-0.000 000 000 000 176 557 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

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

|-0.000 000 000 000 176 557 7| = 0.000 000 000 000 176 557 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: 0.000 000 000 000 176 557 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) 0.000 000 000 000 176 557 7 × 2 = 0 + 0.000 000 000 000 353 115 4;
  • 2) 0.000 000 000 000 353 115 4 × 2 = 0 + 0.000 000 000 000 706 230 8;
  • 3) 0.000 000 000 000 706 230 8 × 2 = 0 + 0.000 000 000 001 412 461 6;
  • 4) 0.000 000 000 001 412 461 6 × 2 = 0 + 0.000 000 000 002 824 923 2;
  • 5) 0.000 000 000 002 824 923 2 × 2 = 0 + 0.000 000 000 005 649 846 4;
  • 6) 0.000 000 000 005 649 846 4 × 2 = 0 + 0.000 000 000 011 299 692 8;
  • 7) 0.000 000 000 011 299 692 8 × 2 = 0 + 0.000 000 000 022 599 385 6;
  • 8) 0.000 000 000 022 599 385 6 × 2 = 0 + 0.000 000 000 045 198 771 2;
  • 9) 0.000 000 000 045 198 771 2 × 2 = 0 + 0.000 000 000 090 397 542 4;
  • 10) 0.000 000 000 090 397 542 4 × 2 = 0 + 0.000 000 000 180 795 084 8;
  • 11) 0.000 000 000 180 795 084 8 × 2 = 0 + 0.000 000 000 361 590 169 6;
  • 12) 0.000 000 000 361 590 169 6 × 2 = 0 + 0.000 000 000 723 180 339 2;
  • 13) 0.000 000 000 723 180 339 2 × 2 = 0 + 0.000 000 001 446 360 678 4;
  • 14) 0.000 000 001 446 360 678 4 × 2 = 0 + 0.000 000 002 892 721 356 8;
  • 15) 0.000 000 002 892 721 356 8 × 2 = 0 + 0.000 000 005 785 442 713 6;
  • 16) 0.000 000 005 785 442 713 6 × 2 = 0 + 0.000 000 011 570 885 427 2;
  • 17) 0.000 000 011 570 885 427 2 × 2 = 0 + 0.000 000 023 141 770 854 4;
  • 18) 0.000 000 023 141 770 854 4 × 2 = 0 + 0.000 000 046 283 541 708 8;
  • 19) 0.000 000 046 283 541 708 8 × 2 = 0 + 0.000 000 092 567 083 417 6;
  • 20) 0.000 000 092 567 083 417 6 × 2 = 0 + 0.000 000 185 134 166 835 2;
  • 21) 0.000 000 185 134 166 835 2 × 2 = 0 + 0.000 000 370 268 333 670 4;
  • 22) 0.000 000 370 268 333 670 4 × 2 = 0 + 0.000 000 740 536 667 340 8;
  • 23) 0.000 000 740 536 667 340 8 × 2 = 0 + 0.000 001 481 073 334 681 6;
  • 24) 0.000 001 481 073 334 681 6 × 2 = 0 + 0.000 002 962 146 669 363 2;
  • 25) 0.000 002 962 146 669 363 2 × 2 = 0 + 0.000 005 924 293 338 726 4;
  • 26) 0.000 005 924 293 338 726 4 × 2 = 0 + 0.000 011 848 586 677 452 8;
  • 27) 0.000 011 848 586 677 452 8 × 2 = 0 + 0.000 023 697 173 354 905 6;
  • 28) 0.000 023 697 173 354 905 6 × 2 = 0 + 0.000 047 394 346 709 811 2;
  • 29) 0.000 047 394 346 709 811 2 × 2 = 0 + 0.000 094 788 693 419 622 4;
  • 30) 0.000 094 788 693 419 622 4 × 2 = 0 + 0.000 189 577 386 839 244 8;
  • 31) 0.000 189 577 386 839 244 8 × 2 = 0 + 0.000 379 154 773 678 489 6;
  • 32) 0.000 379 154 773 678 489 6 × 2 = 0 + 0.000 758 309 547 356 979 2;
  • 33) 0.000 758 309 547 356 979 2 × 2 = 0 + 0.001 516 619 094 713 958 4;
  • 34) 0.001 516 619 094 713 958 4 × 2 = 0 + 0.003 033 238 189 427 916 8;
  • 35) 0.003 033 238 189 427 916 8 × 2 = 0 + 0.006 066 476 378 855 833 6;
  • 36) 0.006 066 476 378 855 833 6 × 2 = 0 + 0.012 132 952 757 711 667 2;
  • 37) 0.012 132 952 757 711 667 2 × 2 = 0 + 0.024 265 905 515 423 334 4;
  • 38) 0.024 265 905 515 423 334 4 × 2 = 0 + 0.048 531 811 030 846 668 8;
  • 39) 0.048 531 811 030 846 668 8 × 2 = 0 + 0.097 063 622 061 693 337 6;
  • 40) 0.097 063 622 061 693 337 6 × 2 = 0 + 0.194 127 244 123 386 675 2;
  • 41) 0.194 127 244 123 386 675 2 × 2 = 0 + 0.388 254 488 246 773 350 4;
  • 42) 0.388 254 488 246 773 350 4 × 2 = 0 + 0.776 508 976 493 546 700 8;
  • 43) 0.776 508 976 493 546 700 8 × 2 = 1 + 0.553 017 952 987 093 401 6;
  • 44) 0.553 017 952 987 093 401 6 × 2 = 1 + 0.106 035 905 974 186 803 2;
  • 45) 0.106 035 905 974 186 803 2 × 2 = 0 + 0.212 071 811 948 373 606 4;
  • 46) 0.212 071 811 948 373 606 4 × 2 = 0 + 0.424 143 623 896 747 212 8;
  • 47) 0.424 143 623 896 747 212 8 × 2 = 0 + 0.848 287 247 793 494 425 6;
  • 48) 0.848 287 247 793 494 425 6 × 2 = 1 + 0.696 574 495 586 988 851 2;
  • 49) 0.696 574 495 586 988 851 2 × 2 = 1 + 0.393 148 991 173 977 702 4;
  • 50) 0.393 148 991 173 977 702 4 × 2 = 0 + 0.786 297 982 347 955 404 8;
  • 51) 0.786 297 982 347 955 404 8 × 2 = 1 + 0.572 595 964 695 910 809 6;
  • 52) 0.572 595 964 695 910 809 6 × 2 = 1 + 0.145 191 929 391 821 619 2;
  • 53) 0.145 191 929 391 821 619 2 × 2 = 0 + 0.290 383 858 783 643 238 4;
  • 54) 0.290 383 858 783 643 238 4 × 2 = 0 + 0.580 767 717 567 286 476 8;
  • 55) 0.580 767 717 567 286 476 8 × 2 = 1 + 0.161 535 435 134 572 953 6;
  • 56) 0.161 535 435 134 572 953 6 × 2 = 0 + 0.323 070 870 269 145 907 2;
  • 57) 0.323 070 870 269 145 907 2 × 2 = 0 + 0.646 141 740 538 291 814 4;
  • 58) 0.646 141 740 538 291 814 4 × 2 = 1 + 0.292 283 481 076 583 628 8;
  • 59) 0.292 283 481 076 583 628 8 × 2 = 0 + 0.584 566 962 153 167 257 6;
  • 60) 0.584 566 962 153 167 257 6 × 2 = 1 + 0.169 133 924 306 334 515 2;
  • 61) 0.169 133 924 306 334 515 2 × 2 = 0 + 0.338 267 848 612 669 030 4;
  • 62) 0.338 267 848 612 669 030 4 × 2 = 0 + 0.676 535 697 225 338 060 8;
  • 63) 0.676 535 697 225 338 060 8 × 2 = 1 + 0.353 071 394 450 676 121 6;
  • 64) 0.353 071 394 450 676 121 6 × 2 = 0 + 0.706 142 788 901 352 243 2;
  • 65) 0.706 142 788 901 352 243 2 × 2 = 1 + 0.412 285 577 802 704 486 4;
  • 66) 0.412 285 577 802 704 486 4 × 2 = 0 + 0.824 571 155 605 408 972 8;
  • 67) 0.824 571 155 605 408 972 8 × 2 = 1 + 0.649 142 311 210 817 945 6;
  • 68) 0.649 142 311 210 817 945 6 × 2 = 1 + 0.298 284 622 421 635 891 2;
  • 69) 0.298 284 622 421 635 891 2 × 2 = 0 + 0.596 569 244 843 271 782 4;
  • 70) 0.596 569 244 843 271 782 4 × 2 = 1 + 0.193 138 489 686 543 564 8;
  • 71) 0.193 138 489 686 543 564 8 × 2 = 0 + 0.386 276 979 373 087 129 6;
  • 72) 0.386 276 979 373 087 129 6 × 2 = 0 + 0.772 553 958 746 174 259 2;
  • 73) 0.772 553 958 746 174 259 2 × 2 = 1 + 0.545 107 917 492 348 518 4;
  • 74) 0.545 107 917 492 348 518 4 × 2 = 1 + 0.090 215 834 984 697 036 8;
  • 75) 0.090 215 834 984 697 036 8 × 2 = 0 + 0.180 431 669 969 394 073 6;
  • 76) 0.180 431 669 969 394 073 6 × 2 = 0 + 0.360 863 339 938 788 147 2;
  • 77) 0.360 863 339 938 788 147 2 × 2 = 0 + 0.721 726 679 877 576 294 4;
  • 78) 0.721 726 679 877 576 294 4 × 2 = 1 + 0.443 453 359 755 152 588 8;
  • 79) 0.443 453 359 755 152 588 8 × 2 = 0 + 0.886 906 719 510 305 177 6;
  • 80) 0.886 906 719 510 305 177 6 × 2 = 1 + 0.773 813 439 020 610 355 2;
  • 81) 0.773 813 439 020 610 355 2 × 2 = 1 + 0.547 626 878 041 220 710 4;
  • 82) 0.547 626 878 041 220 710 4 × 2 = 1 + 0.095 253 756 082 441 420 8;
  • 83) 0.095 253 756 082 441 420 8 × 2 = 0 + 0.190 507 512 164 882 841 6;
  • 84) 0.190 507 512 164 882 841 6 × 2 = 0 + 0.381 015 024 329 765 683 2;
  • 85) 0.381 015 024 329 765 683 2 × 2 = 0 + 0.762 030 048 659 531 366 4;
  • 86) 0.762 030 048 659 531 366 4 × 2 = 1 + 0.524 060 097 319 062 732 8;
  • 87) 0.524 060 097 319 062 732 8 × 2 = 1 + 0.048 120 194 638 125 465 6;
  • 88) 0.048 120 194 638 125 465 6 × 2 = 0 + 0.096 240 389 276 250 931 2;
  • 89) 0.096 240 389 276 250 931 2 × 2 = 0 + 0.192 480 778 552 501 862 4;
  • 90) 0.192 480 778 552 501 862 4 × 2 = 0 + 0.384 961 557 105 003 724 8;
  • 91) 0.384 961 557 105 003 724 8 × 2 = 0 + 0.769 923 114 210 007 449 6;
  • 92) 0.769 923 114 210 007 449 6 × 2 = 1 + 0.539 846 228 420 014 899 2;
  • 93) 0.539 846 228 420 014 899 2 × 2 = 1 + 0.079 692 456 840 029 798 4;
  • 94) 0.079 692 456 840 029 798 4 × 2 = 0 + 0.159 384 913 680 059 596 8;
  • 95) 0.159 384 913 680 059 596 8 × 2 = 0 + 0.318 769 827 360 119 193 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:


0.000 000 000 000 176 557 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 1011 0100 1100 0101 1100 0110 0001 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 1011 0100 1100 0101 1100 0110 0001 100(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 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 1011 0100 1100 0101 1100 0110 0001 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 1011 0100 1100 0101 1100 0110 0001 100(2) × 20 =

1.1000 1101 1001 0010 1001 0101 1010 0110 0010 1110 0011 0000 1100(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 1001 0101 1010 0110 0010 1110 0011 0000 1100


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 1001 0101 1010 0110 0010 1110 0011 0000 1100 =

1000 1101 1001 0010 1001 0101 1010 0110 0010 1110 0011 0000 1100


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 1001 0101 1010 0110 0010 1110 0011 0000 1100


Decimal number -0.000 000 000 000 176 557 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1001 0101 1010 0110 0010 1110 0011 0000 1100

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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