-0.000 000 000 000 014 349 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 349 3(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
-0.000 000 000 000 014 349 3(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:

|-0.000 000 000 000 014 349 3| = 0.000 000 000 000 014 349 3


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: 0.000 000 000 000 014 349 3.

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) 0.000 000 000 000 014 349 3 × 2 = 0 + 0.000 000 000 000 028 698 6;
  • 2) 0.000 000 000 000 028 698 6 × 2 = 0 + 0.000 000 000 000 057 397 2;
  • 3) 0.000 000 000 000 057 397 2 × 2 = 0 + 0.000 000 000 000 114 794 4;
  • 4) 0.000 000 000 000 114 794 4 × 2 = 0 + 0.000 000 000 000 229 588 8;
  • 5) 0.000 000 000 000 229 588 8 × 2 = 0 + 0.000 000 000 000 459 177 6;
  • 6) 0.000 000 000 000 459 177 6 × 2 = 0 + 0.000 000 000 000 918 355 2;
  • 7) 0.000 000 000 000 918 355 2 × 2 = 0 + 0.000 000 000 001 836 710 4;
  • 8) 0.000 000 000 001 836 710 4 × 2 = 0 + 0.000 000 000 003 673 420 8;
  • 9) 0.000 000 000 003 673 420 8 × 2 = 0 + 0.000 000 000 007 346 841 6;
  • 10) 0.000 000 000 007 346 841 6 × 2 = 0 + 0.000 000 000 014 693 683 2;
  • 11) 0.000 000 000 014 693 683 2 × 2 = 0 + 0.000 000 000 029 387 366 4;
  • 12) 0.000 000 000 029 387 366 4 × 2 = 0 + 0.000 000 000 058 774 732 8;
  • 13) 0.000 000 000 058 774 732 8 × 2 = 0 + 0.000 000 000 117 549 465 6;
  • 14) 0.000 000 000 117 549 465 6 × 2 = 0 + 0.000 000 000 235 098 931 2;
  • 15) 0.000 000 000 235 098 931 2 × 2 = 0 + 0.000 000 000 470 197 862 4;
  • 16) 0.000 000 000 470 197 862 4 × 2 = 0 + 0.000 000 000 940 395 724 8;
  • 17) 0.000 000 000 940 395 724 8 × 2 = 0 + 0.000 000 001 880 791 449 6;
  • 18) 0.000 000 001 880 791 449 6 × 2 = 0 + 0.000 000 003 761 582 899 2;
  • 19) 0.000 000 003 761 582 899 2 × 2 = 0 + 0.000 000 007 523 165 798 4;
  • 20) 0.000 000 007 523 165 798 4 × 2 = 0 + 0.000 000 015 046 331 596 8;
  • 21) 0.000 000 015 046 331 596 8 × 2 = 0 + 0.000 000 030 092 663 193 6;
  • 22) 0.000 000 030 092 663 193 6 × 2 = 0 + 0.000 000 060 185 326 387 2;
  • 23) 0.000 000 060 185 326 387 2 × 2 = 0 + 0.000 000 120 370 652 774 4;
  • 24) 0.000 000 120 370 652 774 4 × 2 = 0 + 0.000 000 240 741 305 548 8;
  • 25) 0.000 000 240 741 305 548 8 × 2 = 0 + 0.000 000 481 482 611 097 6;
  • 26) 0.000 000 481 482 611 097 6 × 2 = 0 + 0.000 000 962 965 222 195 2;
  • 27) 0.000 000 962 965 222 195 2 × 2 = 0 + 0.000 001 925 930 444 390 4;
  • 28) 0.000 001 925 930 444 390 4 × 2 = 0 + 0.000 003 851 860 888 780 8;
  • 29) 0.000 003 851 860 888 780 8 × 2 = 0 + 0.000 007 703 721 777 561 6;
  • 30) 0.000 007 703 721 777 561 6 × 2 = 0 + 0.000 015 407 443 555 123 2;
  • 31) 0.000 015 407 443 555 123 2 × 2 = 0 + 0.000 030 814 887 110 246 4;
  • 32) 0.000 030 814 887 110 246 4 × 2 = 0 + 0.000 061 629 774 220 492 8;
  • 33) 0.000 061 629 774 220 492 8 × 2 = 0 + 0.000 123 259 548 440 985 6;
  • 34) 0.000 123 259 548 440 985 6 × 2 = 0 + 0.000 246 519 096 881 971 2;
  • 35) 0.000 246 519 096 881 971 2 × 2 = 0 + 0.000 493 038 193 763 942 4;
  • 36) 0.000 493 038 193 763 942 4 × 2 = 0 + 0.000 986 076 387 527 884 8;
  • 37) 0.000 986 076 387 527 884 8 × 2 = 0 + 0.001 972 152 775 055 769 6;
  • 38) 0.001 972 152 775 055 769 6 × 2 = 0 + 0.003 944 305 550 111 539 2;
  • 39) 0.003 944 305 550 111 539 2 × 2 = 0 + 0.007 888 611 100 223 078 4;
  • 40) 0.007 888 611 100 223 078 4 × 2 = 0 + 0.015 777 222 200 446 156 8;
  • 41) 0.015 777 222 200 446 156 8 × 2 = 0 + 0.031 554 444 400 892 313 6;
  • 42) 0.031 554 444 400 892 313 6 × 2 = 0 + 0.063 108 888 801 784 627 2;
  • 43) 0.063 108 888 801 784 627 2 × 2 = 0 + 0.126 217 777 603 569 254 4;
  • 44) 0.126 217 777 603 569 254 4 × 2 = 0 + 0.252 435 555 207 138 508 8;
  • 45) 0.252 435 555 207 138 508 8 × 2 = 0 + 0.504 871 110 414 277 017 6;
  • 46) 0.504 871 110 414 277 017 6 × 2 = 1 + 0.009 742 220 828 554 035 2;
  • 47) 0.009 742 220 828 554 035 2 × 2 = 0 + 0.019 484 441 657 108 070 4;
  • 48) 0.019 484 441 657 108 070 4 × 2 = 0 + 0.038 968 883 314 216 140 8;
  • 49) 0.038 968 883 314 216 140 8 × 2 = 0 + 0.077 937 766 628 432 281 6;
  • 50) 0.077 937 766 628 432 281 6 × 2 = 0 + 0.155 875 533 256 864 563 2;
  • 51) 0.155 875 533 256 864 563 2 × 2 = 0 + 0.311 751 066 513 729 126 4;
  • 52) 0.311 751 066 513 729 126 4 × 2 = 0 + 0.623 502 133 027 458 252 8;
  • 53) 0.623 502 133 027 458 252 8 × 2 = 1 + 0.247 004 266 054 916 505 6;
  • 54) 0.247 004 266 054 916 505 6 × 2 = 0 + 0.494 008 532 109 833 011 2;
  • 55) 0.494 008 532 109 833 011 2 × 2 = 0 + 0.988 017 064 219 666 022 4;
  • 56) 0.988 017 064 219 666 022 4 × 2 = 1 + 0.976 034 128 439 332 044 8;
  • 57) 0.976 034 128 439 332 044 8 × 2 = 1 + 0.952 068 256 878 664 089 6;
  • 58) 0.952 068 256 878 664 089 6 × 2 = 1 + 0.904 136 513 757 328 179 2;
  • 59) 0.904 136 513 757 328 179 2 × 2 = 1 + 0.808 273 027 514 656 358 4;
  • 60) 0.808 273 027 514 656 358 4 × 2 = 1 + 0.616 546 055 029 312 716 8;
  • 61) 0.616 546 055 029 312 716 8 × 2 = 1 + 0.233 092 110 058 625 433 6;
  • 62) 0.233 092 110 058 625 433 6 × 2 = 0 + 0.466 184 220 117 250 867 2;
  • 63) 0.466 184 220 117 250 867 2 × 2 = 0 + 0.932 368 440 234 501 734 4;
  • 64) 0.932 368 440 234 501 734 4 × 2 = 1 + 0.864 736 880 469 003 468 8;
  • 65) 0.864 736 880 469 003 468 8 × 2 = 1 + 0.729 473 760 938 006 937 6;
  • 66) 0.729 473 760 938 006 937 6 × 2 = 1 + 0.458 947 521 876 013 875 2;
  • 67) 0.458 947 521 876 013 875 2 × 2 = 0 + 0.917 895 043 752 027 750 4;
  • 68) 0.917 895 043 752 027 750 4 × 2 = 1 + 0.835 790 087 504 055 500 8;
  • 69) 0.835 790 087 504 055 500 8 × 2 = 1 + 0.671 580 175 008 111 001 6;
  • 70) 0.671 580 175 008 111 001 6 × 2 = 1 + 0.343 160 350 016 222 003 2;
  • 71) 0.343 160 350 016 222 003 2 × 2 = 0 + 0.686 320 700 032 444 006 4;
  • 72) 0.686 320 700 032 444 006 4 × 2 = 1 + 0.372 641 400 064 888 012 8;
  • 73) 0.372 641 400 064 888 012 8 × 2 = 0 + 0.745 282 800 129 776 025 6;
  • 74) 0.745 282 800 129 776 025 6 × 2 = 1 + 0.490 565 600 259 552 051 2;
  • 75) 0.490 565 600 259 552 051 2 × 2 = 0 + 0.981 131 200 519 104 102 4;
  • 76) 0.981 131 200 519 104 102 4 × 2 = 1 + 0.962 262 401 038 208 204 8;
  • 77) 0.962 262 401 038 208 204 8 × 2 = 1 + 0.924 524 802 076 416 409 6;
  • 78) 0.924 524 802 076 416 409 6 × 2 = 1 + 0.849 049 604 152 832 819 2;
  • 79) 0.849 049 604 152 832 819 2 × 2 = 1 + 0.698 099 208 305 665 638 4;
  • 80) 0.698 099 208 305 665 638 4 × 2 = 1 + 0.396 198 416 611 331 276 8;
  • 81) 0.396 198 416 611 331 276 8 × 2 = 0 + 0.792 396 833 222 662 553 6;
  • 82) 0.792 396 833 222 662 553 6 × 2 = 1 + 0.584 793 666 445 325 107 2;
  • 83) 0.584 793 666 445 325 107 2 × 2 = 1 + 0.169 587 332 890 650 214 4;
  • 84) 0.169 587 332 890 650 214 4 × 2 = 0 + 0.339 174 665 781 300 428 8;
  • 85) 0.339 174 665 781 300 428 8 × 2 = 0 + 0.678 349 331 562 600 857 6;
  • 86) 0.678 349 331 562 600 857 6 × 2 = 1 + 0.356 698 663 125 201 715 2;
  • 87) 0.356 698 663 125 201 715 2 × 2 = 0 + 0.713 397 326 250 403 430 4;
  • 88) 0.713 397 326 250 403 430 4 × 2 = 1 + 0.426 794 652 500 806 860 8;
  • 89) 0.426 794 652 500 806 860 8 × 2 = 0 + 0.853 589 305 001 613 721 6;
  • 90) 0.853 589 305 001 613 721 6 × 2 = 1 + 0.707 178 610 003 227 443 2;
  • 91) 0.707 178 610 003 227 443 2 × 2 = 1 + 0.414 357 220 006 454 886 4;
  • 92) 0.414 357 220 006 454 886 4 × 2 = 0 + 0.828 714 440 012 909 772 8;
  • 93) 0.828 714 440 012 909 772 8 × 2 = 1 + 0.657 428 880 025 819 545 6;
  • 94) 0.657 428 880 025 819 545 6 × 2 = 1 + 0.314 857 760 051 639 091 2;
  • 95) 0.314 857 760 051 639 091 2 × 2 = 0 + 0.629 715 520 103 278 182 4;
  • 96) 0.629 715 520 103 278 182 4 × 2 = 1 + 0.259 431 040 206 556 364 8;
  • 97) 0.259 431 040 206 556 364 8 × 2 = 0 + 0.518 862 080 413 112 729 6;
  • 98) 0.518 862 080 413 112 729 6 × 2 = 1 + 0.037 724 160 826 225 459 2;

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:


0.000 000 000 000 014 349 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1001 1111 1001 1101 1101 0101 1111 0110 0101 0110 1101 01(2)

6. Positive number before normalization:

0.000 000 000 000 014 349 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1001 1111 1001 1101 1101 0101 1111 0110 0101 0110 1101 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:


0.000 000 000 000 014 349 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1001 1111 1001 1101 1101 0101 1111 0110 0101 0110 1101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1001 1111 1001 1101 1101 0101 1111 0110 0101 0110 1101 01(2) × 20 =

1.0000 0010 0111 1110 0111 0111 0101 0111 1101 1001 0101 1011 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 0111 1110 0111 0111 0101 0111 1101 1001 0101 1011 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 0111 1110 0111 0111 0101 0111 1101 1001 0101 1011 0101 =

0000 0010 0111 1110 0111 0111 0101 0111 1101 1001 0101 1011 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 0111 1110 0111 0111 0101 0111 1101 1001 0101 1011 0101


Decimal number -0.000 000 000 000 014 349 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 0111 1110 0111 0111 0101 0111 1101 1001 0101 1011 0101

» Convert decimal -0.000 000 000 000 014 350 7 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