8.000 976 681 723 843 242 366 456 252 057 105 302 810 622 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 8.000 976 681 723 843 242 366 456 252 057 105 302 810 622(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
8.000 976 681 723 843 242 366 456 252 057 105 302 810 622(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: 8.
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;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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.

8(10) =

1000(2)


3. Convert to binary (base 2) the fractional part: 0.000 976 681 723 843 242 366 456 252 057 105 302 810 622.

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) 0.000 976 681 723 843 242 366 456 252 057 105 302 810 622 × 2 = 0 + 0.001 953 363 447 686 484 732 912 504 114 210 605 621 244;
  • 2) 0.001 953 363 447 686 484 732 912 504 114 210 605 621 244 × 2 = 0 + 0.003 906 726 895 372 969 465 825 008 228 421 211 242 488;
  • 3) 0.003 906 726 895 372 969 465 825 008 228 421 211 242 488 × 2 = 0 + 0.007 813 453 790 745 938 931 650 016 456 842 422 484 976;
  • 4) 0.007 813 453 790 745 938 931 650 016 456 842 422 484 976 × 2 = 0 + 0.015 626 907 581 491 877 863 300 032 913 684 844 969 952;
  • 5) 0.015 626 907 581 491 877 863 300 032 913 684 844 969 952 × 2 = 0 + 0.031 253 815 162 983 755 726 600 065 827 369 689 939 904;
  • 6) 0.031 253 815 162 983 755 726 600 065 827 369 689 939 904 × 2 = 0 + 0.062 507 630 325 967 511 453 200 131 654 739 379 879 808;
  • 7) 0.062 507 630 325 967 511 453 200 131 654 739 379 879 808 × 2 = 0 + 0.125 015 260 651 935 022 906 400 263 309 478 759 759 616;
  • 8) 0.125 015 260 651 935 022 906 400 263 309 478 759 759 616 × 2 = 0 + 0.250 030 521 303 870 045 812 800 526 618 957 519 519 232;
  • 9) 0.250 030 521 303 870 045 812 800 526 618 957 519 519 232 × 2 = 0 + 0.500 061 042 607 740 091 625 601 053 237 915 039 038 464;
  • 10) 0.500 061 042 607 740 091 625 601 053 237 915 039 038 464 × 2 = 1 + 0.000 122 085 215 480 183 251 202 106 475 830 078 076 928;
  • 11) 0.000 122 085 215 480 183 251 202 106 475 830 078 076 928 × 2 = 0 + 0.000 244 170 430 960 366 502 404 212 951 660 156 153 856;
  • 12) 0.000 244 170 430 960 366 502 404 212 951 660 156 153 856 × 2 = 0 + 0.000 488 340 861 920 733 004 808 425 903 320 312 307 712;
  • 13) 0.000 488 340 861 920 733 004 808 425 903 320 312 307 712 × 2 = 0 + 0.000 976 681 723 841 466 009 616 851 806 640 624 615 424;
  • 14) 0.000 976 681 723 841 466 009 616 851 806 640 624 615 424 × 2 = 0 + 0.001 953 363 447 682 932 019 233 703 613 281 249 230 848;
  • 15) 0.001 953 363 447 682 932 019 233 703 613 281 249 230 848 × 2 = 0 + 0.003 906 726 895 365 864 038 467 407 226 562 498 461 696;
  • 16) 0.003 906 726 895 365 864 038 467 407 226 562 498 461 696 × 2 = 0 + 0.007 813 453 790 731 728 076 934 814 453 124 996 923 392;
  • 17) 0.007 813 453 790 731 728 076 934 814 453 124 996 923 392 × 2 = 0 + 0.015 626 907 581 463 456 153 869 628 906 249 993 846 784;
  • 18) 0.015 626 907 581 463 456 153 869 628 906 249 993 846 784 × 2 = 0 + 0.031 253 815 162 926 912 307 739 257 812 499 987 693 568;
  • 19) 0.031 253 815 162 926 912 307 739 257 812 499 987 693 568 × 2 = 0 + 0.062 507 630 325 853 824 615 478 515 624 999 975 387 136;
  • 20) 0.062 507 630 325 853 824 615 478 515 624 999 975 387 136 × 2 = 0 + 0.125 015 260 651 707 649 230 957 031 249 999 950 774 272;
  • 21) 0.125 015 260 651 707 649 230 957 031 249 999 950 774 272 × 2 = 0 + 0.250 030 521 303 415 298 461 914 062 499 999 901 548 544;
  • 22) 0.250 030 521 303 415 298 461 914 062 499 999 901 548 544 × 2 = 0 + 0.500 061 042 606 830 596 923 828 124 999 999 803 097 088;
  • 23) 0.500 061 042 606 830 596 923 828 124 999 999 803 097 088 × 2 = 1 + 0.000 122 085 213 661 193 847 656 249 999 999 606 194 176;
  • 24) 0.000 122 085 213 661 193 847 656 249 999 999 606 194 176 × 2 = 0 + 0.000 244 170 427 322 387 695 312 499 999 999 212 388 352;
  • 25) 0.000 244 170 427 322 387 695 312 499 999 999 212 388 352 × 2 = 0 + 0.000 488 340 854 644 775 390 624 999 999 998 424 776 704;
  • 26) 0.000 488 340 854 644 775 390 624 999 999 998 424 776 704 × 2 = 0 + 0.000 976 681 709 289 550 781 249 999 999 996 849 553 408;
  • 27) 0.000 976 681 709 289 550 781 249 999 999 996 849 553 408 × 2 = 0 + 0.001 953 363 418 579 101 562 499 999 999 993 699 106 816;
  • 28) 0.001 953 363 418 579 101 562 499 999 999 993 699 106 816 × 2 = 0 + 0.003 906 726 837 158 203 124 999 999 999 987 398 213 632;
  • 29) 0.003 906 726 837 158 203 124 999 999 999 987 398 213 632 × 2 = 0 + 0.007 813 453 674 316 406 249 999 999 999 974 796 427 264;
  • 30) 0.007 813 453 674 316 406 249 999 999 999 974 796 427 264 × 2 = 0 + 0.015 626 907 348 632 812 499 999 999 999 949 592 854 528;
  • 31) 0.015 626 907 348 632 812 499 999 999 999 949 592 854 528 × 2 = 0 + 0.031 253 814 697 265 624 999 999 999 999 899 185 709 056;
  • 32) 0.031 253 814 697 265 624 999 999 999 999 899 185 709 056 × 2 = 0 + 0.062 507 629 394 531 249 999 999 999 999 798 371 418 112;
  • 33) 0.062 507 629 394 531 249 999 999 999 999 798 371 418 112 × 2 = 0 + 0.125 015 258 789 062 499 999 999 999 999 596 742 836 224;
  • 34) 0.125 015 258 789 062 499 999 999 999 999 596 742 836 224 × 2 = 0 + 0.250 030 517 578 124 999 999 999 999 999 193 485 672 448;
  • 35) 0.250 030 517 578 124 999 999 999 999 999 193 485 672 448 × 2 = 0 + 0.500 061 035 156 249 999 999 999 999 998 386 971 344 896;
  • 36) 0.500 061 035 156 249 999 999 999 999 998 386 971 344 896 × 2 = 1 + 0.000 122 070 312 499 999 999 999 999 996 773 942 689 792;
  • 37) 0.000 122 070 312 499 999 999 999 999 996 773 942 689 792 × 2 = 0 + 0.000 244 140 624 999 999 999 999 999 993 547 885 379 584;
  • 38) 0.000 244 140 624 999 999 999 999 999 993 547 885 379 584 × 2 = 0 + 0.000 488 281 249 999 999 999 999 999 987 095 770 759 168;
  • 39) 0.000 488 281 249 999 999 999 999 999 987 095 770 759 168 × 2 = 0 + 0.000 976 562 499 999 999 999 999 999 974 191 541 518 336;
  • 40) 0.000 976 562 499 999 999 999 999 999 974 191 541 518 336 × 2 = 0 + 0.001 953 124 999 999 999 999 999 999 948 383 083 036 672;
  • 41) 0.001 953 124 999 999 999 999 999 999 948 383 083 036 672 × 2 = 0 + 0.003 906 249 999 999 999 999 999 999 896 766 166 073 344;
  • 42) 0.003 906 249 999 999 999 999 999 999 896 766 166 073 344 × 2 = 0 + 0.007 812 499 999 999 999 999 999 999 793 532 332 146 688;
  • 43) 0.007 812 499 999 999 999 999 999 999 793 532 332 146 688 × 2 = 0 + 0.015 624 999 999 999 999 999 999 999 587 064 664 293 376;
  • 44) 0.015 624 999 999 999 999 999 999 999 587 064 664 293 376 × 2 = 0 + 0.031 249 999 999 999 999 999 999 999 174 129 328 586 752;
  • 45) 0.031 249 999 999 999 999 999 999 999 174 129 328 586 752 × 2 = 0 + 0.062 499 999 999 999 999 999 999 998 348 258 657 173 504;
  • 46) 0.062 499 999 999 999 999 999 999 998 348 258 657 173 504 × 2 = 0 + 0.124 999 999 999 999 999 999 999 996 696 517 314 347 008;
  • 47) 0.124 999 999 999 999 999 999 999 996 696 517 314 347 008 × 2 = 0 + 0.249 999 999 999 999 999 999 999 993 393 034 628 694 016;
  • 48) 0.249 999 999 999 999 999 999 999 993 393 034 628 694 016 × 2 = 0 + 0.499 999 999 999 999 999 999 999 986 786 069 257 388 032;
  • 49) 0.499 999 999 999 999 999 999 999 986 786 069 257 388 032 × 2 = 0 + 0.999 999 999 999 999 999 999 999 973 572 138 514 776 064;
  • 50) 0.999 999 999 999 999 999 999 999 973 572 138 514 776 064 × 2 = 1 + 0.999 999 999 999 999 999 999 999 947 144 277 029 552 128;
  • 51) 0.999 999 999 999 999 999 999 999 947 144 277 029 552 128 × 2 = 1 + 0.999 999 999 999 999 999 999 999 894 288 554 059 104 256;
  • 52) 0.999 999 999 999 999 999 999 999 894 288 554 059 104 256 × 2 = 1 + 0.999 999 999 999 999 999 999 999 788 577 108 118 208 512;
  • 53) 0.999 999 999 999 999 999 999 999 788 577 108 118 208 512 × 2 = 1 + 0.999 999 999 999 999 999 999 999 577 154 216 236 417 024;

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:


0.000 976 681 723 843 242 366 456 252 057 105 302 810 622(10) =

0.0000 0000 0100 0000 0000 0010 0000 0000 0001 0000 0000 0000 0111 1(2)

5. Positive number before normalization:

8.000 976 681 723 843 242 366 456 252 057 105 302 810 622(10) =

1000.0000 0000 0100 0000 0000 0010 0000 0000 0001 0000 0000 0000 0111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


8.000 976 681 723 843 242 366 456 252 057 105 302 810 622(10) =

1000.0000 0000 0100 0000 0000 0010 0000 0000 0001 0000 0000 0000 0111 1(2) =

1000.0000 0000 0100 0000 0000 0010 0000 0000 0001 0000 0000 0000 0111 1(2) × 20 =

1.0000 0000 0000 1000 0000 0000 0100 0000 0000 0010 0000 0000 0000 1111(2) × 23


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): 3

Mantissa (not normalized):
1.0000 0000 0000 1000 0000 0000 0100 0000 0000 0010 0000 0000 0000 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

3 + 2(11-1) - 1 =

(3 + 1 023)(10) =

1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 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) =

1026(10) =

100 0000 0010(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. 0000 0000 0000 1000 0000 0000 0100 0000 0000 0010 0000 0000 0000 1111 =

0000 0000 0000 1000 0000 0000 0100 0000 0000 0010 0000 0000 0000


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 0010

Mantissa (52 bits) =
0000 0000 0000 1000 0000 0000 0100 0000 0000 0010 0000 0000 0000


Decimal number 8.000 976 681 723 843 242 366 456 252 057 105 302 810 622 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0010 - 0000 0000 0000 1000 0000 0000 0100 0000 0000 0010 0000 0000 0000

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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