-0.000 000 000 000 176 557 46 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 46(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 46(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 46| = 0.000 000 000 000 176 557 46


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 46.

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 46 × 2 = 0 + 0.000 000 000 000 353 114 92;
  • 2) 0.000 000 000 000 353 114 92 × 2 = 0 + 0.000 000 000 000 706 229 84;
  • 3) 0.000 000 000 000 706 229 84 × 2 = 0 + 0.000 000 000 001 412 459 68;
  • 4) 0.000 000 000 001 412 459 68 × 2 = 0 + 0.000 000 000 002 824 919 36;
  • 5) 0.000 000 000 002 824 919 36 × 2 = 0 + 0.000 000 000 005 649 838 72;
  • 6) 0.000 000 000 005 649 838 72 × 2 = 0 + 0.000 000 000 011 299 677 44;
  • 7) 0.000 000 000 011 299 677 44 × 2 = 0 + 0.000 000 000 022 599 354 88;
  • 8) 0.000 000 000 022 599 354 88 × 2 = 0 + 0.000 000 000 045 198 709 76;
  • 9) 0.000 000 000 045 198 709 76 × 2 = 0 + 0.000 000 000 090 397 419 52;
  • 10) 0.000 000 000 090 397 419 52 × 2 = 0 + 0.000 000 000 180 794 839 04;
  • 11) 0.000 000 000 180 794 839 04 × 2 = 0 + 0.000 000 000 361 589 678 08;
  • 12) 0.000 000 000 361 589 678 08 × 2 = 0 + 0.000 000 000 723 179 356 16;
  • 13) 0.000 000 000 723 179 356 16 × 2 = 0 + 0.000 000 001 446 358 712 32;
  • 14) 0.000 000 001 446 358 712 32 × 2 = 0 + 0.000 000 002 892 717 424 64;
  • 15) 0.000 000 002 892 717 424 64 × 2 = 0 + 0.000 000 005 785 434 849 28;
  • 16) 0.000 000 005 785 434 849 28 × 2 = 0 + 0.000 000 011 570 869 698 56;
  • 17) 0.000 000 011 570 869 698 56 × 2 = 0 + 0.000 000 023 141 739 397 12;
  • 18) 0.000 000 023 141 739 397 12 × 2 = 0 + 0.000 000 046 283 478 794 24;
  • 19) 0.000 000 046 283 478 794 24 × 2 = 0 + 0.000 000 092 566 957 588 48;
  • 20) 0.000 000 092 566 957 588 48 × 2 = 0 + 0.000 000 185 133 915 176 96;
  • 21) 0.000 000 185 133 915 176 96 × 2 = 0 + 0.000 000 370 267 830 353 92;
  • 22) 0.000 000 370 267 830 353 92 × 2 = 0 + 0.000 000 740 535 660 707 84;
  • 23) 0.000 000 740 535 660 707 84 × 2 = 0 + 0.000 001 481 071 321 415 68;
  • 24) 0.000 001 481 071 321 415 68 × 2 = 0 + 0.000 002 962 142 642 831 36;
  • 25) 0.000 002 962 142 642 831 36 × 2 = 0 + 0.000 005 924 285 285 662 72;
  • 26) 0.000 005 924 285 285 662 72 × 2 = 0 + 0.000 011 848 570 571 325 44;
  • 27) 0.000 011 848 570 571 325 44 × 2 = 0 + 0.000 023 697 141 142 650 88;
  • 28) 0.000 023 697 141 142 650 88 × 2 = 0 + 0.000 047 394 282 285 301 76;
  • 29) 0.000 047 394 282 285 301 76 × 2 = 0 + 0.000 094 788 564 570 603 52;
  • 30) 0.000 094 788 564 570 603 52 × 2 = 0 + 0.000 189 577 129 141 207 04;
  • 31) 0.000 189 577 129 141 207 04 × 2 = 0 + 0.000 379 154 258 282 414 08;
  • 32) 0.000 379 154 258 282 414 08 × 2 = 0 + 0.000 758 308 516 564 828 16;
  • 33) 0.000 758 308 516 564 828 16 × 2 = 0 + 0.001 516 617 033 129 656 32;
  • 34) 0.001 516 617 033 129 656 32 × 2 = 0 + 0.003 033 234 066 259 312 64;
  • 35) 0.003 033 234 066 259 312 64 × 2 = 0 + 0.006 066 468 132 518 625 28;
  • 36) 0.006 066 468 132 518 625 28 × 2 = 0 + 0.012 132 936 265 037 250 56;
  • 37) 0.012 132 936 265 037 250 56 × 2 = 0 + 0.024 265 872 530 074 501 12;
  • 38) 0.024 265 872 530 074 501 12 × 2 = 0 + 0.048 531 745 060 149 002 24;
  • 39) 0.048 531 745 060 149 002 24 × 2 = 0 + 0.097 063 490 120 298 004 48;
  • 40) 0.097 063 490 120 298 004 48 × 2 = 0 + 0.194 126 980 240 596 008 96;
  • 41) 0.194 126 980 240 596 008 96 × 2 = 0 + 0.388 253 960 481 192 017 92;
  • 42) 0.388 253 960 481 192 017 92 × 2 = 0 + 0.776 507 920 962 384 035 84;
  • 43) 0.776 507 920 962 384 035 84 × 2 = 1 + 0.553 015 841 924 768 071 68;
  • 44) 0.553 015 841 924 768 071 68 × 2 = 1 + 0.106 031 683 849 536 143 36;
  • 45) 0.106 031 683 849 536 143 36 × 2 = 0 + 0.212 063 367 699 072 286 72;
  • 46) 0.212 063 367 699 072 286 72 × 2 = 0 + 0.424 126 735 398 144 573 44;
  • 47) 0.424 126 735 398 144 573 44 × 2 = 0 + 0.848 253 470 796 289 146 88;
  • 48) 0.848 253 470 796 289 146 88 × 2 = 1 + 0.696 506 941 592 578 293 76;
  • 49) 0.696 506 941 592 578 293 76 × 2 = 1 + 0.393 013 883 185 156 587 52;
  • 50) 0.393 013 883 185 156 587 52 × 2 = 0 + 0.786 027 766 370 313 175 04;
  • 51) 0.786 027 766 370 313 175 04 × 2 = 1 + 0.572 055 532 740 626 350 08;
  • 52) 0.572 055 532 740 626 350 08 × 2 = 1 + 0.144 111 065 481 252 700 16;
  • 53) 0.144 111 065 481 252 700 16 × 2 = 0 + 0.288 222 130 962 505 400 32;
  • 54) 0.288 222 130 962 505 400 32 × 2 = 0 + 0.576 444 261 925 010 800 64;
  • 55) 0.576 444 261 925 010 800 64 × 2 = 1 + 0.152 888 523 850 021 601 28;
  • 56) 0.152 888 523 850 021 601 28 × 2 = 0 + 0.305 777 047 700 043 202 56;
  • 57) 0.305 777 047 700 043 202 56 × 2 = 0 + 0.611 554 095 400 086 405 12;
  • 58) 0.611 554 095 400 086 405 12 × 2 = 1 + 0.223 108 190 800 172 810 24;
  • 59) 0.223 108 190 800 172 810 24 × 2 = 0 + 0.446 216 381 600 345 620 48;
  • 60) 0.446 216 381 600 345 620 48 × 2 = 0 + 0.892 432 763 200 691 240 96;
  • 61) 0.892 432 763 200 691 240 96 × 2 = 1 + 0.784 865 526 401 382 481 92;
  • 62) 0.784 865 526 401 382 481 92 × 2 = 1 + 0.569 731 052 802 764 963 84;
  • 63) 0.569 731 052 802 764 963 84 × 2 = 1 + 0.139 462 105 605 529 927 68;
  • 64) 0.139 462 105 605 529 927 68 × 2 = 0 + 0.278 924 211 211 059 855 36;
  • 65) 0.278 924 211 211 059 855 36 × 2 = 0 + 0.557 848 422 422 119 710 72;
  • 66) 0.557 848 422 422 119 710 72 × 2 = 1 + 0.115 696 844 844 239 421 44;
  • 67) 0.115 696 844 844 239 421 44 × 2 = 0 + 0.231 393 689 688 478 842 88;
  • 68) 0.231 393 689 688 478 842 88 × 2 = 0 + 0.462 787 379 376 957 685 76;
  • 69) 0.462 787 379 376 957 685 76 × 2 = 0 + 0.925 574 758 753 915 371 52;
  • 70) 0.925 574 758 753 915 371 52 × 2 = 1 + 0.851 149 517 507 830 743 04;
  • 71) 0.851 149 517 507 830 743 04 × 2 = 1 + 0.702 299 035 015 661 486 08;
  • 72) 0.702 299 035 015 661 486 08 × 2 = 1 + 0.404 598 070 031 322 972 16;
  • 73) 0.404 598 070 031 322 972 16 × 2 = 0 + 0.809 196 140 062 645 944 32;
  • 74) 0.809 196 140 062 645 944 32 × 2 = 1 + 0.618 392 280 125 291 888 64;
  • 75) 0.618 392 280 125 291 888 64 × 2 = 1 + 0.236 784 560 250 583 777 28;
  • 76) 0.236 784 560 250 583 777 28 × 2 = 0 + 0.473 569 120 501 167 554 56;
  • 77) 0.473 569 120 501 167 554 56 × 2 = 0 + 0.947 138 241 002 335 109 12;
  • 78) 0.947 138 241 002 335 109 12 × 2 = 1 + 0.894 276 482 004 670 218 24;
  • 79) 0.894 276 482 004 670 218 24 × 2 = 1 + 0.788 552 964 009 340 436 48;
  • 80) 0.788 552 964 009 340 436 48 × 2 = 1 + 0.577 105 928 018 680 872 96;
  • 81) 0.577 105 928 018 680 872 96 × 2 = 1 + 0.154 211 856 037 361 745 92;
  • 82) 0.154 211 856 037 361 745 92 × 2 = 0 + 0.308 423 712 074 723 491 84;
  • 83) 0.308 423 712 074 723 491 84 × 2 = 0 + 0.616 847 424 149 446 983 68;
  • 84) 0.616 847 424 149 446 983 68 × 2 = 1 + 0.233 694 848 298 893 967 36;
  • 85) 0.233 694 848 298 893 967 36 × 2 = 0 + 0.467 389 696 597 787 934 72;
  • 86) 0.467 389 696 597 787 934 72 × 2 = 0 + 0.934 779 393 195 575 869 44;
  • 87) 0.934 779 393 195 575 869 44 × 2 = 1 + 0.869 558 786 391 151 738 88;
  • 88) 0.869 558 786 391 151 738 88 × 2 = 1 + 0.739 117 572 782 303 477 76;
  • 89) 0.739 117 572 782 303 477 76 × 2 = 1 + 0.478 235 145 564 606 955 52;
  • 90) 0.478 235 145 564 606 955 52 × 2 = 0 + 0.956 470 291 129 213 911 04;
  • 91) 0.956 470 291 129 213 911 04 × 2 = 1 + 0.912 940 582 258 427 822 08;
  • 92) 0.912 940 582 258 427 822 08 × 2 = 1 + 0.825 881 164 516 855 644 16;
  • 93) 0.825 881 164 516 855 644 16 × 2 = 1 + 0.651 762 329 033 711 288 32;
  • 94) 0.651 762 329 033 711 288 32 × 2 = 1 + 0.303 524 658 067 422 576 64;
  • 95) 0.303 524 658 067 422 576 64 × 2 = 0 + 0.607 049 316 134 845 153 28;

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 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 0100 0111 0110 0111 1001 0011 1011 110(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 0100 0111 0110 0111 1001 0011 1011 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 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 0100 0111 0110 0111 1001 0011 1011 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 0100 0111 0110 0111 1001 0011 1011 110(2) × 20 =

1.1000 1101 1001 0010 0111 0010 0011 1011 0011 1100 1001 1101 1110(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 0011 1011 0011 1100 1001 1101 1110


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 0011 1011 0011 1100 1001 1101 1110 =

1000 1101 1001 0010 0111 0010 0011 1011 0011 1100 1001 1101 1110


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 0011 1011 0011 1100 1001 1101 1110


Decimal number -0.000 000 000 000 176 557 46 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 0010 0011 1011 0011 1100 1001 1101 1110

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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