-0.000 000 000 000 014 374 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 374 2(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
-0.000 000 000 000 014 374 2(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:

|-0.000 000 000 000 014 374 2| = 0.000 000 000 000 014 374 2


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: 0.000 000 000 000 014 374 2.

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) 0.000 000 000 000 014 374 2 × 2 = 0 + 0.000 000 000 000 028 748 4;
  • 2) 0.000 000 000 000 028 748 4 × 2 = 0 + 0.000 000 000 000 057 496 8;
  • 3) 0.000 000 000 000 057 496 8 × 2 = 0 + 0.000 000 000 000 114 993 6;
  • 4) 0.000 000 000 000 114 993 6 × 2 = 0 + 0.000 000 000 000 229 987 2;
  • 5) 0.000 000 000 000 229 987 2 × 2 = 0 + 0.000 000 000 000 459 974 4;
  • 6) 0.000 000 000 000 459 974 4 × 2 = 0 + 0.000 000 000 000 919 948 8;
  • 7) 0.000 000 000 000 919 948 8 × 2 = 0 + 0.000 000 000 001 839 897 6;
  • 8) 0.000 000 000 001 839 897 6 × 2 = 0 + 0.000 000 000 003 679 795 2;
  • 9) 0.000 000 000 003 679 795 2 × 2 = 0 + 0.000 000 000 007 359 590 4;
  • 10) 0.000 000 000 007 359 590 4 × 2 = 0 + 0.000 000 000 014 719 180 8;
  • 11) 0.000 000 000 014 719 180 8 × 2 = 0 + 0.000 000 000 029 438 361 6;
  • 12) 0.000 000 000 029 438 361 6 × 2 = 0 + 0.000 000 000 058 876 723 2;
  • 13) 0.000 000 000 058 876 723 2 × 2 = 0 + 0.000 000 000 117 753 446 4;
  • 14) 0.000 000 000 117 753 446 4 × 2 = 0 + 0.000 000 000 235 506 892 8;
  • 15) 0.000 000 000 235 506 892 8 × 2 = 0 + 0.000 000 000 471 013 785 6;
  • 16) 0.000 000 000 471 013 785 6 × 2 = 0 + 0.000 000 000 942 027 571 2;
  • 17) 0.000 000 000 942 027 571 2 × 2 = 0 + 0.000 000 001 884 055 142 4;
  • 18) 0.000 000 001 884 055 142 4 × 2 = 0 + 0.000 000 003 768 110 284 8;
  • 19) 0.000 000 003 768 110 284 8 × 2 = 0 + 0.000 000 007 536 220 569 6;
  • 20) 0.000 000 007 536 220 569 6 × 2 = 0 + 0.000 000 015 072 441 139 2;
  • 21) 0.000 000 015 072 441 139 2 × 2 = 0 + 0.000 000 030 144 882 278 4;
  • 22) 0.000 000 030 144 882 278 4 × 2 = 0 + 0.000 000 060 289 764 556 8;
  • 23) 0.000 000 060 289 764 556 8 × 2 = 0 + 0.000 000 120 579 529 113 6;
  • 24) 0.000 000 120 579 529 113 6 × 2 = 0 + 0.000 000 241 159 058 227 2;
  • 25) 0.000 000 241 159 058 227 2 × 2 = 0 + 0.000 000 482 318 116 454 4;
  • 26) 0.000 000 482 318 116 454 4 × 2 = 0 + 0.000 000 964 636 232 908 8;
  • 27) 0.000 000 964 636 232 908 8 × 2 = 0 + 0.000 001 929 272 465 817 6;
  • 28) 0.000 001 929 272 465 817 6 × 2 = 0 + 0.000 003 858 544 931 635 2;
  • 29) 0.000 003 858 544 931 635 2 × 2 = 0 + 0.000 007 717 089 863 270 4;
  • 30) 0.000 007 717 089 863 270 4 × 2 = 0 + 0.000 015 434 179 726 540 8;
  • 31) 0.000 015 434 179 726 540 8 × 2 = 0 + 0.000 030 868 359 453 081 6;
  • 32) 0.000 030 868 359 453 081 6 × 2 = 0 + 0.000 061 736 718 906 163 2;
  • 33) 0.000 061 736 718 906 163 2 × 2 = 0 + 0.000 123 473 437 812 326 4;
  • 34) 0.000 123 473 437 812 326 4 × 2 = 0 + 0.000 246 946 875 624 652 8;
  • 35) 0.000 246 946 875 624 652 8 × 2 = 0 + 0.000 493 893 751 249 305 6;
  • 36) 0.000 493 893 751 249 305 6 × 2 = 0 + 0.000 987 787 502 498 611 2;
  • 37) 0.000 987 787 502 498 611 2 × 2 = 0 + 0.001 975 575 004 997 222 4;
  • 38) 0.001 975 575 004 997 222 4 × 2 = 0 + 0.003 951 150 009 994 444 8;
  • 39) 0.003 951 150 009 994 444 8 × 2 = 0 + 0.007 902 300 019 988 889 6;
  • 40) 0.007 902 300 019 988 889 6 × 2 = 0 + 0.015 804 600 039 977 779 2;
  • 41) 0.015 804 600 039 977 779 2 × 2 = 0 + 0.031 609 200 079 955 558 4;
  • 42) 0.031 609 200 079 955 558 4 × 2 = 0 + 0.063 218 400 159 911 116 8;
  • 43) 0.063 218 400 159 911 116 8 × 2 = 0 + 0.126 436 800 319 822 233 6;
  • 44) 0.126 436 800 319 822 233 6 × 2 = 0 + 0.252 873 600 639 644 467 2;
  • 45) 0.252 873 600 639 644 467 2 × 2 = 0 + 0.505 747 201 279 288 934 4;
  • 46) 0.505 747 201 279 288 934 4 × 2 = 1 + 0.011 494 402 558 577 868 8;
  • 47) 0.011 494 402 558 577 868 8 × 2 = 0 + 0.022 988 805 117 155 737 6;
  • 48) 0.022 988 805 117 155 737 6 × 2 = 0 + 0.045 977 610 234 311 475 2;
  • 49) 0.045 977 610 234 311 475 2 × 2 = 0 + 0.091 955 220 468 622 950 4;
  • 50) 0.091 955 220 468 622 950 4 × 2 = 0 + 0.183 910 440 937 245 900 8;
  • 51) 0.183 910 440 937 245 900 8 × 2 = 0 + 0.367 820 881 874 491 801 6;
  • 52) 0.367 820 881 874 491 801 6 × 2 = 0 + 0.735 641 763 748 983 603 2;
  • 53) 0.735 641 763 748 983 603 2 × 2 = 1 + 0.471 283 527 497 967 206 4;
  • 54) 0.471 283 527 497 967 206 4 × 2 = 0 + 0.942 567 054 995 934 412 8;
  • 55) 0.942 567 054 995 934 412 8 × 2 = 1 + 0.885 134 109 991 868 825 6;
  • 56) 0.885 134 109 991 868 825 6 × 2 = 1 + 0.770 268 219 983 737 651 2;
  • 57) 0.770 268 219 983 737 651 2 × 2 = 1 + 0.540 536 439 967 475 302 4;
  • 58) 0.540 536 439 967 475 302 4 × 2 = 1 + 0.081 072 879 934 950 604 8;
  • 59) 0.081 072 879 934 950 604 8 × 2 = 0 + 0.162 145 759 869 901 209 6;
  • 60) 0.162 145 759 869 901 209 6 × 2 = 0 + 0.324 291 519 739 802 419 2;
  • 61) 0.324 291 519 739 802 419 2 × 2 = 0 + 0.648 583 039 479 604 838 4;
  • 62) 0.648 583 039 479 604 838 4 × 2 = 1 + 0.297 166 078 959 209 676 8;
  • 63) 0.297 166 078 959 209 676 8 × 2 = 0 + 0.594 332 157 918 419 353 6;
  • 64) 0.594 332 157 918 419 353 6 × 2 = 1 + 0.188 664 315 836 838 707 2;
  • 65) 0.188 664 315 836 838 707 2 × 2 = 0 + 0.377 328 631 673 677 414 4;
  • 66) 0.377 328 631 673 677 414 4 × 2 = 0 + 0.754 657 263 347 354 828 8;
  • 67) 0.754 657 263 347 354 828 8 × 2 = 1 + 0.509 314 526 694 709 657 6;
  • 68) 0.509 314 526 694 709 657 6 × 2 = 1 + 0.018 629 053 389 419 315 2;
  • 69) 0.018 629 053 389 419 315 2 × 2 = 0 + 0.037 258 106 778 838 630 4;
  • 70) 0.037 258 106 778 838 630 4 × 2 = 0 + 0.074 516 213 557 677 260 8;
  • 71) 0.074 516 213 557 677 260 8 × 2 = 0 + 0.149 032 427 115 354 521 6;
  • 72) 0.149 032 427 115 354 521 6 × 2 = 0 + 0.298 064 854 230 709 043 2;
  • 73) 0.298 064 854 230 709 043 2 × 2 = 0 + 0.596 129 708 461 418 086 4;
  • 74) 0.596 129 708 461 418 086 4 × 2 = 1 + 0.192 259 416 922 836 172 8;
  • 75) 0.192 259 416 922 836 172 8 × 2 = 0 + 0.384 518 833 845 672 345 6;
  • 76) 0.384 518 833 845 672 345 6 × 2 = 0 + 0.769 037 667 691 344 691 2;
  • 77) 0.769 037 667 691 344 691 2 × 2 = 1 + 0.538 075 335 382 689 382 4;
  • 78) 0.538 075 335 382 689 382 4 × 2 = 1 + 0.076 150 670 765 378 764 8;
  • 79) 0.076 150 670 765 378 764 8 × 2 = 0 + 0.152 301 341 530 757 529 6;
  • 80) 0.152 301 341 530 757 529 6 × 2 = 0 + 0.304 602 683 061 515 059 2;
  • 81) 0.304 602 683 061 515 059 2 × 2 = 0 + 0.609 205 366 123 030 118 4;
  • 82) 0.609 205 366 123 030 118 4 × 2 = 1 + 0.218 410 732 246 060 236 8;
  • 83) 0.218 410 732 246 060 236 8 × 2 = 0 + 0.436 821 464 492 120 473 6;
  • 84) 0.436 821 464 492 120 473 6 × 2 = 0 + 0.873 642 928 984 240 947 2;
  • 85) 0.873 642 928 984 240 947 2 × 2 = 1 + 0.747 285 857 968 481 894 4;
  • 86) 0.747 285 857 968 481 894 4 × 2 = 1 + 0.494 571 715 936 963 788 8;
  • 87) 0.494 571 715 936 963 788 8 × 2 = 0 + 0.989 143 431 873 927 577 6;
  • 88) 0.989 143 431 873 927 577 6 × 2 = 1 + 0.978 286 863 747 855 155 2;
  • 89) 0.978 286 863 747 855 155 2 × 2 = 1 + 0.956 573 727 495 710 310 4;
  • 90) 0.956 573 727 495 710 310 4 × 2 = 1 + 0.913 147 454 991 420 620 8;
  • 91) 0.913 147 454 991 420 620 8 × 2 = 1 + 0.826 294 909 982 841 241 6;
  • 92) 0.826 294 909 982 841 241 6 × 2 = 1 + 0.652 589 819 965 682 483 2;
  • 93) 0.652 589 819 965 682 483 2 × 2 = 1 + 0.305 179 639 931 364 966 4;
  • 94) 0.305 179 639 931 364 966 4 × 2 = 0 + 0.610 359 279 862 729 932 8;
  • 95) 0.610 359 279 862 729 932 8 × 2 = 1 + 0.220 718 559 725 459 865 6;
  • 96) 0.220 718 559 725 459 865 6 × 2 = 0 + 0.441 437 119 450 919 731 2;
  • 97) 0.441 437 119 450 919 731 2 × 2 = 0 + 0.882 874 238 901 839 462 4;
  • 98) 0.882 874 238 901 839 462 4 × 2 = 1 + 0.765 748 477 803 678 924 8;

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:


0.000 000 000 000 014 374 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1100 0101 0011 0000 0100 1100 0100 1101 1111 1010 01(2)

6. Positive number before normalization:

0.000 000 000 000 014 374 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1100 0101 0011 0000 0100 1100 0100 1101 1111 1010 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:


0.000 000 000 000 014 374 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1100 0101 0011 0000 0100 1100 0100 1101 1111 1010 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1100 0101 0011 0000 0100 1100 0100 1101 1111 1010 01(2) × 20 =

1.0000 0010 1111 0001 0100 1100 0001 0011 0001 0011 0111 1110 1001(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 1111 0001 0100 1100 0001 0011 0001 0011 0111 1110 1001


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 1111 0001 0100 1100 0001 0011 0001 0011 0111 1110 1001 =

0000 0010 1111 0001 0100 1100 0001 0011 0001 0011 0111 1110 1001


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 1111 0001 0100 1100 0001 0011 0001 0011 0111 1110 1001


Decimal number -0.000 000 000 000 014 374 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1111 0001 0100 1100 0001 0011 0001 0011 0111 1110 1001

» Convert decimal -0.000 000 000 000 014 369 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