0.528 455 006 375 583 573 510 194 754 408 343 442 292 736 × 2 = 1 + 0.056 910 012 751 167 147 020 389 508 816 686 884 585 472;
  • 39) 0.056 910 012 751 167 147 020 389 508 816 686 884 585 472 × 2 = 0 + 0.113 820 025 502 334 294 040 779 017 633 373 769 170 944;
  • 40) 0.113 820 025 502 334 294 040 779 017 633 373 769 170 944 × 2 = 0 + 0.227 640 051 004 668 588 081 558 035 266 747 538 341 888;
  • 41) 0.227 640 051 004 668 588 081 558 035 266 747 538 341 888 × 2 = 0 + 0.455 280 102 009 337 176 163 116 070 533 495 076 683 776;
  • 42) 0.455 280 102 009 337 176 163 116 070 533 495 076 683 776 × 2 = 0 + 0.910 560 204 018 674 352 326 232 141 066 990 153 367 552;
  • 43) 0.910 560 204 018 674 352 326 232 141 066 990 153 367 552 × 2 = 1 + 0.821 120 408 037 348 704 652 464 282 133 980 306 735 104;
  • 44) 0.821 120 408 037 348 704 652 464 282 133 980 306 735 104 × 2 = 1 + 0.642 240 816 074 697 409 304 928 564 267 960 613 470 208;
  • 45) 0.642 240 816 074 697 409 304 928 564 267 960 613 470 208 × 2 = 1 + 0.284 481 632 149 394 818 609 857 128 535 921 226 940 416;
  • 46) 0.284 481 632 149 394 818 609 857 128 535 921 226 940 416 × 2 = 0 + 0.568 963 264 298 789 637 219 714 257 071 842 453 880 832;
  • 47) 0.568 963 264 298 789 637 219 714 257 071 842 453 880 832 × 2 = 1 + 0.137 926 528 597 579 274 439 428 514 143 684 907 761 664;
  • 48) 0.137 926 528 597 579 274 439 428 514 143 684 907 761 664 × 2 = 0 + 0.275 853 057 195 158 548 878 857 028 287 369 815 523 328;
  • 49) 0.275 853 057 195 158 548 878 857 028 287 369 815 523 328 × 2 = 0 + 0.551 706 114 390 317 097 757 714 056 574 739 631 046 656;
  • 50) 0.551 706 114 390 317 097 757 714 056 574 739 631 046 656 × 2 = 1 + 0.103 412 228 780 634 195 515 428 113 149 479 262 093 312;
  • 51) 0.103 412 228 780 634 195 515 428 113 149 479 262 093 312 × 2 = 0 + 0.206 824 457 561 268 391 030 856 226 298 958 524 186 624;
  • 52) 0.206 824 457 561 268 391 030 856 226 298 958 524 186 624 × 2 = 0 + 0.413 648 915 122 536 782 061 712 452 597 917 048 373 248;
  • 53) 0.413 648 915 122 536 782 061 712 452 597 917 048 373 248 × 2 = 0 + 0.827 297 830 245 073 564 123 424 905 195 834 096 746 496;

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:


0.356 194 490 192 344 928 846 982 537 459 627 163 148 313(10) =

0.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2)

5. Positive number before normalization:

2.356 194 490 192 344 928 846 982 537 459 627 163 148 313(10) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(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:


2.356 194 490 192 344 928 846 982 537 459 627 163 148 313(10) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2) × 20 =

1.0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00(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):
1.0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00


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. 0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00 =

0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010


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) =
0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010


Decimal number 2.356 194 490 192 344 928 846 982 537 459 627 163 148 313 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010

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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