-0.000 000 000 000 293 886 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 293 886 1(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
-0.000 000 000 000 293 886 1(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:

|-0.000 000 000 000 293 886 1| = 0.000 000 000 000 293 886 1


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: 0.000 000 000 000 293 886 1.

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) 0.000 000 000 000 293 886 1 × 2 = 0 + 0.000 000 000 000 587 772 2;
  • 2) 0.000 000 000 000 587 772 2 × 2 = 0 + 0.000 000 000 001 175 544 4;
  • 3) 0.000 000 000 001 175 544 4 × 2 = 0 + 0.000 000 000 002 351 088 8;
  • 4) 0.000 000 000 002 351 088 8 × 2 = 0 + 0.000 000 000 004 702 177 6;
  • 5) 0.000 000 000 004 702 177 6 × 2 = 0 + 0.000 000 000 009 404 355 2;
  • 6) 0.000 000 000 009 404 355 2 × 2 = 0 + 0.000 000 000 018 808 710 4;
  • 7) 0.000 000 000 018 808 710 4 × 2 = 0 + 0.000 000 000 037 617 420 8;
  • 8) 0.000 000 000 037 617 420 8 × 2 = 0 + 0.000 000 000 075 234 841 6;
  • 9) 0.000 000 000 075 234 841 6 × 2 = 0 + 0.000 000 000 150 469 683 2;
  • 10) 0.000 000 000 150 469 683 2 × 2 = 0 + 0.000 000 000 300 939 366 4;
  • 11) 0.000 000 000 300 939 366 4 × 2 = 0 + 0.000 000 000 601 878 732 8;
  • 12) 0.000 000 000 601 878 732 8 × 2 = 0 + 0.000 000 001 203 757 465 6;
  • 13) 0.000 000 001 203 757 465 6 × 2 = 0 + 0.000 000 002 407 514 931 2;
  • 14) 0.000 000 002 407 514 931 2 × 2 = 0 + 0.000 000 004 815 029 862 4;
  • 15) 0.000 000 004 815 029 862 4 × 2 = 0 + 0.000 000 009 630 059 724 8;
  • 16) 0.000 000 009 630 059 724 8 × 2 = 0 + 0.000 000 019 260 119 449 6;
  • 17) 0.000 000 019 260 119 449 6 × 2 = 0 + 0.000 000 038 520 238 899 2;
  • 18) 0.000 000 038 520 238 899 2 × 2 = 0 + 0.000 000 077 040 477 798 4;
  • 19) 0.000 000 077 040 477 798 4 × 2 = 0 + 0.000 000 154 080 955 596 8;
  • 20) 0.000 000 154 080 955 596 8 × 2 = 0 + 0.000 000 308 161 911 193 6;
  • 21) 0.000 000 308 161 911 193 6 × 2 = 0 + 0.000 000 616 323 822 387 2;
  • 22) 0.000 000 616 323 822 387 2 × 2 = 0 + 0.000 001 232 647 644 774 4;
  • 23) 0.000 001 232 647 644 774 4 × 2 = 0 + 0.000 002 465 295 289 548 8;
  • 24) 0.000 002 465 295 289 548 8 × 2 = 0 + 0.000 004 930 590 579 097 6;
  • 25) 0.000 004 930 590 579 097 6 × 2 = 0 + 0.000 009 861 181 158 195 2;
  • 26) 0.000 009 861 181 158 195 2 × 2 = 0 + 0.000 019 722 362 316 390 4;
  • 27) 0.000 019 722 362 316 390 4 × 2 = 0 + 0.000 039 444 724 632 780 8;
  • 28) 0.000 039 444 724 632 780 8 × 2 = 0 + 0.000 078 889 449 265 561 6;
  • 29) 0.000 078 889 449 265 561 6 × 2 = 0 + 0.000 157 778 898 531 123 2;
  • 30) 0.000 157 778 898 531 123 2 × 2 = 0 + 0.000 315 557 797 062 246 4;
  • 31) 0.000 315 557 797 062 246 4 × 2 = 0 + 0.000 631 115 594 124 492 8;
  • 32) 0.000 631 115 594 124 492 8 × 2 = 0 + 0.001 262 231 188 248 985 6;
  • 33) 0.001 262 231 188 248 985 6 × 2 = 0 + 0.002 524 462 376 497 971 2;
  • 34) 0.002 524 462 376 497 971 2 × 2 = 0 + 0.005 048 924 752 995 942 4;
  • 35) 0.005 048 924 752 995 942 4 × 2 = 0 + 0.010 097 849 505 991 884 8;
  • 36) 0.010 097 849 505 991 884 8 × 2 = 0 + 0.020 195 699 011 983 769 6;
  • 37) 0.020 195 699 011 983 769 6 × 2 = 0 + 0.040 391 398 023 967 539 2;
  • 38) 0.040 391 398 023 967 539 2 × 2 = 0 + 0.080 782 796 047 935 078 4;
  • 39) 0.080 782 796 047 935 078 4 × 2 = 0 + 0.161 565 592 095 870 156 8;
  • 40) 0.161 565 592 095 870 156 8 × 2 = 0 + 0.323 131 184 191 740 313 6;
  • 41) 0.323 131 184 191 740 313 6 × 2 = 0 + 0.646 262 368 383 480 627 2;
  • 42) 0.646 262 368 383 480 627 2 × 2 = 1 + 0.292 524 736 766 961 254 4;
  • 43) 0.292 524 736 766 961 254 4 × 2 = 0 + 0.585 049 473 533 922 508 8;
  • 44) 0.585 049 473 533 922 508 8 × 2 = 1 + 0.170 098 947 067 845 017 6;
  • 45) 0.170 098 947 067 845 017 6 × 2 = 0 + 0.340 197 894 135 690 035 2;
  • 46) 0.340 197 894 135 690 035 2 × 2 = 0 + 0.680 395 788 271 380 070 4;
  • 47) 0.680 395 788 271 380 070 4 × 2 = 1 + 0.360 791 576 542 760 140 8;
  • 48) 0.360 791 576 542 760 140 8 × 2 = 0 + 0.721 583 153 085 520 281 6;
  • 49) 0.721 583 153 085 520 281 6 × 2 = 1 + 0.443 166 306 171 040 563 2;
  • 50) 0.443 166 306 171 040 563 2 × 2 = 0 + 0.886 332 612 342 081 126 4;
  • 51) 0.886 332 612 342 081 126 4 × 2 = 1 + 0.772 665 224 684 162 252 8;
  • 52) 0.772 665 224 684 162 252 8 × 2 = 1 + 0.545 330 449 368 324 505 6;
  • 53) 0.545 330 449 368 324 505 6 × 2 = 1 + 0.090 660 898 736 649 011 2;
  • 54) 0.090 660 898 736 649 011 2 × 2 = 0 + 0.181 321 797 473 298 022 4;
  • 55) 0.181 321 797 473 298 022 4 × 2 = 0 + 0.362 643 594 946 596 044 8;
  • 56) 0.362 643 594 946 596 044 8 × 2 = 0 + 0.725 287 189 893 192 089 6;
  • 57) 0.725 287 189 893 192 089 6 × 2 = 1 + 0.450 574 379 786 384 179 2;
  • 58) 0.450 574 379 786 384 179 2 × 2 = 0 + 0.901 148 759 572 768 358 4;
  • 59) 0.901 148 759 572 768 358 4 × 2 = 1 + 0.802 297 519 145 536 716 8;
  • 60) 0.802 297 519 145 536 716 8 × 2 = 1 + 0.604 595 038 291 073 433 6;
  • 61) 0.604 595 038 291 073 433 6 × 2 = 1 + 0.209 190 076 582 146 867 2;
  • 62) 0.209 190 076 582 146 867 2 × 2 = 0 + 0.418 380 153 164 293 734 4;
  • 63) 0.418 380 153 164 293 734 4 × 2 = 0 + 0.836 760 306 328 587 468 8;
  • 64) 0.836 760 306 328 587 468 8 × 2 = 1 + 0.673 520 612 657 174 937 6;
  • 65) 0.673 520 612 657 174 937 6 × 2 = 1 + 0.347 041 225 314 349 875 2;
  • 66) 0.347 041 225 314 349 875 2 × 2 = 0 + 0.694 082 450 628 699 750 4;
  • 67) 0.694 082 450 628 699 750 4 × 2 = 1 + 0.388 164 901 257 399 500 8;
  • 68) 0.388 164 901 257 399 500 8 × 2 = 0 + 0.776 329 802 514 799 001 6;
  • 69) 0.776 329 802 514 799 001 6 × 2 = 1 + 0.552 659 605 029 598 003 2;
  • 70) 0.552 659 605 029 598 003 2 × 2 = 1 + 0.105 319 210 059 196 006 4;
  • 71) 0.105 319 210 059 196 006 4 × 2 = 0 + 0.210 638 420 118 392 012 8;
  • 72) 0.210 638 420 118 392 012 8 × 2 = 0 + 0.421 276 840 236 784 025 6;
  • 73) 0.421 276 840 236 784 025 6 × 2 = 0 + 0.842 553 680 473 568 051 2;
  • 74) 0.842 553 680 473 568 051 2 × 2 = 1 + 0.685 107 360 947 136 102 4;
  • 75) 0.685 107 360 947 136 102 4 × 2 = 1 + 0.370 214 721 894 272 204 8;
  • 76) 0.370 214 721 894 272 204 8 × 2 = 0 + 0.740 429 443 788 544 409 6;
  • 77) 0.740 429 443 788 544 409 6 × 2 = 1 + 0.480 858 887 577 088 819 2;
  • 78) 0.480 858 887 577 088 819 2 × 2 = 0 + 0.961 717 775 154 177 638 4;
  • 79) 0.961 717 775 154 177 638 4 × 2 = 1 + 0.923 435 550 308 355 276 8;
  • 80) 0.923 435 550 308 355 276 8 × 2 = 1 + 0.846 871 100 616 710 553 6;
  • 81) 0.846 871 100 616 710 553 6 × 2 = 1 + 0.693 742 201 233 421 107 2;
  • 82) 0.693 742 201 233 421 107 2 × 2 = 1 + 0.387 484 402 466 842 214 4;
  • 83) 0.387 484 402 466 842 214 4 × 2 = 0 + 0.774 968 804 933 684 428 8;
  • 84) 0.774 968 804 933 684 428 8 × 2 = 1 + 0.549 937 609 867 368 857 6;
  • 85) 0.549 937 609 867 368 857 6 × 2 = 1 + 0.099 875 219 734 737 715 2;
  • 86) 0.099 875 219 734 737 715 2 × 2 = 0 + 0.199 750 439 469 475 430 4;
  • 87) 0.199 750 439 469 475 430 4 × 2 = 0 + 0.399 500 878 938 950 860 8;
  • 88) 0.399 500 878 938 950 860 8 × 2 = 0 + 0.799 001 757 877 901 721 6;
  • 89) 0.799 001 757 877 901 721 6 × 2 = 1 + 0.598 003 515 755 803 443 2;
  • 90) 0.598 003 515 755 803 443 2 × 2 = 1 + 0.196 007 031 511 606 886 4;
  • 91) 0.196 007 031 511 606 886 4 × 2 = 0 + 0.392 014 063 023 213 772 8;
  • 92) 0.392 014 063 023 213 772 8 × 2 = 0 + 0.784 028 126 046 427 545 6;
  • 93) 0.784 028 126 046 427 545 6 × 2 = 1 + 0.568 056 252 092 855 091 2;
  • 94) 0.568 056 252 092 855 091 2 × 2 = 1 + 0.136 112 504 185 710 182 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:


0.000 000 000 000 293 886 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1000 1011 1001 1010 1100 0110 1011 1101 1000 1100 11(2)

6. Positive number before normalization:

0.000 000 000 000 293 886 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1000 1011 1001 1010 1100 0110 1011 1101 1000 1100 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 293 886 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1000 1011 1001 1010 1100 0110 1011 1101 1000 1100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1000 1011 1001 1010 1100 0110 1011 1101 1000 1100 11(2) × 20 =

1.0100 1010 1110 0010 1110 0110 1011 0001 1010 1111 0110 0011 0011(2) × 2-42


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): -42

Mantissa (not normalized):
1.0100 1010 1110 0010 1110 0110 1011 0001 1010 1111 0110 0011 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(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;
  • 981 ÷ 2 = 490 + 1;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

981(10) =

011 1101 0101(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. 0100 1010 1110 0010 1110 0110 1011 0001 1010 1111 0110 0011 0011 =

0100 1010 1110 0010 1110 0110 1011 0001 1010 1111 0110 0011 0011


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 1101 0101

Mantissa (52 bits) =
0100 1010 1110 0010 1110 0110 1011 0001 1010 1111 0110 0011 0011


Decimal number -0.000 000 000 000 293 886 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0101 - 0100 1010 1110 0010 1110 0110 1011 0001 1010 1111 0110 0011 0011

» Convert decimal -0.000 000 000 000 293 885 4 to 64 bit double precision IEEE 754 binary floating point representation standard

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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