2.356 194 490 192 344 928 846 982 537 459 635 08 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.356 194 490 192 344 928 846 982 537 459 635 08(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
2.356 194 490 192 344 928 846 982 537 459 635 08(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: 0.356 194 490 192 344 928 846 982 537 459 635 08.

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) 0.356 194 490 192 344 928 846 982 537 459 635 08 × 2 = 0 + 0.712 388 980 384 689 857 693 965 074 919 270 16;
  • 2) 0.712 388 980 384 689 857 693 965 074 919 270 16 × 2 = 1 + 0.424 777 960 769 379 715 387 930 149 838 540 32;
  • 3) 0.424 777 960 769 379 715 387 930 149 838 540 32 × 2 = 0 + 0.849 555 921 538 759 430 775 860 299 677 080 64;
  • 4) 0.849 555 921 538 759 430 775 860 299 677 080 64 × 2 = 1 + 0.699 111 843 077 518 861 551 720 599 354 161 28;
  • 5) 0.699 111 843 077 518 861 551 720 599 354 161 28 × 2 = 1 + 0.398 223 686 155 037 723 103 441 198 708 322 56;
  • 6) 0.398 223 686 155 037 723 103 441 198 708 322 56 × 2 = 0 + 0.796 447 372 310 075 446 206 882 397 416 645 12;
  • 7) 0.796 447 372 310 075 446 206 882 397 416 645 12 × 2 = 1 + 0.592 894 744 620 150 892 413 764 794 833 290 24;
  • 8) 0.592 894 744 620 150 892 413 764 794 833 290 24 × 2 = 1 + 0.185 789 489 240 301 784 827 529 589 666 580 48;
  • 9) 0.185 789 489 240 301 784 827 529 589 666 580 48 × 2 = 0 + 0.371 578 978 480 603 569 655 059 179 333 160 96;
  • 10) 0.371 578 978 480 603 569 655 059 179 333 160 96 × 2 = 0 + 0.743 157 956 961 207 139 310 118 358 666 321 92;
  • 11) 0.743 157 956 961 207 139 310 118 358 666 321 92 × 2 = 1 + 0.486 315 913 922 414 278 620 236 717 332 643 84;
  • 12) 0.486 315 913 922 414 278 620 236 717 332 643 84 × 2 = 0 + 0.972 631 827 844 828 557 240 473 434 665 287 68;
  • 13) 0.972 631 827 844 828 557 240 473 434 665 287 68 × 2 = 1 + 0.945 263 655 689 657 114 480 946 869 330 575 36;
  • 14) 0.945 263 655 689 657 114 480 946 869 330 575 36 × 2 = 1 + 0.890 527 311 379 314 228 961 893 738 661 150 72;
  • 15) 0.890 527 311 379 314 228 961 893 738 661 150 72 × 2 = 1 + 0.781 054 622 758 628 457 923 787 477 322 301 44;
  • 16) 0.781 054 622 758 628 457 923 787 477 322 301 44 × 2 = 1 + 0.562 109 245 517 256 915 847 574 954 644 602 88;
  • 17) 0.562 109 245 517 256 915 847 574 954 644 602 88 × 2 = 1 + 0.124 218 491 034 513 831 695 149 909 289 205 76;
  • 18) 0.124 218 491 034 513 831 695 149 909 289 205 76 × 2 = 0 + 0.248 436 982 069 027 663 390 299 818 578 411 52;
  • 19) 0.248 436 982 069 027 663 390 299 818 578 411 52 × 2 = 0 + 0.496 873 964 138 055 326 780 599 637 156 823 04;
  • 20) 0.496 873 964 138 055 326 780 599 637 156 823 04 × 2 = 0 + 0.993 747 928 276 110 653 561 199 274 313 646 08;
  • 21) 0.993 747 928 276 110 653 561 199 274 313 646 08 × 2 = 1 + 0.987 495 856 552 221 307 122 398 548 627 292 16;
  • 22) 0.987 495 856 552 221 307 122 398 548 627 292 16 × 2 = 1 + 0.974 991 713 104 442 614 244 797 097 254 584 32;
  • 23) 0.974 991 713 104 442 614 244 797 097 254 584 32 × 2 = 1 + 0.949 983 426 208 885 228 489 594 194 509 168 64;
  • 24) 0.949 983 426 208 885 228 489 594 194 509 168 64 × 2 = 1 + 0.899 966 852 417 770 456 979 188 389 018 337 28;
  • 25) 0.899 966 852 417 770 456 979 188 389 018 337 28 × 2 = 1 + 0.799 933 704 835 540 913 958 376 778 036 674 56;
  • 26) 0.799 933 704 835 540 913 958 376 778 036 674 56 × 2 = 1 + 0.599 867 409 671 081 827 916 753 556 073 349 12;
  • 27) 0.599 867 409 671 081 827 916 753 556 073 349 12 × 2 = 1 + 0.199 734 819 342 163 655 833 507 112 146 698 24;
  • 28) 0.199 734 819 342 163 655 833 507 112 146 698 24 × 2 = 0 + 0.399 469 638 684 327 311 667 014 224 293 396 48;
  • 29) 0.399 469 638 684 327 311 667 014 224 293 396 48 × 2 = 0 + 0.798 939 277 368 654 623 334 028 448 586 792 96;
  • 30) 0.798 939 277 368 654 623 334 028 448 586 792 96 × 2 = 1 + 0.597 878 554 737 309 246 668 056 897 173 585 92;
  • 31) 0.597 878 554 737 309 246 668 056 897 173 585 92 × 2 = 1 + 0.195 757 109 474 618 493 336 113 794 347 171 84;
  • 32) 0.195 757 109 474 618 493 336 113 794 347 171 84 × 2 = 0 + 0.391 514 218 949 236 986 672 227 588 694 343 68;
  • 33) 0.391 514 218 949 236 986 672 227 588 694 343 68 × 2 = 0 + 0.783 028 437 898 473 973 344 455 177 388 687 36;
  • 34) 0.783 028 437 898 473 973 344 455 177 388 687 36 × 2 = 1 + 0.566 056 875 796 947 946 688 910 354 777 374 72;
  • 35) 0.566 056 875 796 947 946 688 910 354 777 374 72 × 2 = 1 + 0.132 113 751 593 895 893 377 820 709 554 749 44;
  • 36) 0.132 113 751 593 895 893 377 820 709 554 749 44 × 2 = 0 + 0.264 227 503 187 791 786 755 641 419 109 498 88;
  • 37) 0.264 227 503 187 791 786 755 641 419 109 498 88 × 2 = 0 + 0.528 455 006 375 583 573 511 282 838 218 997 76;
  • 38) 0.528 455 006 375 583 573 511 282 838 218 997 76 × 2 = 1 + 0.056 910 012 751 167 147 022 565 676 437 995 52;
  • 39) 0.056 910 012 751 167 147 022 565 676 437 995 52 × 2 = 0 + 0.113 820 025 502 334 294 045 131 352 875 991 04;
  • 40) 0.113 820 025 502 334 294 045 131 352 875 991 04 × 2 = 0 + 0.227 640 051 004 668 588 090 262 705 751 982 08;
  • 41) 0.227 640 051 004 668 588 090 262 705 751 982 08 × 2 = 0 + 0.455 280 102 009 337 176 180 525 411 503 964 16;
  • 42) 0.455 280 102 009 337 176 180 525 411 503 964 16 × 2 = 0 + 0.910 560 204 018 674 352 361 050 823 007 928 32;
  • 43) 0.910 560 204 018 674 352 361 050 823 007 928 32 × 2 = 1 + 0.821 120 408 037 348 704 722 101 646 015 856 64;
  • 44) 0.821 120 408 037 348 704 722 101 646 015 856 64 × 2 = 1 + 0.642 240 816 074 697 409 444 203 292 031 713 28;
  • 45) 0.642 240 816 074 697 409 444 203 292 031 713 28 × 2 = 1 + 0.284 481 632 149 394 818 888 406 584 063 426 56;
  • 46) 0.284 481 632 149 394 818 888 406 584 063 426 56 × 2 = 0 + 0.568 963 264 298 789 637 776 813 168 126 853 12;
  • 47) 0.568 963 264 298 789 637 776 813 168 126 853 12 × 2 = 1 + 0.137 926 528 597 579 275 553 626 336 253 706 24;
  • 48) 0.137 926 528 597 579 275 553 626 336 253 706 24 × 2 = 0 + 0.275 853 057 195 158 551 107 252 672 507 412 48;
  • 49) 0.275 853 057 195 158 551 107 252 672 507 412 48 × 2 = 0 + 0.551 706 114 390 317 102 214 505 345 014 824 96;
  • 50) 0.551 706 114 390 317 102 214 505 345 014 824 96 × 2 = 1 + 0.103 412 228 780 634 204 429 010 690 029 649 92;
  • 51) 0.103 412 228 780 634 204 429 010 690 029 649 92 × 2 = 0 + 0.206 824 457 561 268 408 858 021 380 059 299 84;
  • 52) 0.206 824 457 561 268 408 858 021 380 059 299 84 × 2 = 0 + 0.413 648 915 122 536 817 716 042 760 118 599 68;
  • 53) 0.413 648 915 122 536 817 716 042 760 118 599 68 × 2 = 0 + 0.827 297 830 245 073 635 432 085 520 237 199 36;

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:


0.356 194 490 192 344 928 846 982 537 459 635 08(10) =

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

5. Positive number before normalization:

2.356 194 490 192 344 928 846 982 537 459 635 08(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:


2.356 194 490 192 344 928 846 982 537 459 635 08(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 2.356 194 490 192 344 928 846 982 537 459 635 08 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