-0.000 000 000 000 176 557 542 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 542(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
-0.000 000 000 000 176 557 542(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:

|-0.000 000 000 000 176 557 542| = 0.000 000 000 000 176 557 542


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: 0.000 000 000 000 176 557 542.

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) 0.000 000 000 000 176 557 542 × 2 = 0 + 0.000 000 000 000 353 115 084;
  • 2) 0.000 000 000 000 353 115 084 × 2 = 0 + 0.000 000 000 000 706 230 168;
  • 3) 0.000 000 000 000 706 230 168 × 2 = 0 + 0.000 000 000 001 412 460 336;
  • 4) 0.000 000 000 001 412 460 336 × 2 = 0 + 0.000 000 000 002 824 920 672;
  • 5) 0.000 000 000 002 824 920 672 × 2 = 0 + 0.000 000 000 005 649 841 344;
  • 6) 0.000 000 000 005 649 841 344 × 2 = 0 + 0.000 000 000 011 299 682 688;
  • 7) 0.000 000 000 011 299 682 688 × 2 = 0 + 0.000 000 000 022 599 365 376;
  • 8) 0.000 000 000 022 599 365 376 × 2 = 0 + 0.000 000 000 045 198 730 752;
  • 9) 0.000 000 000 045 198 730 752 × 2 = 0 + 0.000 000 000 090 397 461 504;
  • 10) 0.000 000 000 090 397 461 504 × 2 = 0 + 0.000 000 000 180 794 923 008;
  • 11) 0.000 000 000 180 794 923 008 × 2 = 0 + 0.000 000 000 361 589 846 016;
  • 12) 0.000 000 000 361 589 846 016 × 2 = 0 + 0.000 000 000 723 179 692 032;
  • 13) 0.000 000 000 723 179 692 032 × 2 = 0 + 0.000 000 001 446 359 384 064;
  • 14) 0.000 000 001 446 359 384 064 × 2 = 0 + 0.000 000 002 892 718 768 128;
  • 15) 0.000 000 002 892 718 768 128 × 2 = 0 + 0.000 000 005 785 437 536 256;
  • 16) 0.000 000 005 785 437 536 256 × 2 = 0 + 0.000 000 011 570 875 072 512;
  • 17) 0.000 000 011 570 875 072 512 × 2 = 0 + 0.000 000 023 141 750 145 024;
  • 18) 0.000 000 023 141 750 145 024 × 2 = 0 + 0.000 000 046 283 500 290 048;
  • 19) 0.000 000 046 283 500 290 048 × 2 = 0 + 0.000 000 092 567 000 580 096;
  • 20) 0.000 000 092 567 000 580 096 × 2 = 0 + 0.000 000 185 134 001 160 192;
  • 21) 0.000 000 185 134 001 160 192 × 2 = 0 + 0.000 000 370 268 002 320 384;
  • 22) 0.000 000 370 268 002 320 384 × 2 = 0 + 0.000 000 740 536 004 640 768;
  • 23) 0.000 000 740 536 004 640 768 × 2 = 0 + 0.000 001 481 072 009 281 536;
  • 24) 0.000 001 481 072 009 281 536 × 2 = 0 + 0.000 002 962 144 018 563 072;
  • 25) 0.000 002 962 144 018 563 072 × 2 = 0 + 0.000 005 924 288 037 126 144;
  • 26) 0.000 005 924 288 037 126 144 × 2 = 0 + 0.000 011 848 576 074 252 288;
  • 27) 0.000 011 848 576 074 252 288 × 2 = 0 + 0.000 023 697 152 148 504 576;
  • 28) 0.000 023 697 152 148 504 576 × 2 = 0 + 0.000 047 394 304 297 009 152;
  • 29) 0.000 047 394 304 297 009 152 × 2 = 0 + 0.000 094 788 608 594 018 304;
  • 30) 0.000 094 788 608 594 018 304 × 2 = 0 + 0.000 189 577 217 188 036 608;
  • 31) 0.000 189 577 217 188 036 608 × 2 = 0 + 0.000 379 154 434 376 073 216;
  • 32) 0.000 379 154 434 376 073 216 × 2 = 0 + 0.000 758 308 868 752 146 432;
  • 33) 0.000 758 308 868 752 146 432 × 2 = 0 + 0.001 516 617 737 504 292 864;
  • 34) 0.001 516 617 737 504 292 864 × 2 = 0 + 0.003 033 235 475 008 585 728;
  • 35) 0.003 033 235 475 008 585 728 × 2 = 0 + 0.006 066 470 950 017 171 456;
  • 36) 0.006 066 470 950 017 171 456 × 2 = 0 + 0.012 132 941 900 034 342 912;
  • 37) 0.012 132 941 900 034 342 912 × 2 = 0 + 0.024 265 883 800 068 685 824;
  • 38) 0.024 265 883 800 068 685 824 × 2 = 0 + 0.048 531 767 600 137 371 648;
  • 39) 0.048 531 767 600 137 371 648 × 2 = 0 + 0.097 063 535 200 274 743 296;
  • 40) 0.097 063 535 200 274 743 296 × 2 = 0 + 0.194 127 070 400 549 486 592;
  • 41) 0.194 127 070 400 549 486 592 × 2 = 0 + 0.388 254 140 801 098 973 184;
  • 42) 0.388 254 140 801 098 973 184 × 2 = 0 + 0.776 508 281 602 197 946 368;
  • 43) 0.776 508 281 602 197 946 368 × 2 = 1 + 0.553 016 563 204 395 892 736;
  • 44) 0.553 016 563 204 395 892 736 × 2 = 1 + 0.106 033 126 408 791 785 472;
  • 45) 0.106 033 126 408 791 785 472 × 2 = 0 + 0.212 066 252 817 583 570 944;
  • 46) 0.212 066 252 817 583 570 944 × 2 = 0 + 0.424 132 505 635 167 141 888;
  • 47) 0.424 132 505 635 167 141 888 × 2 = 0 + 0.848 265 011 270 334 283 776;
  • 48) 0.848 265 011 270 334 283 776 × 2 = 1 + 0.696 530 022 540 668 567 552;
  • 49) 0.696 530 022 540 668 567 552 × 2 = 1 + 0.393 060 045 081 337 135 104;
  • 50) 0.393 060 045 081 337 135 104 × 2 = 0 + 0.786 120 090 162 674 270 208;
  • 51) 0.786 120 090 162 674 270 208 × 2 = 1 + 0.572 240 180 325 348 540 416;
  • 52) 0.572 240 180 325 348 540 416 × 2 = 1 + 0.144 480 360 650 697 080 832;
  • 53) 0.144 480 360 650 697 080 832 × 2 = 0 + 0.288 960 721 301 394 161 664;
  • 54) 0.288 960 721 301 394 161 664 × 2 = 0 + 0.577 921 442 602 788 323 328;
  • 55) 0.577 921 442 602 788 323 328 × 2 = 1 + 0.155 842 885 205 576 646 656;
  • 56) 0.155 842 885 205 576 646 656 × 2 = 0 + 0.311 685 770 411 153 293 312;
  • 57) 0.311 685 770 411 153 293 312 × 2 = 0 + 0.623 371 540 822 306 586 624;
  • 58) 0.623 371 540 822 306 586 624 × 2 = 1 + 0.246 743 081 644 613 173 248;
  • 59) 0.246 743 081 644 613 173 248 × 2 = 0 + 0.493 486 163 289 226 346 496;
  • 60) 0.493 486 163 289 226 346 496 × 2 = 0 + 0.986 972 326 578 452 692 992;
  • 61) 0.986 972 326 578 452 692 992 × 2 = 1 + 0.973 944 653 156 905 385 984;
  • 62) 0.973 944 653 156 905 385 984 × 2 = 1 + 0.947 889 306 313 810 771 968;
  • 63) 0.947 889 306 313 810 771 968 × 2 = 1 + 0.895 778 612 627 621 543 936;
  • 64) 0.895 778 612 627 621 543 936 × 2 = 1 + 0.791 557 225 255 243 087 872;
  • 65) 0.791 557 225 255 243 087 872 × 2 = 1 + 0.583 114 450 510 486 175 744;
  • 66) 0.583 114 450 510 486 175 744 × 2 = 1 + 0.166 228 901 020 972 351 488;
  • 67) 0.166 228 901 020 972 351 488 × 2 = 0 + 0.332 457 802 041 944 702 976;
  • 68) 0.332 457 802 041 944 702 976 × 2 = 0 + 0.664 915 604 083 889 405 952;
  • 69) 0.664 915 604 083 889 405 952 × 2 = 1 + 0.329 831 208 167 778 811 904;
  • 70) 0.329 831 208 167 778 811 904 × 2 = 0 + 0.659 662 416 335 557 623 808;
  • 71) 0.659 662 416 335 557 623 808 × 2 = 1 + 0.319 324 832 671 115 247 616;
  • 72) 0.319 324 832 671 115 247 616 × 2 = 0 + 0.638 649 665 342 230 495 232;
  • 73) 0.638 649 665 342 230 495 232 × 2 = 1 + 0.277 299 330 684 460 990 464;
  • 74) 0.277 299 330 684 460 990 464 × 2 = 0 + 0.554 598 661 368 921 980 928;
  • 75) 0.554 598 661 368 921 980 928 × 2 = 1 + 0.109 197 322 737 843 961 856;
  • 76) 0.109 197 322 737 843 961 856 × 2 = 0 + 0.218 394 645 475 687 923 712;
  • 77) 0.218 394 645 475 687 923 712 × 2 = 0 + 0.436 789 290 951 375 847 424;
  • 78) 0.436 789 290 951 375 847 424 × 2 = 0 + 0.873 578 581 902 751 694 848;
  • 79) 0.873 578 581 902 751 694 848 × 2 = 1 + 0.747 157 163 805 503 389 696;
  • 80) 0.747 157 163 805 503 389 696 × 2 = 1 + 0.494 314 327 611 006 779 392;
  • 81) 0.494 314 327 611 006 779 392 × 2 = 0 + 0.988 628 655 222 013 558 784;
  • 82) 0.988 628 655 222 013 558 784 × 2 = 1 + 0.977 257 310 444 027 117 568;
  • 83) 0.977 257 310 444 027 117 568 × 2 = 1 + 0.954 514 620 888 054 235 136;
  • 84) 0.954 514 620 888 054 235 136 × 2 = 1 + 0.909 029 241 776 108 470 272;
  • 85) 0.909 029 241 776 108 470 272 × 2 = 1 + 0.818 058 483 552 216 940 544;
  • 86) 0.818 058 483 552 216 940 544 × 2 = 1 + 0.636 116 967 104 433 881 088;
  • 87) 0.636 116 967 104 433 881 088 × 2 = 1 + 0.272 233 934 208 867 762 176;
  • 88) 0.272 233 934 208 867 762 176 × 2 = 0 + 0.544 467 868 417 735 524 352;
  • 89) 0.544 467 868 417 735 524 352 × 2 = 1 + 0.088 935 736 835 471 048 704;
  • 90) 0.088 935 736 835 471 048 704 × 2 = 0 + 0.177 871 473 670 942 097 408;
  • 91) 0.177 871 473 670 942 097 408 × 2 = 0 + 0.355 742 947 341 884 194 816;
  • 92) 0.355 742 947 341 884 194 816 × 2 = 0 + 0.711 485 894 683 768 389 632;
  • 93) 0.711 485 894 683 768 389 632 × 2 = 1 + 0.422 971 789 367 536 779 264;
  • 94) 0.422 971 789 367 536 779 264 × 2 = 0 + 0.845 943 578 735 073 558 528;
  • 95) 0.845 943 578 735 073 558 528 × 2 = 1 + 0.691 887 157 470 147 117 056;

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:


0.000 000 000 000 176 557 542(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 1100 1010 1010 0011 0111 1110 1000 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 542(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 1100 1010 1010 0011 0111 1110 1000 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 557 542(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 1100 1010 1010 0011 0111 1110 1000 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 1100 1010 1010 0011 0111 1110 1000 101(2) × 20 =

1.1000 1101 1001 0010 0111 1110 0101 0101 0001 1011 1111 0100 0101(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0010 0111 1110 0101 0101 0001 1011 1111 0100 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 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) =

980(10) =

011 1101 0100(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 1101 1001 0010 0111 1110 0101 0101 0001 1011 1111 0100 0101 =

1000 1101 1001 0010 0111 1110 0101 0101 0001 1011 1111 0100 0101


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 0100

Mantissa (52 bits) =
1000 1101 1001 0010 0111 1110 0101 0101 0001 1011 1111 0100 0101


Decimal number -0.000 000 000 000 176 557 542 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 1110 0101 0101 0001 1011 1111 0100 0101

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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