-0.000 000 000 000 176 557 517 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 517(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
-0.000 000 000 000 176 557 517(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:

|-0.000 000 000 000 176 557 517| = 0.000 000 000 000 176 557 517


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: 0.000 000 000 000 176 557 517.

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) 0.000 000 000 000 176 557 517 × 2 = 0 + 0.000 000 000 000 353 115 034;
  • 2) 0.000 000 000 000 353 115 034 × 2 = 0 + 0.000 000 000 000 706 230 068;
  • 3) 0.000 000 000 000 706 230 068 × 2 = 0 + 0.000 000 000 001 412 460 136;
  • 4) 0.000 000 000 001 412 460 136 × 2 = 0 + 0.000 000 000 002 824 920 272;
  • 5) 0.000 000 000 002 824 920 272 × 2 = 0 + 0.000 000 000 005 649 840 544;
  • 6) 0.000 000 000 005 649 840 544 × 2 = 0 + 0.000 000 000 011 299 681 088;
  • 7) 0.000 000 000 011 299 681 088 × 2 = 0 + 0.000 000 000 022 599 362 176;
  • 8) 0.000 000 000 022 599 362 176 × 2 = 0 + 0.000 000 000 045 198 724 352;
  • 9) 0.000 000 000 045 198 724 352 × 2 = 0 + 0.000 000 000 090 397 448 704;
  • 10) 0.000 000 000 090 397 448 704 × 2 = 0 + 0.000 000 000 180 794 897 408;
  • 11) 0.000 000 000 180 794 897 408 × 2 = 0 + 0.000 000 000 361 589 794 816;
  • 12) 0.000 000 000 361 589 794 816 × 2 = 0 + 0.000 000 000 723 179 589 632;
  • 13) 0.000 000 000 723 179 589 632 × 2 = 0 + 0.000 000 001 446 359 179 264;
  • 14) 0.000 000 001 446 359 179 264 × 2 = 0 + 0.000 000 002 892 718 358 528;
  • 15) 0.000 000 002 892 718 358 528 × 2 = 0 + 0.000 000 005 785 436 717 056;
  • 16) 0.000 000 005 785 436 717 056 × 2 = 0 + 0.000 000 011 570 873 434 112;
  • 17) 0.000 000 011 570 873 434 112 × 2 = 0 + 0.000 000 023 141 746 868 224;
  • 18) 0.000 000 023 141 746 868 224 × 2 = 0 + 0.000 000 046 283 493 736 448;
  • 19) 0.000 000 046 283 493 736 448 × 2 = 0 + 0.000 000 092 566 987 472 896;
  • 20) 0.000 000 092 566 987 472 896 × 2 = 0 + 0.000 000 185 133 974 945 792;
  • 21) 0.000 000 185 133 974 945 792 × 2 = 0 + 0.000 000 370 267 949 891 584;
  • 22) 0.000 000 370 267 949 891 584 × 2 = 0 + 0.000 000 740 535 899 783 168;
  • 23) 0.000 000 740 535 899 783 168 × 2 = 0 + 0.000 001 481 071 799 566 336;
  • 24) 0.000 001 481 071 799 566 336 × 2 = 0 + 0.000 002 962 143 599 132 672;
  • 25) 0.000 002 962 143 599 132 672 × 2 = 0 + 0.000 005 924 287 198 265 344;
  • 26) 0.000 005 924 287 198 265 344 × 2 = 0 + 0.000 011 848 574 396 530 688;
  • 27) 0.000 011 848 574 396 530 688 × 2 = 0 + 0.000 023 697 148 793 061 376;
  • 28) 0.000 023 697 148 793 061 376 × 2 = 0 + 0.000 047 394 297 586 122 752;
  • 29) 0.000 047 394 297 586 122 752 × 2 = 0 + 0.000 094 788 595 172 245 504;
  • 30) 0.000 094 788 595 172 245 504 × 2 = 0 + 0.000 189 577 190 344 491 008;
  • 31) 0.000 189 577 190 344 491 008 × 2 = 0 + 0.000 379 154 380 688 982 016;
  • 32) 0.000 379 154 380 688 982 016 × 2 = 0 + 0.000 758 308 761 377 964 032;
  • 33) 0.000 758 308 761 377 964 032 × 2 = 0 + 0.001 516 617 522 755 928 064;
  • 34) 0.001 516 617 522 755 928 064 × 2 = 0 + 0.003 033 235 045 511 856 128;
  • 35) 0.003 033 235 045 511 856 128 × 2 = 0 + 0.006 066 470 091 023 712 256;
  • 36) 0.006 066 470 091 023 712 256 × 2 = 0 + 0.012 132 940 182 047 424 512;
  • 37) 0.012 132 940 182 047 424 512 × 2 = 0 + 0.024 265 880 364 094 849 024;
  • 38) 0.024 265 880 364 094 849 024 × 2 = 0 + 0.048 531 760 728 189 698 048;
  • 39) 0.048 531 760 728 189 698 048 × 2 = 0 + 0.097 063 521 456 379 396 096;
  • 40) 0.097 063 521 456 379 396 096 × 2 = 0 + 0.194 127 042 912 758 792 192;
  • 41) 0.194 127 042 912 758 792 192 × 2 = 0 + 0.388 254 085 825 517 584 384;
  • 42) 0.388 254 085 825 517 584 384 × 2 = 0 + 0.776 508 171 651 035 168 768;
  • 43) 0.776 508 171 651 035 168 768 × 2 = 1 + 0.553 016 343 302 070 337 536;
  • 44) 0.553 016 343 302 070 337 536 × 2 = 1 + 0.106 032 686 604 140 675 072;
  • 45) 0.106 032 686 604 140 675 072 × 2 = 0 + 0.212 065 373 208 281 350 144;
  • 46) 0.212 065 373 208 281 350 144 × 2 = 0 + 0.424 130 746 416 562 700 288;
  • 47) 0.424 130 746 416 562 700 288 × 2 = 0 + 0.848 261 492 833 125 400 576;
  • 48) 0.848 261 492 833 125 400 576 × 2 = 1 + 0.696 522 985 666 250 801 152;
  • 49) 0.696 522 985 666 250 801 152 × 2 = 1 + 0.393 045 971 332 501 602 304;
  • 50) 0.393 045 971 332 501 602 304 × 2 = 0 + 0.786 091 942 665 003 204 608;
  • 51) 0.786 091 942 665 003 204 608 × 2 = 1 + 0.572 183 885 330 006 409 216;
  • 52) 0.572 183 885 330 006 409 216 × 2 = 1 + 0.144 367 770 660 012 818 432;
  • 53) 0.144 367 770 660 012 818 432 × 2 = 0 + 0.288 735 541 320 025 636 864;
  • 54) 0.288 735 541 320 025 636 864 × 2 = 0 + 0.577 471 082 640 051 273 728;
  • 55) 0.577 471 082 640 051 273 728 × 2 = 1 + 0.154 942 165 280 102 547 456;
  • 56) 0.154 942 165 280 102 547 456 × 2 = 0 + 0.309 884 330 560 205 094 912;
  • 57) 0.309 884 330 560 205 094 912 × 2 = 0 + 0.619 768 661 120 410 189 824;
  • 58) 0.619 768 661 120 410 189 824 × 2 = 1 + 0.239 537 322 240 820 379 648;
  • 59) 0.239 537 322 240 820 379 648 × 2 = 0 + 0.479 074 644 481 640 759 296;
  • 60) 0.479 074 644 481 640 759 296 × 2 = 0 + 0.958 149 288 963 281 518 592;
  • 61) 0.958 149 288 963 281 518 592 × 2 = 1 + 0.916 298 577 926 563 037 184;
  • 62) 0.916 298 577 926 563 037 184 × 2 = 1 + 0.832 597 155 853 126 074 368;
  • 63) 0.832 597 155 853 126 074 368 × 2 = 1 + 0.665 194 311 706 252 148 736;
  • 64) 0.665 194 311 706 252 148 736 × 2 = 1 + 0.330 388 623 412 504 297 472;
  • 65) 0.330 388 623 412 504 297 472 × 2 = 0 + 0.660 777 246 825 008 594 944;
  • 66) 0.660 777 246 825 008 594 944 × 2 = 1 + 0.321 554 493 650 017 189 888;
  • 67) 0.321 554 493 650 017 189 888 × 2 = 0 + 0.643 108 987 300 034 379 776;
  • 68) 0.643 108 987 300 034 379 776 × 2 = 1 + 0.286 217 974 600 068 759 552;
  • 69) 0.286 217 974 600 068 759 552 × 2 = 0 + 0.572 435 949 200 137 519 104;
  • 70) 0.572 435 949 200 137 519 104 × 2 = 1 + 0.144 871 898 400 275 038 208;
  • 71) 0.144 871 898 400 275 038 208 × 2 = 0 + 0.289 743 796 800 550 076 416;
  • 72) 0.289 743 796 800 550 076 416 × 2 = 0 + 0.579 487 593 601 100 152 832;
  • 73) 0.579 487 593 601 100 152 832 × 2 = 1 + 0.158 975 187 202 200 305 664;
  • 74) 0.158 975 187 202 200 305 664 × 2 = 0 + 0.317 950 374 404 400 611 328;
  • 75) 0.317 950 374 404 400 611 328 × 2 = 0 + 0.635 900 748 808 801 222 656;
  • 76) 0.635 900 748 808 801 222 656 × 2 = 1 + 0.271 801 497 617 602 445 312;
  • 77) 0.271 801 497 617 602 445 312 × 2 = 0 + 0.543 602 995 235 204 890 624;
  • 78) 0.543 602 995 235 204 890 624 × 2 = 1 + 0.087 205 990 470 409 781 248;
  • 79) 0.087 205 990 470 409 781 248 × 2 = 0 + 0.174 411 980 940 819 562 496;
  • 80) 0.174 411 980 940 819 562 496 × 2 = 0 + 0.348 823 961 881 639 124 992;
  • 81) 0.348 823 961 881 639 124 992 × 2 = 0 + 0.697 647 923 763 278 249 984;
  • 82) 0.697 647 923 763 278 249 984 × 2 = 1 + 0.395 295 847 526 556 499 968;
  • 83) 0.395 295 847 526 556 499 968 × 2 = 0 + 0.790 591 695 053 112 999 936;
  • 84) 0.790 591 695 053 112 999 936 × 2 = 1 + 0.581 183 390 106 225 999 872;
  • 85) 0.581 183 390 106 225 999 872 × 2 = 1 + 0.162 366 780 212 451 999 744;
  • 86) 0.162 366 780 212 451 999 744 × 2 = 0 + 0.324 733 560 424 903 999 488;
  • 87) 0.324 733 560 424 903 999 488 × 2 = 0 + 0.649 467 120 849 807 998 976;
  • 88) 0.649 467 120 849 807 998 976 × 2 = 1 + 0.298 934 241 699 615 997 952;
  • 89) 0.298 934 241 699 615 997 952 × 2 = 0 + 0.597 868 483 399 231 995 904;
  • 90) 0.597 868 483 399 231 995 904 × 2 = 1 + 0.195 736 966 798 463 991 808;
  • 91) 0.195 736 966 798 463 991 808 × 2 = 0 + 0.391 473 933 596 927 983 616;
  • 92) 0.391 473 933 596 927 983 616 × 2 = 0 + 0.782 947 867 193 855 967 232;
  • 93) 0.782 947 867 193 855 967 232 × 2 = 1 + 0.565 895 734 387 711 934 464;
  • 94) 0.565 895 734 387 711 934 464 × 2 = 1 + 0.131 791 468 775 423 868 928;
  • 95) 0.131 791 468 775 423 868 928 × 2 = 0 + 0.263 582 937 550 847 737 856;

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:


0.000 000 000 000 176 557 517(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0101 0100 1001 0100 0101 1001 0100 110(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 517(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0101 0100 1001 0100 0101 1001 0100 110(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:


0.000 000 000 000 176 557 517(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0101 0100 1001 0100 0101 1001 0100 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0101 0100 1001 0100 0101 1001 0100 110(2) × 20 =

1.1000 1101 1001 0010 0111 1010 1010 0100 1010 0010 1100 1010 0110(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 1010 1010 0100 1010 0010 1100 1010 0110


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 1010 1010 0100 1010 0010 1100 1010 0110 =

1000 1101 1001 0010 0111 1010 1010 0100 1010 0010 1100 1010 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 1101 0100

Mantissa (52 bits) =
1000 1101 1001 0010 0111 1010 1010 0100 1010 0010 1100 1010 0110


Decimal number -0.000 000 000 000 176 557 517 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 1010 1010 0100 1010 0010 1100 1010 0110

Share

» Convert decimal -0.000 000 000 000 176 557 574 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 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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