0.000 000 000 029 103 830 456 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 029 103 830 456 12(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
0.000 000 000 029 103 830 456 12(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 029 103 830 456 12.

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) 0.000 000 000 029 103 830 456 12 × 2 = 0 + 0.000 000 000 058 207 660 912 24;
  • 2) 0.000 000 000 058 207 660 912 24 × 2 = 0 + 0.000 000 000 116 415 321 824 48;
  • 3) 0.000 000 000 116 415 321 824 48 × 2 = 0 + 0.000 000 000 232 830 643 648 96;
  • 4) 0.000 000 000 232 830 643 648 96 × 2 = 0 + 0.000 000 000 465 661 287 297 92;
  • 5) 0.000 000 000 465 661 287 297 92 × 2 = 0 + 0.000 000 000 931 322 574 595 84;
  • 6) 0.000 000 000 931 322 574 595 84 × 2 = 0 + 0.000 000 001 862 645 149 191 68;
  • 7) 0.000 000 001 862 645 149 191 68 × 2 = 0 + 0.000 000 003 725 290 298 383 36;
  • 8) 0.000 000 003 725 290 298 383 36 × 2 = 0 + 0.000 000 007 450 580 596 766 72;
  • 9) 0.000 000 007 450 580 596 766 72 × 2 = 0 + 0.000 000 014 901 161 193 533 44;
  • 10) 0.000 000 014 901 161 193 533 44 × 2 = 0 + 0.000 000 029 802 322 387 066 88;
  • 11) 0.000 000 029 802 322 387 066 88 × 2 = 0 + 0.000 000 059 604 644 774 133 76;
  • 12) 0.000 000 059 604 644 774 133 76 × 2 = 0 + 0.000 000 119 209 289 548 267 52;
  • 13) 0.000 000 119 209 289 548 267 52 × 2 = 0 + 0.000 000 238 418 579 096 535 04;
  • 14) 0.000 000 238 418 579 096 535 04 × 2 = 0 + 0.000 000 476 837 158 193 070 08;
  • 15) 0.000 000 476 837 158 193 070 08 × 2 = 0 + 0.000 000 953 674 316 386 140 16;
  • 16) 0.000 000 953 674 316 386 140 16 × 2 = 0 + 0.000 001 907 348 632 772 280 32;
  • 17) 0.000 001 907 348 632 772 280 32 × 2 = 0 + 0.000 003 814 697 265 544 560 64;
  • 18) 0.000 003 814 697 265 544 560 64 × 2 = 0 + 0.000 007 629 394 531 089 121 28;
  • 19) 0.000 007 629 394 531 089 121 28 × 2 = 0 + 0.000 015 258 789 062 178 242 56;
  • 20) 0.000 015 258 789 062 178 242 56 × 2 = 0 + 0.000 030 517 578 124 356 485 12;
  • 21) 0.000 030 517 578 124 356 485 12 × 2 = 0 + 0.000 061 035 156 248 712 970 24;
  • 22) 0.000 061 035 156 248 712 970 24 × 2 = 0 + 0.000 122 070 312 497 425 940 48;
  • 23) 0.000 122 070 312 497 425 940 48 × 2 = 0 + 0.000 244 140 624 994 851 880 96;
  • 24) 0.000 244 140 624 994 851 880 96 × 2 = 0 + 0.000 488 281 249 989 703 761 92;
  • 25) 0.000 488 281 249 989 703 761 92 × 2 = 0 + 0.000 976 562 499 979 407 523 84;
  • 26) 0.000 976 562 499 979 407 523 84 × 2 = 0 + 0.001 953 124 999 958 815 047 68;
  • 27) 0.001 953 124 999 958 815 047 68 × 2 = 0 + 0.003 906 249 999 917 630 095 36;
  • 28) 0.003 906 249 999 917 630 095 36 × 2 = 0 + 0.007 812 499 999 835 260 190 72;
  • 29) 0.007 812 499 999 835 260 190 72 × 2 = 0 + 0.015 624 999 999 670 520 381 44;
  • 30) 0.015 624 999 999 670 520 381 44 × 2 = 0 + 0.031 249 999 999 341 040 762 88;
  • 31) 0.031 249 999 999 341 040 762 88 × 2 = 0 + 0.062 499 999 998 682 081 525 76;
  • 32) 0.062 499 999 998 682 081 525 76 × 2 = 0 + 0.124 999 999 997 364 163 051 52;
  • 33) 0.124 999 999 997 364 163 051 52 × 2 = 0 + 0.249 999 999 994 728 326 103 04;
  • 34) 0.249 999 999 994 728 326 103 04 × 2 = 0 + 0.499 999 999 989 456 652 206 08;
  • 35) 0.499 999 999 989 456 652 206 08 × 2 = 0 + 0.999 999 999 978 913 304 412 16;
  • 36) 0.999 999 999 978 913 304 412 16 × 2 = 1 + 0.999 999 999 957 826 608 824 32;
  • 37) 0.999 999 999 957 826 608 824 32 × 2 = 1 + 0.999 999 999 915 653 217 648 64;
  • 38) 0.999 999 999 915 653 217 648 64 × 2 = 1 + 0.999 999 999 831 306 435 297 28;
  • 39) 0.999 999 999 831 306 435 297 28 × 2 = 1 + 0.999 999 999 662 612 870 594 56;
  • 40) 0.999 999 999 662 612 870 594 56 × 2 = 1 + 0.999 999 999 325 225 741 189 12;
  • 41) 0.999 999 999 325 225 741 189 12 × 2 = 1 + 0.999 999 998 650 451 482 378 24;
  • 42) 0.999 999 998 650 451 482 378 24 × 2 = 1 + 0.999 999 997 300 902 964 756 48;
  • 43) 0.999 999 997 300 902 964 756 48 × 2 = 1 + 0.999 999 994 601 805 929 512 96;
  • 44) 0.999 999 994 601 805 929 512 96 × 2 = 1 + 0.999 999 989 203 611 859 025 92;
  • 45) 0.999 999 989 203 611 859 025 92 × 2 = 1 + 0.999 999 978 407 223 718 051 84;
  • 46) 0.999 999 978 407 223 718 051 84 × 2 = 1 + 0.999 999 956 814 447 436 103 68;
  • 47) 0.999 999 956 814 447 436 103 68 × 2 = 1 + 0.999 999 913 628 894 872 207 36;
  • 48) 0.999 999 913 628 894 872 207 36 × 2 = 1 + 0.999 999 827 257 789 744 414 72;
  • 49) 0.999 999 827 257 789 744 414 72 × 2 = 1 + 0.999 999 654 515 579 488 829 44;
  • 50) 0.999 999 654 515 579 488 829 44 × 2 = 1 + 0.999 999 309 031 158 977 658 88;
  • 51) 0.999 999 309 031 158 977 658 88 × 2 = 1 + 0.999 998 618 062 317 955 317 76;
  • 52) 0.999 998 618 062 317 955 317 76 × 2 = 1 + 0.999 997 236 124 635 910 635 52;
  • 53) 0.999 997 236 124 635 910 635 52 × 2 = 1 + 0.999 994 472 249 271 821 271 04;
  • 54) 0.999 994 472 249 271 821 271 04 × 2 = 1 + 0.999 988 944 498 543 642 542 08;
  • 55) 0.999 988 944 498 543 642 542 08 × 2 = 1 + 0.999 977 888 997 087 285 084 16;
  • 56) 0.999 977 888 997 087 285 084 16 × 2 = 1 + 0.999 955 777 994 174 570 168 32;
  • 57) 0.999 955 777 994 174 570 168 32 × 2 = 1 + 0.999 911 555 988 349 140 336 64;
  • 58) 0.999 911 555 988 349 140 336 64 × 2 = 1 + 0.999 823 111 976 698 280 673 28;
  • 59) 0.999 823 111 976 698 280 673 28 × 2 = 1 + 0.999 646 223 953 396 561 346 56;
  • 60) 0.999 646 223 953 396 561 346 56 × 2 = 1 + 0.999 292 447 906 793 122 693 12;
  • 61) 0.999 292 447 906 793 122 693 12 × 2 = 1 + 0.998 584 895 813 586 245 386 24;
  • 62) 0.998 584 895 813 586 245 386 24 × 2 = 1 + 0.997 169 791 627 172 490 772 48;
  • 63) 0.997 169 791 627 172 490 772 48 × 2 = 1 + 0.994 339 583 254 344 981 544 96;
  • 64) 0.994 339 583 254 344 981 544 96 × 2 = 1 + 0.988 679 166 508 689 963 089 92;
  • 65) 0.988 679 166 508 689 963 089 92 × 2 = 1 + 0.977 358 333 017 379 926 179 84;
  • 66) 0.977 358 333 017 379 926 179 84 × 2 = 1 + 0.954 716 666 034 759 852 359 68;
  • 67) 0.954 716 666 034 759 852 359 68 × 2 = 1 + 0.909 433 332 069 519 704 719 36;
  • 68) 0.909 433 332 069 519 704 719 36 × 2 = 1 + 0.818 866 664 139 039 409 438 72;
  • 69) 0.818 866 664 139 039 409 438 72 × 2 = 1 + 0.637 733 328 278 078 818 877 44;
  • 70) 0.637 733 328 278 078 818 877 44 × 2 = 1 + 0.275 466 656 556 157 637 754 88;
  • 71) 0.275 466 656 556 157 637 754 88 × 2 = 0 + 0.550 933 313 112 315 275 509 76;
  • 72) 0.550 933 313 112 315 275 509 76 × 2 = 1 + 0.101 866 626 224 630 551 019 52;
  • 73) 0.101 866 626 224 630 551 019 52 × 2 = 0 + 0.203 733 252 449 261 102 039 04;
  • 74) 0.203 733 252 449 261 102 039 04 × 2 = 0 + 0.407 466 504 898 522 204 078 08;
  • 75) 0.407 466 504 898 522 204 078 08 × 2 = 0 + 0.814 933 009 797 044 408 156 16;
  • 76) 0.814 933 009 797 044 408 156 16 × 2 = 1 + 0.629 866 019 594 088 816 312 32;
  • 77) 0.629 866 019 594 088 816 312 32 × 2 = 1 + 0.259 732 039 188 177 632 624 64;
  • 78) 0.259 732 039 188 177 632 624 64 × 2 = 0 + 0.519 464 078 376 355 265 249 28;
  • 79) 0.519 464 078 376 355 265 249 28 × 2 = 1 + 0.038 928 156 752 710 530 498 56;
  • 80) 0.038 928 156 752 710 530 498 56 × 2 = 0 + 0.077 856 313 505 421 060 997 12;
  • 81) 0.077 856 313 505 421 060 997 12 × 2 = 0 + 0.155 712 627 010 842 121 994 24;
  • 82) 0.155 712 627 010 842 121 994 24 × 2 = 0 + 0.311 425 254 021 684 243 988 48;
  • 83) 0.311 425 254 021 684 243 988 48 × 2 = 0 + 0.622 850 508 043 368 487 976 96;
  • 84) 0.622 850 508 043 368 487 976 96 × 2 = 1 + 0.245 701 016 086 736 975 953 92;
  • 85) 0.245 701 016 086 736 975 953 92 × 2 = 0 + 0.491 402 032 173 473 951 907 84;
  • 86) 0.491 402 032 173 473 951 907 84 × 2 = 0 + 0.982 804 064 346 947 903 815 68;
  • 87) 0.982 804 064 346 947 903 815 68 × 2 = 1 + 0.965 608 128 693 895 807 631 36;
  • 88) 0.965 608 128 693 895 807 631 36 × 2 = 1 + 0.931 216 257 387 791 615 262 72;

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).


4. 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:


0.000 000 000 029 103 830 456 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011(2)

5. Positive number before normalization:

0.000 000 000 029 103 830 456 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 36 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 029 103 830 456 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011(2) × 2-36


7. 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 0 (a positive number)

Exponent (unadjusted): -36

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-36 + 2(11-1) - 1 =

(-36 + 1 023)(10) =

987(10)


9. 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;
  • 987 ÷ 2 = 493 + 1;
  • 493 ÷ 2 = 246 + 1;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

987(10) =

011 1101 1011(2)


11. 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. 1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011 =

1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1101 1011

Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011


Decimal number 0.000 000 000 029 103 830 456 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 1011 - 1111 1111 1111 1111 1111 1111 1111 1111 1101 0001 1010 0001 0011

Share

» Convert decimal 0.000 000 000 029 103 830 456 66 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 minutes

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