-0.000 000 000 000 176 557 459 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 459(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
-0.000 000 000 000 176 557 459(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:

|-0.000 000 000 000 176 557 459| = 0.000 000 000 000 176 557 459


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: 0.000 000 000 000 176 557 459.

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) 0.000 000 000 000 176 557 459 × 2 = 0 + 0.000 000 000 000 353 114 918;
  • 2) 0.000 000 000 000 353 114 918 × 2 = 0 + 0.000 000 000 000 706 229 836;
  • 3) 0.000 000 000 000 706 229 836 × 2 = 0 + 0.000 000 000 001 412 459 672;
  • 4) 0.000 000 000 001 412 459 672 × 2 = 0 + 0.000 000 000 002 824 919 344;
  • 5) 0.000 000 000 002 824 919 344 × 2 = 0 + 0.000 000 000 005 649 838 688;
  • 6) 0.000 000 000 005 649 838 688 × 2 = 0 + 0.000 000 000 011 299 677 376;
  • 7) 0.000 000 000 011 299 677 376 × 2 = 0 + 0.000 000 000 022 599 354 752;
  • 8) 0.000 000 000 022 599 354 752 × 2 = 0 + 0.000 000 000 045 198 709 504;
  • 9) 0.000 000 000 045 198 709 504 × 2 = 0 + 0.000 000 000 090 397 419 008;
  • 10) 0.000 000 000 090 397 419 008 × 2 = 0 + 0.000 000 000 180 794 838 016;
  • 11) 0.000 000 000 180 794 838 016 × 2 = 0 + 0.000 000 000 361 589 676 032;
  • 12) 0.000 000 000 361 589 676 032 × 2 = 0 + 0.000 000 000 723 179 352 064;
  • 13) 0.000 000 000 723 179 352 064 × 2 = 0 + 0.000 000 001 446 358 704 128;
  • 14) 0.000 000 001 446 358 704 128 × 2 = 0 + 0.000 000 002 892 717 408 256;
  • 15) 0.000 000 002 892 717 408 256 × 2 = 0 + 0.000 000 005 785 434 816 512;
  • 16) 0.000 000 005 785 434 816 512 × 2 = 0 + 0.000 000 011 570 869 633 024;
  • 17) 0.000 000 011 570 869 633 024 × 2 = 0 + 0.000 000 023 141 739 266 048;
  • 18) 0.000 000 023 141 739 266 048 × 2 = 0 + 0.000 000 046 283 478 532 096;
  • 19) 0.000 000 046 283 478 532 096 × 2 = 0 + 0.000 000 092 566 957 064 192;
  • 20) 0.000 000 092 566 957 064 192 × 2 = 0 + 0.000 000 185 133 914 128 384;
  • 21) 0.000 000 185 133 914 128 384 × 2 = 0 + 0.000 000 370 267 828 256 768;
  • 22) 0.000 000 370 267 828 256 768 × 2 = 0 + 0.000 000 740 535 656 513 536;
  • 23) 0.000 000 740 535 656 513 536 × 2 = 0 + 0.000 001 481 071 313 027 072;
  • 24) 0.000 001 481 071 313 027 072 × 2 = 0 + 0.000 002 962 142 626 054 144;
  • 25) 0.000 002 962 142 626 054 144 × 2 = 0 + 0.000 005 924 285 252 108 288;
  • 26) 0.000 005 924 285 252 108 288 × 2 = 0 + 0.000 011 848 570 504 216 576;
  • 27) 0.000 011 848 570 504 216 576 × 2 = 0 + 0.000 023 697 141 008 433 152;
  • 28) 0.000 023 697 141 008 433 152 × 2 = 0 + 0.000 047 394 282 016 866 304;
  • 29) 0.000 047 394 282 016 866 304 × 2 = 0 + 0.000 094 788 564 033 732 608;
  • 30) 0.000 094 788 564 033 732 608 × 2 = 0 + 0.000 189 577 128 067 465 216;
  • 31) 0.000 189 577 128 067 465 216 × 2 = 0 + 0.000 379 154 256 134 930 432;
  • 32) 0.000 379 154 256 134 930 432 × 2 = 0 + 0.000 758 308 512 269 860 864;
  • 33) 0.000 758 308 512 269 860 864 × 2 = 0 + 0.001 516 617 024 539 721 728;
  • 34) 0.001 516 617 024 539 721 728 × 2 = 0 + 0.003 033 234 049 079 443 456;
  • 35) 0.003 033 234 049 079 443 456 × 2 = 0 + 0.006 066 468 098 158 886 912;
  • 36) 0.006 066 468 098 158 886 912 × 2 = 0 + 0.012 132 936 196 317 773 824;
  • 37) 0.012 132 936 196 317 773 824 × 2 = 0 + 0.024 265 872 392 635 547 648;
  • 38) 0.024 265 872 392 635 547 648 × 2 = 0 + 0.048 531 744 785 271 095 296;
  • 39) 0.048 531 744 785 271 095 296 × 2 = 0 + 0.097 063 489 570 542 190 592;
  • 40) 0.097 063 489 570 542 190 592 × 2 = 0 + 0.194 126 979 141 084 381 184;
  • 41) 0.194 126 979 141 084 381 184 × 2 = 0 + 0.388 253 958 282 168 762 368;
  • 42) 0.388 253 958 282 168 762 368 × 2 = 0 + 0.776 507 916 564 337 524 736;
  • 43) 0.776 507 916 564 337 524 736 × 2 = 1 + 0.553 015 833 128 675 049 472;
  • 44) 0.553 015 833 128 675 049 472 × 2 = 1 + 0.106 031 666 257 350 098 944;
  • 45) 0.106 031 666 257 350 098 944 × 2 = 0 + 0.212 063 332 514 700 197 888;
  • 46) 0.212 063 332 514 700 197 888 × 2 = 0 + 0.424 126 665 029 400 395 776;
  • 47) 0.424 126 665 029 400 395 776 × 2 = 0 + 0.848 253 330 058 800 791 552;
  • 48) 0.848 253 330 058 800 791 552 × 2 = 1 + 0.696 506 660 117 601 583 104;
  • 49) 0.696 506 660 117 601 583 104 × 2 = 1 + 0.393 013 320 235 203 166 208;
  • 50) 0.393 013 320 235 203 166 208 × 2 = 0 + 0.786 026 640 470 406 332 416;
  • 51) 0.786 026 640 470 406 332 416 × 2 = 1 + 0.572 053 280 940 812 664 832;
  • 52) 0.572 053 280 940 812 664 832 × 2 = 1 + 0.144 106 561 881 625 329 664;
  • 53) 0.144 106 561 881 625 329 664 × 2 = 0 + 0.288 213 123 763 250 659 328;
  • 54) 0.288 213 123 763 250 659 328 × 2 = 0 + 0.576 426 247 526 501 318 656;
  • 55) 0.576 426 247 526 501 318 656 × 2 = 1 + 0.152 852 495 053 002 637 312;
  • 56) 0.152 852 495 053 002 637 312 × 2 = 0 + 0.305 704 990 106 005 274 624;
  • 57) 0.305 704 990 106 005 274 624 × 2 = 0 + 0.611 409 980 212 010 549 248;
  • 58) 0.611 409 980 212 010 549 248 × 2 = 1 + 0.222 819 960 424 021 098 496;
  • 59) 0.222 819 960 424 021 098 496 × 2 = 0 + 0.445 639 920 848 042 196 992;
  • 60) 0.445 639 920 848 042 196 992 × 2 = 0 + 0.891 279 841 696 084 393 984;
  • 61) 0.891 279 841 696 084 393 984 × 2 = 1 + 0.782 559 683 392 168 787 968;
  • 62) 0.782 559 683 392 168 787 968 × 2 = 1 + 0.565 119 366 784 337 575 936;
  • 63) 0.565 119 366 784 337 575 936 × 2 = 1 + 0.130 238 733 568 675 151 872;
  • 64) 0.130 238 733 568 675 151 872 × 2 = 0 + 0.260 477 467 137 350 303 744;
  • 65) 0.260 477 467 137 350 303 744 × 2 = 0 + 0.520 954 934 274 700 607 488;
  • 66) 0.520 954 934 274 700 607 488 × 2 = 1 + 0.041 909 868 549 401 214 976;
  • 67) 0.041 909 868 549 401 214 976 × 2 = 0 + 0.083 819 737 098 802 429 952;
  • 68) 0.083 819 737 098 802 429 952 × 2 = 0 + 0.167 639 474 197 604 859 904;
  • 69) 0.167 639 474 197 604 859 904 × 2 = 0 + 0.335 278 948 395 209 719 808;
  • 70) 0.335 278 948 395 209 719 808 × 2 = 0 + 0.670 557 896 790 419 439 616;
  • 71) 0.670 557 896 790 419 439 616 × 2 = 1 + 0.341 115 793 580 838 879 232;
  • 72) 0.341 115 793 580 838 879 232 × 2 = 0 + 0.682 231 587 161 677 758 464;
  • 73) 0.682 231 587 161 677 758 464 × 2 = 1 + 0.364 463 174 323 355 516 928;
  • 74) 0.364 463 174 323 355 516 928 × 2 = 0 + 0.728 926 348 646 711 033 856;
  • 75) 0.728 926 348 646 711 033 856 × 2 = 1 + 0.457 852 697 293 422 067 712;
  • 76) 0.457 852 697 293 422 067 712 × 2 = 0 + 0.915 705 394 586 844 135 424;
  • 77) 0.915 705 394 586 844 135 424 × 2 = 1 + 0.831 410 789 173 688 270 848;
  • 78) 0.831 410 789 173 688 270 848 × 2 = 1 + 0.662 821 578 347 376 541 696;
  • 79) 0.662 821 578 347 376 541 696 × 2 = 1 + 0.325 643 156 694 753 083 392;
  • 80) 0.325 643 156 694 753 083 392 × 2 = 0 + 0.651 286 313 389 506 166 784;
  • 81) 0.651 286 313 389 506 166 784 × 2 = 1 + 0.302 572 626 779 012 333 568;
  • 82) 0.302 572 626 779 012 333 568 × 2 = 0 + 0.605 145 253 558 024 667 136;
  • 83) 0.605 145 253 558 024 667 136 × 2 = 1 + 0.210 290 507 116 049 334 272;
  • 84) 0.210 290 507 116 049 334 272 × 2 = 0 + 0.420 581 014 232 098 668 544;
  • 85) 0.420 581 014 232 098 668 544 × 2 = 0 + 0.841 162 028 464 197 337 088;
  • 86) 0.841 162 028 464 197 337 088 × 2 = 1 + 0.682 324 056 928 394 674 176;
  • 87) 0.682 324 056 928 394 674 176 × 2 = 1 + 0.364 648 113 856 789 348 352;
  • 88) 0.364 648 113 856 789 348 352 × 2 = 0 + 0.729 296 227 713 578 696 704;
  • 89) 0.729 296 227 713 578 696 704 × 2 = 1 + 0.458 592 455 427 157 393 408;
  • 90) 0.458 592 455 427 157 393 408 × 2 = 0 + 0.917 184 910 854 314 786 816;
  • 91) 0.917 184 910 854 314 786 816 × 2 = 1 + 0.834 369 821 708 629 573 632;
  • 92) 0.834 369 821 708 629 573 632 × 2 = 1 + 0.668 739 643 417 259 147 264;
  • 93) 0.668 739 643 417 259 147 264 × 2 = 1 + 0.337 479 286 834 518 294 528;
  • 94) 0.337 479 286 834 518 294 528 × 2 = 0 + 0.674 958 573 669 036 589 056;
  • 95) 0.674 958 573 669 036 589 056 × 2 = 1 + 0.349 917 147 338 073 178 112;

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:


0.000 000 000 000 176 557 459(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 0100 0010 1010 1110 1010 0110 1011 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 459(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 0100 0010 1010 1110 1010 0110 1011 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:


0.000 000 000 000 176 557 459(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 0100 0010 1010 1110 1010 0110 1011 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 0100 0010 1010 1110 1010 0110 1011 101(2) × 20 =

1.1000 1101 1001 0010 0111 0010 0001 0101 0111 0101 0011 0101 1101(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 0010 0001 0101 0111 0101 0011 0101 1101


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 0010 0001 0101 0111 0101 0011 0101 1101 =

1000 1101 1001 0010 0111 0010 0001 0101 0111 0101 0011 0101 1101


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 0010 0001 0101 0111 0101 0011 0101 1101


Decimal number -0.000 000 000 000 176 557 459 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 0010 0001 0101 0111 0101 0011 0101 1101

Share

» Convert decimal -0.000 000 000 000 176 557 546 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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