-0.000 000 000 000 014 370 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 370 9(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
-0.000 000 000 000 014 370 9(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:

|-0.000 000 000 000 014 370 9| = 0.000 000 000 000 014 370 9


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: 0.000 000 000 000 014 370 9.

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) 0.000 000 000 000 014 370 9 × 2 = 0 + 0.000 000 000 000 028 741 8;
  • 2) 0.000 000 000 000 028 741 8 × 2 = 0 + 0.000 000 000 000 057 483 6;
  • 3) 0.000 000 000 000 057 483 6 × 2 = 0 + 0.000 000 000 000 114 967 2;
  • 4) 0.000 000 000 000 114 967 2 × 2 = 0 + 0.000 000 000 000 229 934 4;
  • 5) 0.000 000 000 000 229 934 4 × 2 = 0 + 0.000 000 000 000 459 868 8;
  • 6) 0.000 000 000 000 459 868 8 × 2 = 0 + 0.000 000 000 000 919 737 6;
  • 7) 0.000 000 000 000 919 737 6 × 2 = 0 + 0.000 000 000 001 839 475 2;
  • 8) 0.000 000 000 001 839 475 2 × 2 = 0 + 0.000 000 000 003 678 950 4;
  • 9) 0.000 000 000 003 678 950 4 × 2 = 0 + 0.000 000 000 007 357 900 8;
  • 10) 0.000 000 000 007 357 900 8 × 2 = 0 + 0.000 000 000 014 715 801 6;
  • 11) 0.000 000 000 014 715 801 6 × 2 = 0 + 0.000 000 000 029 431 603 2;
  • 12) 0.000 000 000 029 431 603 2 × 2 = 0 + 0.000 000 000 058 863 206 4;
  • 13) 0.000 000 000 058 863 206 4 × 2 = 0 + 0.000 000 000 117 726 412 8;
  • 14) 0.000 000 000 117 726 412 8 × 2 = 0 + 0.000 000 000 235 452 825 6;
  • 15) 0.000 000 000 235 452 825 6 × 2 = 0 + 0.000 000 000 470 905 651 2;
  • 16) 0.000 000 000 470 905 651 2 × 2 = 0 + 0.000 000 000 941 811 302 4;
  • 17) 0.000 000 000 941 811 302 4 × 2 = 0 + 0.000 000 001 883 622 604 8;
  • 18) 0.000 000 001 883 622 604 8 × 2 = 0 + 0.000 000 003 767 245 209 6;
  • 19) 0.000 000 003 767 245 209 6 × 2 = 0 + 0.000 000 007 534 490 419 2;
  • 20) 0.000 000 007 534 490 419 2 × 2 = 0 + 0.000 000 015 068 980 838 4;
  • 21) 0.000 000 015 068 980 838 4 × 2 = 0 + 0.000 000 030 137 961 676 8;
  • 22) 0.000 000 030 137 961 676 8 × 2 = 0 + 0.000 000 060 275 923 353 6;
  • 23) 0.000 000 060 275 923 353 6 × 2 = 0 + 0.000 000 120 551 846 707 2;
  • 24) 0.000 000 120 551 846 707 2 × 2 = 0 + 0.000 000 241 103 693 414 4;
  • 25) 0.000 000 241 103 693 414 4 × 2 = 0 + 0.000 000 482 207 386 828 8;
  • 26) 0.000 000 482 207 386 828 8 × 2 = 0 + 0.000 000 964 414 773 657 6;
  • 27) 0.000 000 964 414 773 657 6 × 2 = 0 + 0.000 001 928 829 547 315 2;
  • 28) 0.000 001 928 829 547 315 2 × 2 = 0 + 0.000 003 857 659 094 630 4;
  • 29) 0.000 003 857 659 094 630 4 × 2 = 0 + 0.000 007 715 318 189 260 8;
  • 30) 0.000 007 715 318 189 260 8 × 2 = 0 + 0.000 015 430 636 378 521 6;
  • 31) 0.000 015 430 636 378 521 6 × 2 = 0 + 0.000 030 861 272 757 043 2;
  • 32) 0.000 030 861 272 757 043 2 × 2 = 0 + 0.000 061 722 545 514 086 4;
  • 33) 0.000 061 722 545 514 086 4 × 2 = 0 + 0.000 123 445 091 028 172 8;
  • 34) 0.000 123 445 091 028 172 8 × 2 = 0 + 0.000 246 890 182 056 345 6;
  • 35) 0.000 246 890 182 056 345 6 × 2 = 0 + 0.000 493 780 364 112 691 2;
  • 36) 0.000 493 780 364 112 691 2 × 2 = 0 + 0.000 987 560 728 225 382 4;
  • 37) 0.000 987 560 728 225 382 4 × 2 = 0 + 0.001 975 121 456 450 764 8;
  • 38) 0.001 975 121 456 450 764 8 × 2 = 0 + 0.003 950 242 912 901 529 6;
  • 39) 0.003 950 242 912 901 529 6 × 2 = 0 + 0.007 900 485 825 803 059 2;
  • 40) 0.007 900 485 825 803 059 2 × 2 = 0 + 0.015 800 971 651 606 118 4;
  • 41) 0.015 800 971 651 606 118 4 × 2 = 0 + 0.031 601 943 303 212 236 8;
  • 42) 0.031 601 943 303 212 236 8 × 2 = 0 + 0.063 203 886 606 424 473 6;
  • 43) 0.063 203 886 606 424 473 6 × 2 = 0 + 0.126 407 773 212 848 947 2;
  • 44) 0.126 407 773 212 848 947 2 × 2 = 0 + 0.252 815 546 425 697 894 4;
  • 45) 0.252 815 546 425 697 894 4 × 2 = 0 + 0.505 631 092 851 395 788 8;
  • 46) 0.505 631 092 851 395 788 8 × 2 = 1 + 0.011 262 185 702 791 577 6;
  • 47) 0.011 262 185 702 791 577 6 × 2 = 0 + 0.022 524 371 405 583 155 2;
  • 48) 0.022 524 371 405 583 155 2 × 2 = 0 + 0.045 048 742 811 166 310 4;
  • 49) 0.045 048 742 811 166 310 4 × 2 = 0 + 0.090 097 485 622 332 620 8;
  • 50) 0.090 097 485 622 332 620 8 × 2 = 0 + 0.180 194 971 244 665 241 6;
  • 51) 0.180 194 971 244 665 241 6 × 2 = 0 + 0.360 389 942 489 330 483 2;
  • 52) 0.360 389 942 489 330 483 2 × 2 = 0 + 0.720 779 884 978 660 966 4;
  • 53) 0.720 779 884 978 660 966 4 × 2 = 1 + 0.441 559 769 957 321 932 8;
  • 54) 0.441 559 769 957 321 932 8 × 2 = 0 + 0.883 119 539 914 643 865 6;
  • 55) 0.883 119 539 914 643 865 6 × 2 = 1 + 0.766 239 079 829 287 731 2;
  • 56) 0.766 239 079 829 287 731 2 × 2 = 1 + 0.532 478 159 658 575 462 4;
  • 57) 0.532 478 159 658 575 462 4 × 2 = 1 + 0.064 956 319 317 150 924 8;
  • 58) 0.064 956 319 317 150 924 8 × 2 = 0 + 0.129 912 638 634 301 849 6;
  • 59) 0.129 912 638 634 301 849 6 × 2 = 0 + 0.259 825 277 268 603 699 2;
  • 60) 0.259 825 277 268 603 699 2 × 2 = 0 + 0.519 650 554 537 207 398 4;
  • 61) 0.519 650 554 537 207 398 4 × 2 = 1 + 0.039 301 109 074 414 796 8;
  • 62) 0.039 301 109 074 414 796 8 × 2 = 0 + 0.078 602 218 148 829 593 6;
  • 63) 0.078 602 218 148 829 593 6 × 2 = 0 + 0.157 204 436 297 659 187 2;
  • 64) 0.157 204 436 297 659 187 2 × 2 = 0 + 0.314 408 872 595 318 374 4;
  • 65) 0.314 408 872 595 318 374 4 × 2 = 0 + 0.628 817 745 190 636 748 8;
  • 66) 0.628 817 745 190 636 748 8 × 2 = 1 + 0.257 635 490 381 273 497 6;
  • 67) 0.257 635 490 381 273 497 6 × 2 = 0 + 0.515 270 980 762 546 995 2;
  • 68) 0.515 270 980 762 546 995 2 × 2 = 1 + 0.030 541 961 525 093 990 4;
  • 69) 0.030 541 961 525 093 990 4 × 2 = 0 + 0.061 083 923 050 187 980 8;
  • 70) 0.061 083 923 050 187 980 8 × 2 = 0 + 0.122 167 846 100 375 961 6;
  • 71) 0.122 167 846 100 375 961 6 × 2 = 0 + 0.244 335 692 200 751 923 2;
  • 72) 0.244 335 692 200 751 923 2 × 2 = 0 + 0.488 671 384 401 503 846 4;
  • 73) 0.488 671 384 401 503 846 4 × 2 = 0 + 0.977 342 768 803 007 692 8;
  • 74) 0.977 342 768 803 007 692 8 × 2 = 1 + 0.954 685 537 606 015 385 6;
  • 75) 0.954 685 537 606 015 385 6 × 2 = 1 + 0.909 371 075 212 030 771 2;
  • 76) 0.909 371 075 212 030 771 2 × 2 = 1 + 0.818 742 150 424 061 542 4;
  • 77) 0.818 742 150 424 061 542 4 × 2 = 1 + 0.637 484 300 848 123 084 8;
  • 78) 0.637 484 300 848 123 084 8 × 2 = 1 + 0.274 968 601 696 246 169 6;
  • 79) 0.274 968 601 696 246 169 6 × 2 = 0 + 0.549 937 203 392 492 339 2;
  • 80) 0.549 937 203 392 492 339 2 × 2 = 1 + 0.099 874 406 784 984 678 4;
  • 81) 0.099 874 406 784 984 678 4 × 2 = 0 + 0.199 748 813 569 969 356 8;
  • 82) 0.199 748 813 569 969 356 8 × 2 = 0 + 0.399 497 627 139 938 713 6;
  • 83) 0.399 497 627 139 938 713 6 × 2 = 0 + 0.798 995 254 279 877 427 2;
  • 84) 0.798 995 254 279 877 427 2 × 2 = 1 + 0.597 990 508 559 754 854 4;
  • 85) 0.597 990 508 559 754 854 4 × 2 = 1 + 0.195 981 017 119 509 708 8;
  • 86) 0.195 981 017 119 509 708 8 × 2 = 0 + 0.391 962 034 239 019 417 6;
  • 87) 0.391 962 034 239 019 417 6 × 2 = 0 + 0.783 924 068 478 038 835 2;
  • 88) 0.783 924 068 478 038 835 2 × 2 = 1 + 0.567 848 136 956 077 670 4;
  • 89) 0.567 848 136 956 077 670 4 × 2 = 1 + 0.135 696 273 912 155 340 8;
  • 90) 0.135 696 273 912 155 340 8 × 2 = 0 + 0.271 392 547 824 310 681 6;
  • 91) 0.271 392 547 824 310 681 6 × 2 = 0 + 0.542 785 095 648 621 363 2;
  • 92) 0.542 785 095 648 621 363 2 × 2 = 1 + 0.085 570 191 297 242 726 4;
  • 93) 0.085 570 191 297 242 726 4 × 2 = 0 + 0.171 140 382 594 485 452 8;
  • 94) 0.171 140 382 594 485 452 8 × 2 = 0 + 0.342 280 765 188 970 905 6;
  • 95) 0.342 280 765 188 970 905 6 × 2 = 0 + 0.684 561 530 377 941 811 2;
  • 96) 0.684 561 530 377 941 811 2 × 2 = 1 + 0.369 123 060 755 883 622 4;
  • 97) 0.369 123 060 755 883 622 4 × 2 = 0 + 0.738 246 121 511 767 244 8;
  • 98) 0.738 246 121 511 767 244 8 × 2 = 1 + 0.476 492 243 023 534 489 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:


0.000 000 000 000 014 370 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1000 1000 0101 0000 0111 1101 0001 1001 1001 0001 01(2)

6. Positive number before normalization:

0.000 000 000 000 014 370 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1000 1000 0101 0000 0111 1101 0001 1001 1001 0001 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 014 370 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1000 1000 0101 0000 0111 1101 0001 1001 1001 0001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1000 1000 0101 0000 0111 1101 0001 1001 1001 0001 01(2) × 20 =

1.0000 0010 1110 0010 0001 0100 0001 1111 0100 0110 0110 0100 0101(2) × 2-46


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): -46

Mantissa (not normalized):
1.0000 0010 1110 0010 0001 0100 0001 1111 0100 0110 0110 0100 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

977(10) =

011 1101 0001(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. 0000 0010 1110 0010 0001 0100 0001 1111 0100 0110 0110 0100 0101 =

0000 0010 1110 0010 0001 0100 0001 1111 0100 0110 0110 0100 0101


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 0001

Mantissa (52 bits) =
0000 0010 1110 0010 0001 0100 0001 1111 0100 0110 0110 0100 0101


Decimal number -0.000 000 000 000 014 370 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1110 0010 0001 0100 0001 1111 0100 0110 0110 0100 0101

» Convert decimal -0.000 000 000 000 014 369 3 to 64 bit double precision IEEE 754 binary floating point representation standard

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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