-0.000 000 045 419 381 005 632 302 94 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 045 419 381 005 632 302 94(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
-0.000 000 045 419 381 005 632 302 94(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. Start with the positive version of the number:

|-0.000 000 045 419 381 005 632 302 94| = 0.000 000 045 419 381 005 632 302 94


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 045 419 381 005 632 302 94.

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) 0.000 000 045 419 381 005 632 302 94 × 2 = 0 + 0.000 000 090 838 762 011 264 605 88;
  • 2) 0.000 000 090 838 762 011 264 605 88 × 2 = 0 + 0.000 000 181 677 524 022 529 211 76;
  • 3) 0.000 000 181 677 524 022 529 211 76 × 2 = 0 + 0.000 000 363 355 048 045 058 423 52;
  • 4) 0.000 000 363 355 048 045 058 423 52 × 2 = 0 + 0.000 000 726 710 096 090 116 847 04;
  • 5) 0.000 000 726 710 096 090 116 847 04 × 2 = 0 + 0.000 001 453 420 192 180 233 694 08;
  • 6) 0.000 001 453 420 192 180 233 694 08 × 2 = 0 + 0.000 002 906 840 384 360 467 388 16;
  • 7) 0.000 002 906 840 384 360 467 388 16 × 2 = 0 + 0.000 005 813 680 768 720 934 776 32;
  • 8) 0.000 005 813 680 768 720 934 776 32 × 2 = 0 + 0.000 011 627 361 537 441 869 552 64;
  • 9) 0.000 011 627 361 537 441 869 552 64 × 2 = 0 + 0.000 023 254 723 074 883 739 105 28;
  • 10) 0.000 023 254 723 074 883 739 105 28 × 2 = 0 + 0.000 046 509 446 149 767 478 210 56;
  • 11) 0.000 046 509 446 149 767 478 210 56 × 2 = 0 + 0.000 093 018 892 299 534 956 421 12;
  • 12) 0.000 093 018 892 299 534 956 421 12 × 2 = 0 + 0.000 186 037 784 599 069 912 842 24;
  • 13) 0.000 186 037 784 599 069 912 842 24 × 2 = 0 + 0.000 372 075 569 198 139 825 684 48;
  • 14) 0.000 372 075 569 198 139 825 684 48 × 2 = 0 + 0.000 744 151 138 396 279 651 368 96;
  • 15) 0.000 744 151 138 396 279 651 368 96 × 2 = 0 + 0.001 488 302 276 792 559 302 737 92;
  • 16) 0.001 488 302 276 792 559 302 737 92 × 2 = 0 + 0.002 976 604 553 585 118 605 475 84;
  • 17) 0.002 976 604 553 585 118 605 475 84 × 2 = 0 + 0.005 953 209 107 170 237 210 951 68;
  • 18) 0.005 953 209 107 170 237 210 951 68 × 2 = 0 + 0.011 906 418 214 340 474 421 903 36;
  • 19) 0.011 906 418 214 340 474 421 903 36 × 2 = 0 + 0.023 812 836 428 680 948 843 806 72;
  • 20) 0.023 812 836 428 680 948 843 806 72 × 2 = 0 + 0.047 625 672 857 361 897 687 613 44;
  • 21) 0.047 625 672 857 361 897 687 613 44 × 2 = 0 + 0.095 251 345 714 723 795 375 226 88;
  • 22) 0.095 251 345 714 723 795 375 226 88 × 2 = 0 + 0.190 502 691 429 447 590 750 453 76;
  • 23) 0.190 502 691 429 447 590 750 453 76 × 2 = 0 + 0.381 005 382 858 895 181 500 907 52;
  • 24) 0.381 005 382 858 895 181 500 907 52 × 2 = 0 + 0.762 010 765 717 790 363 001 815 04;
  • 25) 0.762 010 765 717 790 363 001 815 04 × 2 = 1 + 0.524 021 531 435 580 726 003 630 08;
  • 26) 0.524 021 531 435 580 726 003 630 08 × 2 = 1 + 0.048 043 062 871 161 452 007 260 16;
  • 27) 0.048 043 062 871 161 452 007 260 16 × 2 = 0 + 0.096 086 125 742 322 904 014 520 32;
  • 28) 0.096 086 125 742 322 904 014 520 32 × 2 = 0 + 0.192 172 251 484 645 808 029 040 64;
  • 29) 0.192 172 251 484 645 808 029 040 64 × 2 = 0 + 0.384 344 502 969 291 616 058 081 28;
  • 30) 0.384 344 502 969 291 616 058 081 28 × 2 = 0 + 0.768 689 005 938 583 232 116 162 56;
  • 31) 0.768 689 005 938 583 232 116 162 56 × 2 = 1 + 0.537 378 011 877 166 464 232 325 12;
  • 32) 0.537 378 011 877 166 464 232 325 12 × 2 = 1 + 0.074 756 023 754 332 928 464 650 24;
  • 33) 0.074 756 023 754 332 928 464 650 24 × 2 = 0 + 0.149 512 047 508 665 856 929 300 48;
  • 34) 0.149 512 047 508 665 856 929 300 48 × 2 = 0 + 0.299 024 095 017 331 713 858 600 96;
  • 35) 0.299 024 095 017 331 713 858 600 96 × 2 = 0 + 0.598 048 190 034 663 427 717 201 92;
  • 36) 0.598 048 190 034 663 427 717 201 92 × 2 = 1 + 0.196 096 380 069 326 855 434 403 84;
  • 37) 0.196 096 380 069 326 855 434 403 84 × 2 = 0 + 0.392 192 760 138 653 710 868 807 68;
  • 38) 0.392 192 760 138 653 710 868 807 68 × 2 = 0 + 0.784 385 520 277 307 421 737 615 36;
  • 39) 0.784 385 520 277 307 421 737 615 36 × 2 = 1 + 0.568 771 040 554 614 843 475 230 72;
  • 40) 0.568 771 040 554 614 843 475 230 72 × 2 = 1 + 0.137 542 081 109 229 686 950 461 44;
  • 41) 0.137 542 081 109 229 686 950 461 44 × 2 = 0 + 0.275 084 162 218 459 373 900 922 88;
  • 42) 0.275 084 162 218 459 373 900 922 88 × 2 = 0 + 0.550 168 324 436 918 747 801 845 76;
  • 43) 0.550 168 324 436 918 747 801 845 76 × 2 = 1 + 0.100 336 648 873 837 495 603 691 52;
  • 44) 0.100 336 648 873 837 495 603 691 52 × 2 = 0 + 0.200 673 297 747 674 991 207 383 04;
  • 45) 0.200 673 297 747 674 991 207 383 04 × 2 = 0 + 0.401 346 595 495 349 982 414 766 08;
  • 46) 0.401 346 595 495 349 982 414 766 08 × 2 = 0 + 0.802 693 190 990 699 964 829 532 16;
  • 47) 0.802 693 190 990 699 964 829 532 16 × 2 = 1 + 0.605 386 381 981 399 929 659 064 32;
  • 48) 0.605 386 381 981 399 929 659 064 32 × 2 = 1 + 0.210 772 763 962 799 859 318 128 64;
  • 49) 0.210 772 763 962 799 859 318 128 64 × 2 = 0 + 0.421 545 527 925 599 718 636 257 28;
  • 50) 0.421 545 527 925 599 718 636 257 28 × 2 = 0 + 0.843 091 055 851 199 437 272 514 56;
  • 51) 0.843 091 055 851 199 437 272 514 56 × 2 = 1 + 0.686 182 111 702 398 874 545 029 12;
  • 52) 0.686 182 111 702 398 874 545 029 12 × 2 = 1 + 0.372 364 223 404 797 749 090 058 24;
  • 53) 0.372 364 223 404 797 749 090 058 24 × 2 = 0 + 0.744 728 446 809 595 498 180 116 48;
  • 54) 0.744 728 446 809 595 498 180 116 48 × 2 = 1 + 0.489 456 893 619 190 996 360 232 96;
  • 55) 0.489 456 893 619 190 996 360 232 96 × 2 = 0 + 0.978 913 787 238 381 992 720 465 92;
  • 56) 0.978 913 787 238 381 992 720 465 92 × 2 = 1 + 0.957 827 574 476 763 985 440 931 84;
  • 57) 0.957 827 574 476 763 985 440 931 84 × 2 = 1 + 0.915 655 148 953 527 970 881 863 68;
  • 58) 0.915 655 148 953 527 970 881 863 68 × 2 = 1 + 0.831 310 297 907 055 941 763 727 36;
  • 59) 0.831 310 297 907 055 941 763 727 36 × 2 = 1 + 0.662 620 595 814 111 883 527 454 72;
  • 60) 0.662 620 595 814 111 883 527 454 72 × 2 = 1 + 0.325 241 191 628 223 767 054 909 44;
  • 61) 0.325 241 191 628 223 767 054 909 44 × 2 = 0 + 0.650 482 383 256 447 534 109 818 88;
  • 62) 0.650 482 383 256 447 534 109 818 88 × 2 = 1 + 0.300 964 766 512 895 068 219 637 76;
  • 63) 0.300 964 766 512 895 068 219 637 76 × 2 = 0 + 0.601 929 533 025 790 136 439 275 52;
  • 64) 0.601 929 533 025 790 136 439 275 52 × 2 = 1 + 0.203 859 066 051 580 272 878 551 04;
  • 65) 0.203 859 066 051 580 272 878 551 04 × 2 = 0 + 0.407 718 132 103 160 545 757 102 08;
  • 66) 0.407 718 132 103 160 545 757 102 08 × 2 = 0 + 0.815 436 264 206 321 091 514 204 16;
  • 67) 0.815 436 264 206 321 091 514 204 16 × 2 = 1 + 0.630 872 528 412 642 183 028 408 32;
  • 68) 0.630 872 528 412 642 183 028 408 32 × 2 = 1 + 0.261 745 056 825 284 366 056 816 64;
  • 69) 0.261 745 056 825 284 366 056 816 64 × 2 = 0 + 0.523 490 113 650 568 732 113 633 28;
  • 70) 0.523 490 113 650 568 732 113 633 28 × 2 = 1 + 0.046 980 227 301 137 464 227 266 56;
  • 71) 0.046 980 227 301 137 464 227 266 56 × 2 = 0 + 0.093 960 454 602 274 928 454 533 12;
  • 72) 0.093 960 454 602 274 928 454 533 12 × 2 = 0 + 0.187 920 909 204 549 856 909 066 24;
  • 73) 0.187 920 909 204 549 856 909 066 24 × 2 = 0 + 0.375 841 818 409 099 713 818 132 48;
  • 74) 0.375 841 818 409 099 713 818 132 48 × 2 = 0 + 0.751 683 636 818 199 427 636 264 96;
  • 75) 0.751 683 636 818 199 427 636 264 96 × 2 = 1 + 0.503 367 273 636 398 855 272 529 92;
  • 76) 0.503 367 273 636 398 855 272 529 92 × 2 = 1 + 0.006 734 547 272 797 710 545 059 84;
  • 77) 0.006 734 547 272 797 710 545 059 84 × 2 = 0 + 0.013 469 094 545 595 421 090 119 68;

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).


5. 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:


0.000 000 045 419 381 005 632 302 94(10) =

0.0000 0000 0000 0000 0000 0000 1100 0011 0001 0011 0010 0011 0011 0101 1111 0101 0011 0100 0011 0(2)

6. Positive number before normalization:

0.000 000 045 419 381 005 632 302 94(10) =

0.0000 0000 0000 0000 0000 0000 1100 0011 0001 0011 0010 0011 0011 0101 1111 0101 0011 0100 0011 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 045 419 381 005 632 302 94(10) =

0.0000 0000 0000 0000 0000 0000 1100 0011 0001 0011 0010 0011 0011 0101 1111 0101 0011 0100 0011 0(2) =

0.0000 0000 0000 0000 0000 0000 1100 0011 0001 0011 0010 0011 0011 0101 1111 0101 0011 0100 0011 0(2) × 20 =

1.1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110(2) × 2-25


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): -25

Mantissa (not normalized):
1.1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(11-1) - 1 =

(-25 + 1 023)(10) =

998(10)


10. 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;
  • 998 ÷ 2 = 499 + 0;
  • 499 ÷ 2 = 249 + 1;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

998(10) =

011 1110 0110(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110 =

1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1110 0110

Mantissa (52 bits) =
1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110


Decimal number -0.000 000 045 419 381 005 632 302 94 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0110 - 1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110

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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