3.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063(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
3.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063.

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) 0.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063 × 2 = 0 + 0.283 185 307 179 586 476 925 286 766 559 005 768 394 340 126;
  • 2) 0.283 185 307 179 586 476 925 286 766 559 005 768 394 340 126 × 2 = 0 + 0.566 370 614 359 172 953 850 573 533 118 011 536 788 680 252;
  • 3) 0.566 370 614 359 172 953 850 573 533 118 011 536 788 680 252 × 2 = 1 + 0.132 741 228 718 345 907 701 147 066 236 023 073 577 360 504;
  • 4) 0.132 741 228 718 345 907 701 147 066 236 023 073 577 360 504 × 2 = 0 + 0.265 482 457 436 691 815 402 294 132 472 046 147 154 721 008;
  • 5) 0.265 482 457 436 691 815 402 294 132 472 046 147 154 721 008 × 2 = 0 + 0.530 964 914 873 383 630 804 588 264 944 092 294 309 442 016;
  • 6) 0.530 964 914 873 383 630 804 588 264 944 092 294 309 442 016 × 2 = 1 + 0.061 929 829 746 767 261 609 176 529 888 184 588 618 884 032;
  • 7) 0.061 929 829 746 767 261 609 176 529 888 184 588 618 884 032 × 2 = 0 + 0.123 859 659 493 534 523 218 353 059 776 369 177 237 768 064;
  • 8) 0.123 859 659 493 534 523 218 353 059 776 369 177 237 768 064 × 2 = 0 + 0.247 719 318 987 069 046 436 706 119 552 738 354 475 536 128;
  • 9) 0.247 719 318 987 069 046 436 706 119 552 738 354 475 536 128 × 2 = 0 + 0.495 438 637 974 138 092 873 412 239 105 476 708 951 072 256;
  • 10) 0.495 438 637 974 138 092 873 412 239 105 476 708 951 072 256 × 2 = 0 + 0.990 877 275 948 276 185 746 824 478 210 953 417 902 144 512;
  • 11) 0.990 877 275 948 276 185 746 824 478 210 953 417 902 144 512 × 2 = 1 + 0.981 754 551 896 552 371 493 648 956 421 906 835 804 289 024;
  • 12) 0.981 754 551 896 552 371 493 648 956 421 906 835 804 289 024 × 2 = 1 + 0.963 509 103 793 104 742 987 297 912 843 813 671 608 578 048;
  • 13) 0.963 509 103 793 104 742 987 297 912 843 813 671 608 578 048 × 2 = 1 + 0.927 018 207 586 209 485 974 595 825 687 627 343 217 156 096;
  • 14) 0.927 018 207 586 209 485 974 595 825 687 627 343 217 156 096 × 2 = 1 + 0.854 036 415 172 418 971 949 191 651 375 254 686 434 312 192;
  • 15) 0.854 036 415 172 418 971 949 191 651 375 254 686 434 312 192 × 2 = 1 + 0.708 072 830 344 837 943 898 383 302 750 509 372 868 624 384;
  • 16) 0.708 072 830 344 837 943 898 383 302 750 509 372 868 624 384 × 2 = 1 + 0.416 145 660 689 675 887 796 766 605 501 018 745 737 248 768;
  • 17) 0.416 145 660 689 675 887 796 766 605 501 018 745 737 248 768 × 2 = 0 + 0.832 291 321 379 351 775 593 533 211 002 037 491 474 497 536;
  • 18) 0.832 291 321 379 351 775 593 533 211 002 037 491 474 497 536 × 2 = 1 + 0.664 582 642 758 703 551 187 066 422 004 074 982 948 995 072;
  • 19) 0.664 582 642 758 703 551 187 066 422 004 074 982 948 995 072 × 2 = 1 + 0.329 165 285 517 407 102 374 132 844 008 149 965 897 990 144;
  • 20) 0.329 165 285 517 407 102 374 132 844 008 149 965 897 990 144 × 2 = 0 + 0.658 330 571 034 814 204 748 265 688 016 299 931 795 980 288;
  • 21) 0.658 330 571 034 814 204 748 265 688 016 299 931 795 980 288 × 2 = 1 + 0.316 661 142 069 628 409 496 531 376 032 599 863 591 960 576;
  • 22) 0.316 661 142 069 628 409 496 531 376 032 599 863 591 960 576 × 2 = 0 + 0.633 322 284 139 256 818 993 062 752 065 199 727 183 921 152;
  • 23) 0.633 322 284 139 256 818 993 062 752 065 199 727 183 921 152 × 2 = 1 + 0.266 644 568 278 513 637 986 125 504 130 399 454 367 842 304;
  • 24) 0.266 644 568 278 513 637 986 125 504 130 399 454 367 842 304 × 2 = 0 + 0.533 289 136 557 027 275 972 251 008 260 798 908 735 684 608;
  • 25) 0.533 289 136 557 027 275 972 251 008 260 798 908 735 684 608 × 2 = 1 + 0.066 578 273 114 054 551 944 502 016 521 597 817 471 369 216;
  • 26) 0.066 578 273 114 054 551 944 502 016 521 597 817 471 369 216 × 2 = 0 + 0.133 156 546 228 109 103 889 004 033 043 195 634 942 738 432;
  • 27) 0.133 156 546 228 109 103 889 004 033 043 195 634 942 738 432 × 2 = 0 + 0.266 313 092 456 218 207 778 008 066 086 391 269 885 476 864;
  • 28) 0.266 313 092 456 218 207 778 008 066 086 391 269 885 476 864 × 2 = 0 + 0.532 626 184 912 436 415 556 016 132 172 782 539 770 953 728;
  • 29) 0.532 626 184 912 436 415 556 016 132 172 782 539 770 953 728 × 2 = 1 + 0.065 252 369 824 872 831 112 032 264 345 565 079 541 907 456;
  • 30) 0.065 252 369 824 872 831 112 032 264 345 565 079 541 907 456 × 2 = 0 + 0.130 504 739 649 745 662 224 064 528 691 130 159 083 814 912;
  • 31) 0.130 504 739 649 745 662 224 064 528 691 130 159 083 814 912 × 2 = 0 + 0.261 009 479 299 491 324 448 129 057 382 260 318 167 629 824;
  • 32) 0.261 009 479 299 491 324 448 129 057 382 260 318 167 629 824 × 2 = 0 + 0.522 018 958 598 982 648 896 258 114 764 520 636 335 259 648;
  • 33) 0.522 018 958 598 982 648 896 258 114 764 520 636 335 259 648 × 2 = 1 + 0.044 037 917 197 965 297 792 516 229 529 041 272 670 519 296;
  • 34) 0.044 037 917 197 965 297 792 516 229 529 041 272 670 519 296 × 2 = 0 + 0.088 075 834 395 930 595 585 032 459 058 082 545 341 038 592;
  • 35) 0.088 075 834 395 930 595 585 032 459 058 082 545 341 038 592 × 2 = 0 + 0.176 151 668 791 861 191 170 064 918 116 165 090 682 077 184;
  • 36) 0.176 151 668 791 861 191 170 064 918 116 165 090 682 077 184 × 2 = 0 + 0.352 303 337 583 722 382 340 129 836 232 330 181 364 154 368;
  • 37) 0.352 303 337 583 722 382 340 129 836 232 330 181 364 154 368 × 2 = 0 + 0.704 606 675 167 444 764 680 259 672 464 660 362 728 308 736;
  • 38) 0.704 606 675 167 444 764 680 259 672 464 660 362 728 308 736 × 2 = 1 + 0.409 213 350 334 889 529 360 519 344 929 320 725 456 617 472;
  • 39) 0.409 213 350 334 889 529 360 519 344 929 320 725 456 617 472 × 2 = 0 + 0.818 426 700 669 779 058 721 038 689 858 641 450 913 234 944;
  • 40) 0.818 426 700 669 779 058 721 038 689 858 641 450 913 234 944 × 2 = 1 + 0.636 853 401 339 558 117 442 077 379 717 282 901 826 469 888;
  • 41) 0.636 853 401 339 558 117 442 077 379 717 282 901 826 469 888 × 2 = 1 + 0.273 706 802 679 116 234 884 154 759 434 565 803 652 939 776;
  • 42) 0.273 706 802 679 116 234 884 154 759 434 565 803 652 939 776 × 2 = 0 + 0.547 413 605 358 232 469 768 309 518 869 131 607 305 879 552;
  • 43) 0.547 413 605 358 232 469 768 309 518 869 131 607 305 879 552 × 2 = 1 + 0.094 827 210 716 464 939 536 619 037 738 263 214 611 759 104;
  • 44) 0.094 827 210 716 464 939 536 619 037 738 263 214 611 759 104 × 2 = 0 + 0.189 654 421 432 929 879 073 238 075 476 526 429 223 518 208;
  • 45) 0.189 654 421 432 929 879 073 238 075 476 526 429 223 518 208 × 2 = 0 + 0.379 308 842 865 859 758 146 476 150 953 052 858 447 036 416;
  • 46) 0.379 308 842 865 859 758 146 476 150 953 052 858 447 036 416 × 2 = 0 + 0.758 617 685 731 719 516 292 952 301 906 105 716 894 072 832;
  • 47) 0.758 617 685 731 719 516 292 952 301 906 105 716 894 072 832 × 2 = 1 + 0.517 235 371 463 439 032 585 904 603 812 211 433 788 145 664;
  • 48) 0.517 235 371 463 439 032 585 904 603 812 211 433 788 145 664 × 2 = 1 + 0.034 470 742 926 878 065 171 809 207 624 422 867 576 291 328;
  • 49) 0.034 470 742 926 878 065 171 809 207 624 422 867 576 291 328 × 2 = 0 + 0.068 941 485 853 756 130 343 618 415 248 845 735 152 582 656;
  • 50) 0.068 941 485 853 756 130 343 618 415 248 845 735 152 582 656 × 2 = 0 + 0.137 882 971 707 512 260 687 236 830 497 691 470 305 165 312;
  • 51) 0.137 882 971 707 512 260 687 236 830 497 691 470 305 165 312 × 2 = 0 + 0.275 765 943 415 024 521 374 473 660 995 382 940 610 330 624;
  • 52) 0.275 765 943 415 024 521 374 473 660 995 382 940 610 330 624 × 2 = 0 + 0.551 531 886 830 049 042 748 947 321 990 765 881 220 661 248;
  • 53) 0.551 531 886 830 049 042 748 947 321 990 765 881 220 661 248 × 2 = 1 + 0.103 063 773 660 098 085 497 894 643 981 531 762 441 322 496;

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:


0.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2)

5. Positive number before normalization:

3.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 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:


3.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 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):
1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 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. 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 01 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000


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) =
1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000


Decimal number 3.141 592 653 589 793 238 462 643 383 279 502 884 197 170 063 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000

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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