-0.000 000 045 419 381 005 632 302 95 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 045 419 381 005 632 302 95(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 95(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 95| = 0.000 000 045 419 381 005 632 302 95


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 95.

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 95 × 2 = 0 + 0.000 000 090 838 762 011 264 605 9;
  • 2) 0.000 000 090 838 762 011 264 605 9 × 2 = 0 + 0.000 000 181 677 524 022 529 211 8;
  • 3) 0.000 000 181 677 524 022 529 211 8 × 2 = 0 + 0.000 000 363 355 048 045 058 423 6;
  • 4) 0.000 000 363 355 048 045 058 423 6 × 2 = 0 + 0.000 000 726 710 096 090 116 847 2;
  • 5) 0.000 000 726 710 096 090 116 847 2 × 2 = 0 + 0.000 001 453 420 192 180 233 694 4;
  • 6) 0.000 001 453 420 192 180 233 694 4 × 2 = 0 + 0.000 002 906 840 384 360 467 388 8;
  • 7) 0.000 002 906 840 384 360 467 388 8 × 2 = 0 + 0.000 005 813 680 768 720 934 777 6;
  • 8) 0.000 005 813 680 768 720 934 777 6 × 2 = 0 + 0.000 011 627 361 537 441 869 555 2;
  • 9) 0.000 011 627 361 537 441 869 555 2 × 2 = 0 + 0.000 023 254 723 074 883 739 110 4;
  • 10) 0.000 023 254 723 074 883 739 110 4 × 2 = 0 + 0.000 046 509 446 149 767 478 220 8;
  • 11) 0.000 046 509 446 149 767 478 220 8 × 2 = 0 + 0.000 093 018 892 299 534 956 441 6;
  • 12) 0.000 093 018 892 299 534 956 441 6 × 2 = 0 + 0.000 186 037 784 599 069 912 883 2;
  • 13) 0.000 186 037 784 599 069 912 883 2 × 2 = 0 + 0.000 372 075 569 198 139 825 766 4;
  • 14) 0.000 372 075 569 198 139 825 766 4 × 2 = 0 + 0.000 744 151 138 396 279 651 532 8;
  • 15) 0.000 744 151 138 396 279 651 532 8 × 2 = 0 + 0.001 488 302 276 792 559 303 065 6;
  • 16) 0.001 488 302 276 792 559 303 065 6 × 2 = 0 + 0.002 976 604 553 585 118 606 131 2;
  • 17) 0.002 976 604 553 585 118 606 131 2 × 2 = 0 + 0.005 953 209 107 170 237 212 262 4;
  • 18) 0.005 953 209 107 170 237 212 262 4 × 2 = 0 + 0.011 906 418 214 340 474 424 524 8;
  • 19) 0.011 906 418 214 340 474 424 524 8 × 2 = 0 + 0.023 812 836 428 680 948 849 049 6;
  • 20) 0.023 812 836 428 680 948 849 049 6 × 2 = 0 + 0.047 625 672 857 361 897 698 099 2;
  • 21) 0.047 625 672 857 361 897 698 099 2 × 2 = 0 + 0.095 251 345 714 723 795 396 198 4;
  • 22) 0.095 251 345 714 723 795 396 198 4 × 2 = 0 + 0.190 502 691 429 447 590 792 396 8;
  • 23) 0.190 502 691 429 447 590 792 396 8 × 2 = 0 + 0.381 005 382 858 895 181 584 793 6;
  • 24) 0.381 005 382 858 895 181 584 793 6 × 2 = 0 + 0.762 010 765 717 790 363 169 587 2;
  • 25) 0.762 010 765 717 790 363 169 587 2 × 2 = 1 + 0.524 021 531 435 580 726 339 174 4;
  • 26) 0.524 021 531 435 580 726 339 174 4 × 2 = 1 + 0.048 043 062 871 161 452 678 348 8;
  • 27) 0.048 043 062 871 161 452 678 348 8 × 2 = 0 + 0.096 086 125 742 322 905 356 697 6;
  • 28) 0.096 086 125 742 322 905 356 697 6 × 2 = 0 + 0.192 172 251 484 645 810 713 395 2;
  • 29) 0.192 172 251 484 645 810 713 395 2 × 2 = 0 + 0.384 344 502 969 291 621 426 790 4;
  • 30) 0.384 344 502 969 291 621 426 790 4 × 2 = 0 + 0.768 689 005 938 583 242 853 580 8;
  • 31) 0.768 689 005 938 583 242 853 580 8 × 2 = 1 + 0.537 378 011 877 166 485 707 161 6;
  • 32) 0.537 378 011 877 166 485 707 161 6 × 2 = 1 + 0.074 756 023 754 332 971 414 323 2;
  • 33) 0.074 756 023 754 332 971 414 323 2 × 2 = 0 + 0.149 512 047 508 665 942 828 646 4;
  • 34) 0.149 512 047 508 665 942 828 646 4 × 2 = 0 + 0.299 024 095 017 331 885 657 292 8;
  • 35) 0.299 024 095 017 331 885 657 292 8 × 2 = 0 + 0.598 048 190 034 663 771 314 585 6;
  • 36) 0.598 048 190 034 663 771 314 585 6 × 2 = 1 + 0.196 096 380 069 327 542 629 171 2;
  • 37) 0.196 096 380 069 327 542 629 171 2 × 2 = 0 + 0.392 192 760 138 655 085 258 342 4;
  • 38) 0.392 192 760 138 655 085 258 342 4 × 2 = 0 + 0.784 385 520 277 310 170 516 684 8;
  • 39) 0.784 385 520 277 310 170 516 684 8 × 2 = 1 + 0.568 771 040 554 620 341 033 369 6;
  • 40) 0.568 771 040 554 620 341 033 369 6 × 2 = 1 + 0.137 542 081 109 240 682 066 739 2;
  • 41) 0.137 542 081 109 240 682 066 739 2 × 2 = 0 + 0.275 084 162 218 481 364 133 478 4;
  • 42) 0.275 084 162 218 481 364 133 478 4 × 2 = 0 + 0.550 168 324 436 962 728 266 956 8;
  • 43) 0.550 168 324 436 962 728 266 956 8 × 2 = 1 + 0.100 336 648 873 925 456 533 913 6;
  • 44) 0.100 336 648 873 925 456 533 913 6 × 2 = 0 + 0.200 673 297 747 850 913 067 827 2;
  • 45) 0.200 673 297 747 850 913 067 827 2 × 2 = 0 + 0.401 346 595 495 701 826 135 654 4;
  • 46) 0.401 346 595 495 701 826 135 654 4 × 2 = 0 + 0.802 693 190 991 403 652 271 308 8;
  • 47) 0.802 693 190 991 403 652 271 308 8 × 2 = 1 + 0.605 386 381 982 807 304 542 617 6;
  • 48) 0.605 386 381 982 807 304 542 617 6 × 2 = 1 + 0.210 772 763 965 614 609 085 235 2;
  • 49) 0.210 772 763 965 614 609 085 235 2 × 2 = 0 + 0.421 545 527 931 229 218 170 470 4;
  • 50) 0.421 545 527 931 229 218 170 470 4 × 2 = 0 + 0.843 091 055 862 458 436 340 940 8;
  • 51) 0.843 091 055 862 458 436 340 940 8 × 2 = 1 + 0.686 182 111 724 916 872 681 881 6;
  • 52) 0.686 182 111 724 916 872 681 881 6 × 2 = 1 + 0.372 364 223 449 833 745 363 763 2;
  • 53) 0.372 364 223 449 833 745 363 763 2 × 2 = 0 + 0.744 728 446 899 667 490 727 526 4;
  • 54) 0.744 728 446 899 667 490 727 526 4 × 2 = 1 + 0.489 456 893 799 334 981 455 052 8;
  • 55) 0.489 456 893 799 334 981 455 052 8 × 2 = 0 + 0.978 913 787 598 669 962 910 105 6;
  • 56) 0.978 913 787 598 669 962 910 105 6 × 2 = 1 + 0.957 827 575 197 339 925 820 211 2;
  • 57) 0.957 827 575 197 339 925 820 211 2 × 2 = 1 + 0.915 655 150 394 679 851 640 422 4;
  • 58) 0.915 655 150 394 679 851 640 422 4 × 2 = 1 + 0.831 310 300 789 359 703 280 844 8;
  • 59) 0.831 310 300 789 359 703 280 844 8 × 2 = 1 + 0.662 620 601 578 719 406 561 689 6;
  • 60) 0.662 620 601 578 719 406 561 689 6 × 2 = 1 + 0.325 241 203 157 438 813 123 379 2;
  • 61) 0.325 241 203 157 438 813 123 379 2 × 2 = 0 + 0.650 482 406 314 877 626 246 758 4;
  • 62) 0.650 482 406 314 877 626 246 758 4 × 2 = 1 + 0.300 964 812 629 755 252 493 516 8;
  • 63) 0.300 964 812 629 755 252 493 516 8 × 2 = 0 + 0.601 929 625 259 510 504 987 033 6;
  • 64) 0.601 929 625 259 510 504 987 033 6 × 2 = 1 + 0.203 859 250 519 021 009 974 067 2;
  • 65) 0.203 859 250 519 021 009 974 067 2 × 2 = 0 + 0.407 718 501 038 042 019 948 134 4;
  • 66) 0.407 718 501 038 042 019 948 134 4 × 2 = 0 + 0.815 437 002 076 084 039 896 268 8;
  • 67) 0.815 437 002 076 084 039 896 268 8 × 2 = 1 + 0.630 874 004 152 168 079 792 537 6;
  • 68) 0.630 874 004 152 168 079 792 537 6 × 2 = 1 + 0.261 748 008 304 336 159 585 075 2;
  • 69) 0.261 748 008 304 336 159 585 075 2 × 2 = 0 + 0.523 496 016 608 672 319 170 150 4;
  • 70) 0.523 496 016 608 672 319 170 150 4 × 2 = 1 + 0.046 992 033 217 344 638 340 300 8;
  • 71) 0.046 992 033 217 344 638 340 300 8 × 2 = 0 + 0.093 984 066 434 689 276 680 601 6;
  • 72) 0.093 984 066 434 689 276 680 601 6 × 2 = 0 + 0.187 968 132 869 378 553 361 203 2;
  • 73) 0.187 968 132 869 378 553 361 203 2 × 2 = 0 + 0.375 936 265 738 757 106 722 406 4;
  • 74) 0.375 936 265 738 757 106 722 406 4 × 2 = 0 + 0.751 872 531 477 514 213 444 812 8;
  • 75) 0.751 872 531 477 514 213 444 812 8 × 2 = 1 + 0.503 745 062 955 028 426 889 625 6;
  • 76) 0.503 745 062 955 028 426 889 625 6 × 2 = 1 + 0.007 490 125 910 056 853 779 251 2;
  • 77) 0.007 490 125 910 056 853 779 251 2 × 2 = 0 + 0.014 980 251 820 113 707 558 502 4;

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 95(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 95(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 95(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 95 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

Share

» Convert decimal -0.000 000 045 419 381 005 632 303 53 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 minutes

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