0.000 000 000 029 103 830 456 49 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 029 103 830 456 49(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 49(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 49.

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 49 × 2 = 0 + 0.000 000 000 058 207 660 912 98;
  • 2) 0.000 000 000 058 207 660 912 98 × 2 = 0 + 0.000 000 000 116 415 321 825 96;
  • 3) 0.000 000 000 116 415 321 825 96 × 2 = 0 + 0.000 000 000 232 830 643 651 92;
  • 4) 0.000 000 000 232 830 643 651 92 × 2 = 0 + 0.000 000 000 465 661 287 303 84;
  • 5) 0.000 000 000 465 661 287 303 84 × 2 = 0 + 0.000 000 000 931 322 574 607 68;
  • 6) 0.000 000 000 931 322 574 607 68 × 2 = 0 + 0.000 000 001 862 645 149 215 36;
  • 7) 0.000 000 001 862 645 149 215 36 × 2 = 0 + 0.000 000 003 725 290 298 430 72;
  • 8) 0.000 000 003 725 290 298 430 72 × 2 = 0 + 0.000 000 007 450 580 596 861 44;
  • 9) 0.000 000 007 450 580 596 861 44 × 2 = 0 + 0.000 000 014 901 161 193 722 88;
  • 10) 0.000 000 014 901 161 193 722 88 × 2 = 0 + 0.000 000 029 802 322 387 445 76;
  • 11) 0.000 000 029 802 322 387 445 76 × 2 = 0 + 0.000 000 059 604 644 774 891 52;
  • 12) 0.000 000 059 604 644 774 891 52 × 2 = 0 + 0.000 000 119 209 289 549 783 04;
  • 13) 0.000 000 119 209 289 549 783 04 × 2 = 0 + 0.000 000 238 418 579 099 566 08;
  • 14) 0.000 000 238 418 579 099 566 08 × 2 = 0 + 0.000 000 476 837 158 199 132 16;
  • 15) 0.000 000 476 837 158 199 132 16 × 2 = 0 + 0.000 000 953 674 316 398 264 32;
  • 16) 0.000 000 953 674 316 398 264 32 × 2 = 0 + 0.000 001 907 348 632 796 528 64;
  • 17) 0.000 001 907 348 632 796 528 64 × 2 = 0 + 0.000 003 814 697 265 593 057 28;
  • 18) 0.000 003 814 697 265 593 057 28 × 2 = 0 + 0.000 007 629 394 531 186 114 56;
  • 19) 0.000 007 629 394 531 186 114 56 × 2 = 0 + 0.000 015 258 789 062 372 229 12;
  • 20) 0.000 015 258 789 062 372 229 12 × 2 = 0 + 0.000 030 517 578 124 744 458 24;
  • 21) 0.000 030 517 578 124 744 458 24 × 2 = 0 + 0.000 061 035 156 249 488 916 48;
  • 22) 0.000 061 035 156 249 488 916 48 × 2 = 0 + 0.000 122 070 312 498 977 832 96;
  • 23) 0.000 122 070 312 498 977 832 96 × 2 = 0 + 0.000 244 140 624 997 955 665 92;
  • 24) 0.000 244 140 624 997 955 665 92 × 2 = 0 + 0.000 488 281 249 995 911 331 84;
  • 25) 0.000 488 281 249 995 911 331 84 × 2 = 0 + 0.000 976 562 499 991 822 663 68;
  • 26) 0.000 976 562 499 991 822 663 68 × 2 = 0 + 0.001 953 124 999 983 645 327 36;
  • 27) 0.001 953 124 999 983 645 327 36 × 2 = 0 + 0.003 906 249 999 967 290 654 72;
  • 28) 0.003 906 249 999 967 290 654 72 × 2 = 0 + 0.007 812 499 999 934 581 309 44;
  • 29) 0.007 812 499 999 934 581 309 44 × 2 = 0 + 0.015 624 999 999 869 162 618 88;
  • 30) 0.015 624 999 999 869 162 618 88 × 2 = 0 + 0.031 249 999 999 738 325 237 76;
  • 31) 0.031 249 999 999 738 325 237 76 × 2 = 0 + 0.062 499 999 999 476 650 475 52;
  • 32) 0.062 499 999 999 476 650 475 52 × 2 = 0 + 0.124 999 999 998 953 300 951 04;
  • 33) 0.124 999 999 998 953 300 951 04 × 2 = 0 + 0.249 999 999 997 906 601 902 08;
  • 34) 0.249 999 999 997 906 601 902 08 × 2 = 0 + 0.499 999 999 995 813 203 804 16;
  • 35) 0.499 999 999 995 813 203 804 16 × 2 = 0 + 0.999 999 999 991 626 407 608 32;
  • 36) 0.999 999 999 991 626 407 608 32 × 2 = 1 + 0.999 999 999 983 252 815 216 64;
  • 37) 0.999 999 999 983 252 815 216 64 × 2 = 1 + 0.999 999 999 966 505 630 433 28;
  • 38) 0.999 999 999 966 505 630 433 28 × 2 = 1 + 0.999 999 999 933 011 260 866 56;
  • 39) 0.999 999 999 933 011 260 866 56 × 2 = 1 + 0.999 999 999 866 022 521 733 12;
  • 40) 0.999 999 999 866 022 521 733 12 × 2 = 1 + 0.999 999 999 732 045 043 466 24;
  • 41) 0.999 999 999 732 045 043 466 24 × 2 = 1 + 0.999 999 999 464 090 086 932 48;
  • 42) 0.999 999 999 464 090 086 932 48 × 2 = 1 + 0.999 999 998 928 180 173 864 96;
  • 43) 0.999 999 998 928 180 173 864 96 × 2 = 1 + 0.999 999 997 856 360 347 729 92;
  • 44) 0.999 999 997 856 360 347 729 92 × 2 = 1 + 0.999 999 995 712 720 695 459 84;
  • 45) 0.999 999 995 712 720 695 459 84 × 2 = 1 + 0.999 999 991 425 441 390 919 68;
  • 46) 0.999 999 991 425 441 390 919 68 × 2 = 1 + 0.999 999 982 850 882 781 839 36;
  • 47) 0.999 999 982 850 882 781 839 36 × 2 = 1 + 0.999 999 965 701 765 563 678 72;
  • 48) 0.999 999 965 701 765 563 678 72 × 2 = 1 + 0.999 999 931 403 531 127 357 44;
  • 49) 0.999 999 931 403 531 127 357 44 × 2 = 1 + 0.999 999 862 807 062 254 714 88;
  • 50) 0.999 999 862 807 062 254 714 88 × 2 = 1 + 0.999 999 725 614 124 509 429 76;
  • 51) 0.999 999 725 614 124 509 429 76 × 2 = 1 + 0.999 999 451 228 249 018 859 52;
  • 52) 0.999 999 451 228 249 018 859 52 × 2 = 1 + 0.999 998 902 456 498 037 719 04;
  • 53) 0.999 998 902 456 498 037 719 04 × 2 = 1 + 0.999 997 804 912 996 075 438 08;
  • 54) 0.999 997 804 912 996 075 438 08 × 2 = 1 + 0.999 995 609 825 992 150 876 16;
  • 55) 0.999 995 609 825 992 150 876 16 × 2 = 1 + 0.999 991 219 651 984 301 752 32;
  • 56) 0.999 991 219 651 984 301 752 32 × 2 = 1 + 0.999 982 439 303 968 603 504 64;
  • 57) 0.999 982 439 303 968 603 504 64 × 2 = 1 + 0.999 964 878 607 937 207 009 28;
  • 58) 0.999 964 878 607 937 207 009 28 × 2 = 1 + 0.999 929 757 215 874 414 018 56;
  • 59) 0.999 929 757 215 874 414 018 56 × 2 = 1 + 0.999 859 514 431 748 828 037 12;
  • 60) 0.999 859 514 431 748 828 037 12 × 2 = 1 + 0.999 719 028 863 497 656 074 24;
  • 61) 0.999 719 028 863 497 656 074 24 × 2 = 1 + 0.999 438 057 726 995 312 148 48;
  • 62) 0.999 438 057 726 995 312 148 48 × 2 = 1 + 0.998 876 115 453 990 624 296 96;
  • 63) 0.998 876 115 453 990 624 296 96 × 2 = 1 + 0.997 752 230 907 981 248 593 92;
  • 64) 0.997 752 230 907 981 248 593 92 × 2 = 1 + 0.995 504 461 815 962 497 187 84;
  • 65) 0.995 504 461 815 962 497 187 84 × 2 = 1 + 0.991 008 923 631 924 994 375 68;
  • 66) 0.991 008 923 631 924 994 375 68 × 2 = 1 + 0.982 017 847 263 849 988 751 36;
  • 67) 0.982 017 847 263 849 988 751 36 × 2 = 1 + 0.964 035 694 527 699 977 502 72;
  • 68) 0.964 035 694 527 699 977 502 72 × 2 = 1 + 0.928 071 389 055 399 955 005 44;
  • 69) 0.928 071 389 055 399 955 005 44 × 2 = 1 + 0.856 142 778 110 799 910 010 88;
  • 70) 0.856 142 778 110 799 910 010 88 × 2 = 1 + 0.712 285 556 221 599 820 021 76;
  • 71) 0.712 285 556 221 599 820 021 76 × 2 = 1 + 0.424 571 112 443 199 640 043 52;
  • 72) 0.424 571 112 443 199 640 043 52 × 2 = 0 + 0.849 142 224 886 399 280 087 04;
  • 73) 0.849 142 224 886 399 280 087 04 × 2 = 1 + 0.698 284 449 772 798 560 174 08;
  • 74) 0.698 284 449 772 798 560 174 08 × 2 = 1 + 0.396 568 899 545 597 120 348 16;
  • 75) 0.396 568 899 545 597 120 348 16 × 2 = 0 + 0.793 137 799 091 194 240 696 32;
  • 76) 0.793 137 799 091 194 240 696 32 × 2 = 1 + 0.586 275 598 182 388 481 392 64;
  • 77) 0.586 275 598 182 388 481 392 64 × 2 = 1 + 0.172 551 196 364 776 962 785 28;
  • 78) 0.172 551 196 364 776 962 785 28 × 2 = 0 + 0.345 102 392 729 553 925 570 56;
  • 79) 0.345 102 392 729 553 925 570 56 × 2 = 0 + 0.690 204 785 459 107 851 141 12;
  • 80) 0.690 204 785 459 107 851 141 12 × 2 = 1 + 0.380 409 570 918 215 702 282 24;
  • 81) 0.380 409 570 918 215 702 282 24 × 2 = 0 + 0.760 819 141 836 431 404 564 48;
  • 82) 0.760 819 141 836 431 404 564 48 × 2 = 1 + 0.521 638 283 672 862 809 128 96;
  • 83) 0.521 638 283 672 862 809 128 96 × 2 = 1 + 0.043 276 567 345 725 618 257 92;
  • 84) 0.043 276 567 345 725 618 257 92 × 2 = 0 + 0.086 553 134 691 451 236 515 84;
  • 85) 0.086 553 134 691 451 236 515 84 × 2 = 0 + 0.173 106 269 382 902 473 031 68;
  • 86) 0.173 106 269 382 902 473 031 68 × 2 = 0 + 0.346 212 538 765 804 946 063 36;
  • 87) 0.346 212 538 765 804 946 063 36 × 2 = 0 + 0.692 425 077 531 609 892 126 72;
  • 88) 0.692 425 077 531 609 892 126 72 × 2 = 1 + 0.384 850 155 063 219 784 253 44;

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 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1111 1110 1101 1001 0110 0001(2)

5. Positive number before normalization:

0.000 000 000 029 103 830 456 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1111 1110 1101 1001 0110 0001(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 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1111 1110 1101 1001 0110 0001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1111 1110 1101 1001 0110 0001(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1111 1111 1110 1101 1001 0110 0001(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 1110 1101 1001 0110 0001


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 1110 1101 1001 0110 0001 =

1111 1111 1111 1111 1111 1111 1111 1111 1110 1101 1001 0110 0001


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 1110 1101 1001 0110 0001


Decimal number 0.000 000 000 029 103 830 456 49 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 1011 - 1111 1111 1111 1111 1111 1111 1111 1111 1110 1101 1001 0110 0001

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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