-0.000 000 000 000 176 546 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 546(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
-0.000 000 000 000 176 546(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:

|-0.000 000 000 000 176 546| = 0.000 000 000 000 176 546


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: 0.000 000 000 000 176 546.

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) 0.000 000 000 000 176 546 × 2 = 0 + 0.000 000 000 000 353 092;
  • 2) 0.000 000 000 000 353 092 × 2 = 0 + 0.000 000 000 000 706 184;
  • 3) 0.000 000 000 000 706 184 × 2 = 0 + 0.000 000 000 001 412 368;
  • 4) 0.000 000 000 001 412 368 × 2 = 0 + 0.000 000 000 002 824 736;
  • 5) 0.000 000 000 002 824 736 × 2 = 0 + 0.000 000 000 005 649 472;
  • 6) 0.000 000 000 005 649 472 × 2 = 0 + 0.000 000 000 011 298 944;
  • 7) 0.000 000 000 011 298 944 × 2 = 0 + 0.000 000 000 022 597 888;
  • 8) 0.000 000 000 022 597 888 × 2 = 0 + 0.000 000 000 045 195 776;
  • 9) 0.000 000 000 045 195 776 × 2 = 0 + 0.000 000 000 090 391 552;
  • 10) 0.000 000 000 090 391 552 × 2 = 0 + 0.000 000 000 180 783 104;
  • 11) 0.000 000 000 180 783 104 × 2 = 0 + 0.000 000 000 361 566 208;
  • 12) 0.000 000 000 361 566 208 × 2 = 0 + 0.000 000 000 723 132 416;
  • 13) 0.000 000 000 723 132 416 × 2 = 0 + 0.000 000 001 446 264 832;
  • 14) 0.000 000 001 446 264 832 × 2 = 0 + 0.000 000 002 892 529 664;
  • 15) 0.000 000 002 892 529 664 × 2 = 0 + 0.000 000 005 785 059 328;
  • 16) 0.000 000 005 785 059 328 × 2 = 0 + 0.000 000 011 570 118 656;
  • 17) 0.000 000 011 570 118 656 × 2 = 0 + 0.000 000 023 140 237 312;
  • 18) 0.000 000 023 140 237 312 × 2 = 0 + 0.000 000 046 280 474 624;
  • 19) 0.000 000 046 280 474 624 × 2 = 0 + 0.000 000 092 560 949 248;
  • 20) 0.000 000 092 560 949 248 × 2 = 0 + 0.000 000 185 121 898 496;
  • 21) 0.000 000 185 121 898 496 × 2 = 0 + 0.000 000 370 243 796 992;
  • 22) 0.000 000 370 243 796 992 × 2 = 0 + 0.000 000 740 487 593 984;
  • 23) 0.000 000 740 487 593 984 × 2 = 0 + 0.000 001 480 975 187 968;
  • 24) 0.000 001 480 975 187 968 × 2 = 0 + 0.000 002 961 950 375 936;
  • 25) 0.000 002 961 950 375 936 × 2 = 0 + 0.000 005 923 900 751 872;
  • 26) 0.000 005 923 900 751 872 × 2 = 0 + 0.000 011 847 801 503 744;
  • 27) 0.000 011 847 801 503 744 × 2 = 0 + 0.000 023 695 603 007 488;
  • 28) 0.000 023 695 603 007 488 × 2 = 0 + 0.000 047 391 206 014 976;
  • 29) 0.000 047 391 206 014 976 × 2 = 0 + 0.000 094 782 412 029 952;
  • 30) 0.000 094 782 412 029 952 × 2 = 0 + 0.000 189 564 824 059 904;
  • 31) 0.000 189 564 824 059 904 × 2 = 0 + 0.000 379 129 648 119 808;
  • 32) 0.000 379 129 648 119 808 × 2 = 0 + 0.000 758 259 296 239 616;
  • 33) 0.000 758 259 296 239 616 × 2 = 0 + 0.001 516 518 592 479 232;
  • 34) 0.001 516 518 592 479 232 × 2 = 0 + 0.003 033 037 184 958 464;
  • 35) 0.003 033 037 184 958 464 × 2 = 0 + 0.006 066 074 369 916 928;
  • 36) 0.006 066 074 369 916 928 × 2 = 0 + 0.012 132 148 739 833 856;
  • 37) 0.012 132 148 739 833 856 × 2 = 0 + 0.024 264 297 479 667 712;
  • 38) 0.024 264 297 479 667 712 × 2 = 0 + 0.048 528 594 959 335 424;
  • 39) 0.048 528 594 959 335 424 × 2 = 0 + 0.097 057 189 918 670 848;
  • 40) 0.097 057 189 918 670 848 × 2 = 0 + 0.194 114 379 837 341 696;
  • 41) 0.194 114 379 837 341 696 × 2 = 0 + 0.388 228 759 674 683 392;
  • 42) 0.388 228 759 674 683 392 × 2 = 0 + 0.776 457 519 349 366 784;
  • 43) 0.776 457 519 349 366 784 × 2 = 1 + 0.552 915 038 698 733 568;
  • 44) 0.552 915 038 698 733 568 × 2 = 1 + 0.105 830 077 397 467 136;
  • 45) 0.105 830 077 397 467 136 × 2 = 0 + 0.211 660 154 794 934 272;
  • 46) 0.211 660 154 794 934 272 × 2 = 0 + 0.423 320 309 589 868 544;
  • 47) 0.423 320 309 589 868 544 × 2 = 0 + 0.846 640 619 179 737 088;
  • 48) 0.846 640 619 179 737 088 × 2 = 1 + 0.693 281 238 359 474 176;
  • 49) 0.693 281 238 359 474 176 × 2 = 1 + 0.386 562 476 718 948 352;
  • 50) 0.386 562 476 718 948 352 × 2 = 0 + 0.773 124 953 437 896 704;
  • 51) 0.773 124 953 437 896 704 × 2 = 1 + 0.546 249 906 875 793 408;
  • 52) 0.546 249 906 875 793 408 × 2 = 1 + 0.092 499 813 751 586 816;
  • 53) 0.092 499 813 751 586 816 × 2 = 0 + 0.184 999 627 503 173 632;
  • 54) 0.184 999 627 503 173 632 × 2 = 0 + 0.369 999 255 006 347 264;
  • 55) 0.369 999 255 006 347 264 × 2 = 0 + 0.739 998 510 012 694 528;
  • 56) 0.739 998 510 012 694 528 × 2 = 1 + 0.479 997 020 025 389 056;
  • 57) 0.479 997 020 025 389 056 × 2 = 0 + 0.959 994 040 050 778 112;
  • 58) 0.959 994 040 050 778 112 × 2 = 1 + 0.919 988 080 101 556 224;
  • 59) 0.919 988 080 101 556 224 × 2 = 1 + 0.839 976 160 203 112 448;
  • 60) 0.839 976 160 203 112 448 × 2 = 1 + 0.679 952 320 406 224 896;
  • 61) 0.679 952 320 406 224 896 × 2 = 1 + 0.359 904 640 812 449 792;
  • 62) 0.359 904 640 812 449 792 × 2 = 0 + 0.719 809 281 624 899 584;
  • 63) 0.719 809 281 624 899 584 × 2 = 1 + 0.439 618 563 249 799 168;
  • 64) 0.439 618 563 249 799 168 × 2 = 0 + 0.879 237 126 499 598 336;
  • 65) 0.879 237 126 499 598 336 × 2 = 1 + 0.758 474 252 999 196 672;
  • 66) 0.758 474 252 999 196 672 × 2 = 1 + 0.516 948 505 998 393 344;
  • 67) 0.516 948 505 998 393 344 × 2 = 1 + 0.033 897 011 996 786 688;
  • 68) 0.033 897 011 996 786 688 × 2 = 0 + 0.067 794 023 993 573 376;
  • 69) 0.067 794 023 993 573 376 × 2 = 0 + 0.135 588 047 987 146 752;
  • 70) 0.135 588 047 987 146 752 × 2 = 0 + 0.271 176 095 974 293 504;
  • 71) 0.271 176 095 974 293 504 × 2 = 0 + 0.542 352 191 948 587 008;
  • 72) 0.542 352 191 948 587 008 × 2 = 1 + 0.084 704 383 897 174 016;
  • 73) 0.084 704 383 897 174 016 × 2 = 0 + 0.169 408 767 794 348 032;
  • 74) 0.169 408 767 794 348 032 × 2 = 0 + 0.338 817 535 588 696 064;
  • 75) 0.338 817 535 588 696 064 × 2 = 0 + 0.677 635 071 177 392 128;
  • 76) 0.677 635 071 177 392 128 × 2 = 1 + 0.355 270 142 354 784 256;
  • 77) 0.355 270 142 354 784 256 × 2 = 0 + 0.710 540 284 709 568 512;
  • 78) 0.710 540 284 709 568 512 × 2 = 1 + 0.421 080 569 419 137 024;
  • 79) 0.421 080 569 419 137 024 × 2 = 0 + 0.842 161 138 838 274 048;
  • 80) 0.842 161 138 838 274 048 × 2 = 1 + 0.684 322 277 676 548 096;
  • 81) 0.684 322 277 676 548 096 × 2 = 1 + 0.368 644 555 353 096 192;
  • 82) 0.368 644 555 353 096 192 × 2 = 0 + 0.737 289 110 706 192 384;
  • 83) 0.737 289 110 706 192 384 × 2 = 1 + 0.474 578 221 412 384 768;
  • 84) 0.474 578 221 412 384 768 × 2 = 0 + 0.949 156 442 824 769 536;
  • 85) 0.949 156 442 824 769 536 × 2 = 1 + 0.898 312 885 649 539 072;
  • 86) 0.898 312 885 649 539 072 × 2 = 1 + 0.796 625 771 299 078 144;
  • 87) 0.796 625 771 299 078 144 × 2 = 1 + 0.593 251 542 598 156 288;
  • 88) 0.593 251 542 598 156 288 × 2 = 1 + 0.186 503 085 196 312 576;
  • 89) 0.186 503 085 196 312 576 × 2 = 0 + 0.373 006 170 392 625 152;
  • 90) 0.373 006 170 392 625 152 × 2 = 0 + 0.746 012 340 785 250 304;
  • 91) 0.746 012 340 785 250 304 × 2 = 1 + 0.492 024 681 570 500 608;
  • 92) 0.492 024 681 570 500 608 × 2 = 0 + 0.984 049 363 141 001 216;
  • 93) 0.984 049 363 141 001 216 × 2 = 1 + 0.968 098 726 282 002 432;
  • 94) 0.968 098 726 282 002 432 × 2 = 1 + 0.936 197 452 564 004 864;
  • 95) 0.936 197 452 564 004 864 × 2 = 1 + 0.872 394 905 128 009 728;

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:


0.000 000 000 000 176 546(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0111 1010 1110 0001 0001 0101 1010 1111 0010 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 546(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0111 1010 1110 0001 0001 0101 1010 1111 0010 111(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:


0.000 000 000 000 176 546(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0111 1010 1110 0001 0001 0101 1010 1111 0010 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0111 1010 1110 0001 0001 0101 1010 1111 0010 111(2) × 20 =

1.1000 1101 1000 1011 1101 0111 0000 1000 1010 1101 0111 1001 0111(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 1000 1011 1101 0111 0000 1000 1010 1101 0111 1001 0111


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 1000 1011 1101 0111 0000 1000 1010 1101 0111 1001 0111 =

1000 1101 1000 1011 1101 0111 0000 1000 1010 1101 0111 1001 0111


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 1000 1011 1101 0111 0000 1000 1010 1101 0111 1001 0111


Decimal number -0.000 000 000 000 176 546 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1011 1101 0111 0000 1000 1010 1101 0111 1001 0111

Share

» Convert decimal -0.000 000 000 000 176 514 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: My manager only says "we" when referring to "my work". NotebookLM YouTube Video of 11.13 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