-0.000 000 000 000 176 553 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 553 7(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
-0.000 000 000 000 176 553 7(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:

|-0.000 000 000 000 176 553 7| = 0.000 000 000 000 176 553 7


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: 0.000 000 000 000 176 553 7.

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) 0.000 000 000 000 176 553 7 × 2 = 0 + 0.000 000 000 000 353 107 4;
  • 2) 0.000 000 000 000 353 107 4 × 2 = 0 + 0.000 000 000 000 706 214 8;
  • 3) 0.000 000 000 000 706 214 8 × 2 = 0 + 0.000 000 000 001 412 429 6;
  • 4) 0.000 000 000 001 412 429 6 × 2 = 0 + 0.000 000 000 002 824 859 2;
  • 5) 0.000 000 000 002 824 859 2 × 2 = 0 + 0.000 000 000 005 649 718 4;
  • 6) 0.000 000 000 005 649 718 4 × 2 = 0 + 0.000 000 000 011 299 436 8;
  • 7) 0.000 000 000 011 299 436 8 × 2 = 0 + 0.000 000 000 022 598 873 6;
  • 8) 0.000 000 000 022 598 873 6 × 2 = 0 + 0.000 000 000 045 197 747 2;
  • 9) 0.000 000 000 045 197 747 2 × 2 = 0 + 0.000 000 000 090 395 494 4;
  • 10) 0.000 000 000 090 395 494 4 × 2 = 0 + 0.000 000 000 180 790 988 8;
  • 11) 0.000 000 000 180 790 988 8 × 2 = 0 + 0.000 000 000 361 581 977 6;
  • 12) 0.000 000 000 361 581 977 6 × 2 = 0 + 0.000 000 000 723 163 955 2;
  • 13) 0.000 000 000 723 163 955 2 × 2 = 0 + 0.000 000 001 446 327 910 4;
  • 14) 0.000 000 001 446 327 910 4 × 2 = 0 + 0.000 000 002 892 655 820 8;
  • 15) 0.000 000 002 892 655 820 8 × 2 = 0 + 0.000 000 005 785 311 641 6;
  • 16) 0.000 000 005 785 311 641 6 × 2 = 0 + 0.000 000 011 570 623 283 2;
  • 17) 0.000 000 011 570 623 283 2 × 2 = 0 + 0.000 000 023 141 246 566 4;
  • 18) 0.000 000 023 141 246 566 4 × 2 = 0 + 0.000 000 046 282 493 132 8;
  • 19) 0.000 000 046 282 493 132 8 × 2 = 0 + 0.000 000 092 564 986 265 6;
  • 20) 0.000 000 092 564 986 265 6 × 2 = 0 + 0.000 000 185 129 972 531 2;
  • 21) 0.000 000 185 129 972 531 2 × 2 = 0 + 0.000 000 370 259 945 062 4;
  • 22) 0.000 000 370 259 945 062 4 × 2 = 0 + 0.000 000 740 519 890 124 8;
  • 23) 0.000 000 740 519 890 124 8 × 2 = 0 + 0.000 001 481 039 780 249 6;
  • 24) 0.000 001 481 039 780 249 6 × 2 = 0 + 0.000 002 962 079 560 499 2;
  • 25) 0.000 002 962 079 560 499 2 × 2 = 0 + 0.000 005 924 159 120 998 4;
  • 26) 0.000 005 924 159 120 998 4 × 2 = 0 + 0.000 011 848 318 241 996 8;
  • 27) 0.000 011 848 318 241 996 8 × 2 = 0 + 0.000 023 696 636 483 993 6;
  • 28) 0.000 023 696 636 483 993 6 × 2 = 0 + 0.000 047 393 272 967 987 2;
  • 29) 0.000 047 393 272 967 987 2 × 2 = 0 + 0.000 094 786 545 935 974 4;
  • 30) 0.000 094 786 545 935 974 4 × 2 = 0 + 0.000 189 573 091 871 948 8;
  • 31) 0.000 189 573 091 871 948 8 × 2 = 0 + 0.000 379 146 183 743 897 6;
  • 32) 0.000 379 146 183 743 897 6 × 2 = 0 + 0.000 758 292 367 487 795 2;
  • 33) 0.000 758 292 367 487 795 2 × 2 = 0 + 0.001 516 584 734 975 590 4;
  • 34) 0.001 516 584 734 975 590 4 × 2 = 0 + 0.003 033 169 469 951 180 8;
  • 35) 0.003 033 169 469 951 180 8 × 2 = 0 + 0.006 066 338 939 902 361 6;
  • 36) 0.006 066 338 939 902 361 6 × 2 = 0 + 0.012 132 677 879 804 723 2;
  • 37) 0.012 132 677 879 804 723 2 × 2 = 0 + 0.024 265 355 759 609 446 4;
  • 38) 0.024 265 355 759 609 446 4 × 2 = 0 + 0.048 530 711 519 218 892 8;
  • 39) 0.048 530 711 519 218 892 8 × 2 = 0 + 0.097 061 423 038 437 785 6;
  • 40) 0.097 061 423 038 437 785 6 × 2 = 0 + 0.194 122 846 076 875 571 2;
  • 41) 0.194 122 846 076 875 571 2 × 2 = 0 + 0.388 245 692 153 751 142 4;
  • 42) 0.388 245 692 153 751 142 4 × 2 = 0 + 0.776 491 384 307 502 284 8;
  • 43) 0.776 491 384 307 502 284 8 × 2 = 1 + 0.552 982 768 615 004 569 6;
  • 44) 0.552 982 768 615 004 569 6 × 2 = 1 + 0.105 965 537 230 009 139 2;
  • 45) 0.105 965 537 230 009 139 2 × 2 = 0 + 0.211 931 074 460 018 278 4;
  • 46) 0.211 931 074 460 018 278 4 × 2 = 0 + 0.423 862 148 920 036 556 8;
  • 47) 0.423 862 148 920 036 556 8 × 2 = 0 + 0.847 724 297 840 073 113 6;
  • 48) 0.847 724 297 840 073 113 6 × 2 = 1 + 0.695 448 595 680 146 227 2;
  • 49) 0.695 448 595 680 146 227 2 × 2 = 1 + 0.390 897 191 360 292 454 4;
  • 50) 0.390 897 191 360 292 454 4 × 2 = 0 + 0.781 794 382 720 584 908 8;
  • 51) 0.781 794 382 720 584 908 8 × 2 = 1 + 0.563 588 765 441 169 817 6;
  • 52) 0.563 588 765 441 169 817 6 × 2 = 1 + 0.127 177 530 882 339 635 2;
  • 53) 0.127 177 530 882 339 635 2 × 2 = 0 + 0.254 355 061 764 679 270 4;
  • 54) 0.254 355 061 764 679 270 4 × 2 = 0 + 0.508 710 123 529 358 540 8;
  • 55) 0.508 710 123 529 358 540 8 × 2 = 1 + 0.017 420 247 058 717 081 6;
  • 56) 0.017 420 247 058 717 081 6 × 2 = 0 + 0.034 840 494 117 434 163 2;
  • 57) 0.034 840 494 117 434 163 2 × 2 = 0 + 0.069 680 988 234 868 326 4;
  • 58) 0.069 680 988 234 868 326 4 × 2 = 0 + 0.139 361 976 469 736 652 8;
  • 59) 0.139 361 976 469 736 652 8 × 2 = 0 + 0.278 723 952 939 473 305 6;
  • 60) 0.278 723 952 939 473 305 6 × 2 = 0 + 0.557 447 905 878 946 611 2;
  • 61) 0.557 447 905 878 946 611 2 × 2 = 1 + 0.114 895 811 757 893 222 4;
  • 62) 0.114 895 811 757 893 222 4 × 2 = 0 + 0.229 791 623 515 786 444 8;
  • 63) 0.229 791 623 515 786 444 8 × 2 = 0 + 0.459 583 247 031 572 889 6;
  • 64) 0.459 583 247 031 572 889 6 × 2 = 0 + 0.919 166 494 063 145 779 2;
  • 65) 0.919 166 494 063 145 779 2 × 2 = 1 + 0.838 332 988 126 291 558 4;
  • 66) 0.838 332 988 126 291 558 4 × 2 = 1 + 0.676 665 976 252 583 116 8;
  • 67) 0.676 665 976 252 583 116 8 × 2 = 1 + 0.353 331 952 505 166 233 6;
  • 68) 0.353 331 952 505 166 233 6 × 2 = 0 + 0.706 663 905 010 332 467 2;
  • 69) 0.706 663 905 010 332 467 2 × 2 = 1 + 0.413 327 810 020 664 934 4;
  • 70) 0.413 327 810 020 664 934 4 × 2 = 0 + 0.826 655 620 041 329 868 8;
  • 71) 0.826 655 620 041 329 868 8 × 2 = 1 + 0.653 311 240 082 659 737 6;
  • 72) 0.653 311 240 082 659 737 6 × 2 = 1 + 0.306 622 480 165 319 475 2;
  • 73) 0.306 622 480 165 319 475 2 × 2 = 0 + 0.613 244 960 330 638 950 4;
  • 74) 0.613 244 960 330 638 950 4 × 2 = 1 + 0.226 489 920 661 277 900 8;
  • 75) 0.226 489 920 661 277 900 8 × 2 = 0 + 0.452 979 841 322 555 801 6;
  • 76) 0.452 979 841 322 555 801 6 × 2 = 0 + 0.905 959 682 645 111 603 2;
  • 77) 0.905 959 682 645 111 603 2 × 2 = 1 + 0.811 919 365 290 223 206 4;
  • 78) 0.811 919 365 290 223 206 4 × 2 = 1 + 0.623 838 730 580 446 412 8;
  • 79) 0.623 838 730 580 446 412 8 × 2 = 1 + 0.247 677 461 160 892 825 6;
  • 80) 0.247 677 461 160 892 825 6 × 2 = 0 + 0.495 354 922 321 785 651 2;
  • 81) 0.495 354 922 321 785 651 2 × 2 = 0 + 0.990 709 844 643 571 302 4;
  • 82) 0.990 709 844 643 571 302 4 × 2 = 1 + 0.981 419 689 287 142 604 8;
  • 83) 0.981 419 689 287 142 604 8 × 2 = 1 + 0.962 839 378 574 285 209 6;
  • 84) 0.962 839 378 574 285 209 6 × 2 = 1 + 0.925 678 757 148 570 419 2;
  • 85) 0.925 678 757 148 570 419 2 × 2 = 1 + 0.851 357 514 297 140 838 4;
  • 86) 0.851 357 514 297 140 838 4 × 2 = 1 + 0.702 715 028 594 281 676 8;
  • 87) 0.702 715 028 594 281 676 8 × 2 = 1 + 0.405 430 057 188 563 353 6;
  • 88) 0.405 430 057 188 563 353 6 × 2 = 0 + 0.810 860 114 377 126 707 2;
  • 89) 0.810 860 114 377 126 707 2 × 2 = 1 + 0.621 720 228 754 253 414 4;
  • 90) 0.621 720 228 754 253 414 4 × 2 = 1 + 0.243 440 457 508 506 828 8;
  • 91) 0.243 440 457 508 506 828 8 × 2 = 0 + 0.486 880 915 017 013 657 6;
  • 92) 0.486 880 915 017 013 657 6 × 2 = 0 + 0.973 761 830 034 027 315 2;
  • 93) 0.973 761 830 034 027 315 2 × 2 = 1 + 0.947 523 660 068 054 630 4;
  • 94) 0.947 523 660 068 054 630 4 × 2 = 1 + 0.895 047 320 136 109 260 8;
  • 95) 0.895 047 320 136 109 260 8 × 2 = 1 + 0.790 094 640 272 218 521 6;

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:


0.000 000 000 000 176 553 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0000 1000 1110 1011 0100 1110 0111 1110 1100 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 553 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0000 1000 1110 1011 0100 1110 0111 1110 1100 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:


0.000 000 000 000 176 553 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0000 1000 1110 1011 0100 1110 0111 1110 1100 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0000 1000 1110 1011 0100 1110 0111 1110 1100 111(2) × 20 =

1.1000 1101 1001 0000 0100 0111 0101 1010 0111 0011 1111 0110 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 1001 0000 0100 0111 0101 1010 0111 0011 1111 0110 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 1001 0000 0100 0111 0101 1010 0111 0011 1111 0110 0111 =

1000 1101 1001 0000 0100 0111 0101 1010 0111 0011 1111 0110 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 1001 0000 0100 0111 0101 1010 0111 0011 1111 0110 0111


Decimal number -0.000 000 000 000 176 553 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0000 0100 0111 0101 1010 0111 0011 1111 0110 0111

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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