-0.000 000 000 010 068 461 037 56 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 010 068 461 037 56(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
-0.000 000 000 010 068 461 037 56(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:

|-0.000 000 000 010 068 461 037 56| = 0.000 000 000 010 068 461 037 56


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: 0.000 000 000 010 068 461 037 56.

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) 0.000 000 000 010 068 461 037 56 × 2 = 0 + 0.000 000 000 020 136 922 075 12;
  • 2) 0.000 000 000 020 136 922 075 12 × 2 = 0 + 0.000 000 000 040 273 844 150 24;
  • 3) 0.000 000 000 040 273 844 150 24 × 2 = 0 + 0.000 000 000 080 547 688 300 48;
  • 4) 0.000 000 000 080 547 688 300 48 × 2 = 0 + 0.000 000 000 161 095 376 600 96;
  • 5) 0.000 000 000 161 095 376 600 96 × 2 = 0 + 0.000 000 000 322 190 753 201 92;
  • 6) 0.000 000 000 322 190 753 201 92 × 2 = 0 + 0.000 000 000 644 381 506 403 84;
  • 7) 0.000 000 000 644 381 506 403 84 × 2 = 0 + 0.000 000 001 288 763 012 807 68;
  • 8) 0.000 000 001 288 763 012 807 68 × 2 = 0 + 0.000 000 002 577 526 025 615 36;
  • 9) 0.000 000 002 577 526 025 615 36 × 2 = 0 + 0.000 000 005 155 052 051 230 72;
  • 10) 0.000 000 005 155 052 051 230 72 × 2 = 0 + 0.000 000 010 310 104 102 461 44;
  • 11) 0.000 000 010 310 104 102 461 44 × 2 = 0 + 0.000 000 020 620 208 204 922 88;
  • 12) 0.000 000 020 620 208 204 922 88 × 2 = 0 + 0.000 000 041 240 416 409 845 76;
  • 13) 0.000 000 041 240 416 409 845 76 × 2 = 0 + 0.000 000 082 480 832 819 691 52;
  • 14) 0.000 000 082 480 832 819 691 52 × 2 = 0 + 0.000 000 164 961 665 639 383 04;
  • 15) 0.000 000 164 961 665 639 383 04 × 2 = 0 + 0.000 000 329 923 331 278 766 08;
  • 16) 0.000 000 329 923 331 278 766 08 × 2 = 0 + 0.000 000 659 846 662 557 532 16;
  • 17) 0.000 000 659 846 662 557 532 16 × 2 = 0 + 0.000 001 319 693 325 115 064 32;
  • 18) 0.000 001 319 693 325 115 064 32 × 2 = 0 + 0.000 002 639 386 650 230 128 64;
  • 19) 0.000 002 639 386 650 230 128 64 × 2 = 0 + 0.000 005 278 773 300 460 257 28;
  • 20) 0.000 005 278 773 300 460 257 28 × 2 = 0 + 0.000 010 557 546 600 920 514 56;
  • 21) 0.000 010 557 546 600 920 514 56 × 2 = 0 + 0.000 021 115 093 201 841 029 12;
  • 22) 0.000 021 115 093 201 841 029 12 × 2 = 0 + 0.000 042 230 186 403 682 058 24;
  • 23) 0.000 042 230 186 403 682 058 24 × 2 = 0 + 0.000 084 460 372 807 364 116 48;
  • 24) 0.000 084 460 372 807 364 116 48 × 2 = 0 + 0.000 168 920 745 614 728 232 96;
  • 25) 0.000 168 920 745 614 728 232 96 × 2 = 0 + 0.000 337 841 491 229 456 465 92;
  • 26) 0.000 337 841 491 229 456 465 92 × 2 = 0 + 0.000 675 682 982 458 912 931 84;
  • 27) 0.000 675 682 982 458 912 931 84 × 2 = 0 + 0.001 351 365 964 917 825 863 68;
  • 28) 0.001 351 365 964 917 825 863 68 × 2 = 0 + 0.002 702 731 929 835 651 727 36;
  • 29) 0.002 702 731 929 835 651 727 36 × 2 = 0 + 0.005 405 463 859 671 303 454 72;
  • 30) 0.005 405 463 859 671 303 454 72 × 2 = 0 + 0.010 810 927 719 342 606 909 44;
  • 31) 0.010 810 927 719 342 606 909 44 × 2 = 0 + 0.021 621 855 438 685 213 818 88;
  • 32) 0.021 621 855 438 685 213 818 88 × 2 = 0 + 0.043 243 710 877 370 427 637 76;
  • 33) 0.043 243 710 877 370 427 637 76 × 2 = 0 + 0.086 487 421 754 740 855 275 52;
  • 34) 0.086 487 421 754 740 855 275 52 × 2 = 0 + 0.172 974 843 509 481 710 551 04;
  • 35) 0.172 974 843 509 481 710 551 04 × 2 = 0 + 0.345 949 687 018 963 421 102 08;
  • 36) 0.345 949 687 018 963 421 102 08 × 2 = 0 + 0.691 899 374 037 926 842 204 16;
  • 37) 0.691 899 374 037 926 842 204 16 × 2 = 1 + 0.383 798 748 075 853 684 408 32;
  • 38) 0.383 798 748 075 853 684 408 32 × 2 = 0 + 0.767 597 496 151 707 368 816 64;
  • 39) 0.767 597 496 151 707 368 816 64 × 2 = 1 + 0.535 194 992 303 414 737 633 28;
  • 40) 0.535 194 992 303 414 737 633 28 × 2 = 1 + 0.070 389 984 606 829 475 266 56;
  • 41) 0.070 389 984 606 829 475 266 56 × 2 = 0 + 0.140 779 969 213 658 950 533 12;
  • 42) 0.140 779 969 213 658 950 533 12 × 2 = 0 + 0.281 559 938 427 317 901 066 24;
  • 43) 0.281 559 938 427 317 901 066 24 × 2 = 0 + 0.563 119 876 854 635 802 132 48;
  • 44) 0.563 119 876 854 635 802 132 48 × 2 = 1 + 0.126 239 753 709 271 604 264 96;
  • 45) 0.126 239 753 709 271 604 264 96 × 2 = 0 + 0.252 479 507 418 543 208 529 92;
  • 46) 0.252 479 507 418 543 208 529 92 × 2 = 0 + 0.504 959 014 837 086 417 059 84;
  • 47) 0.504 959 014 837 086 417 059 84 × 2 = 1 + 0.009 918 029 674 172 834 119 68;
  • 48) 0.009 918 029 674 172 834 119 68 × 2 = 0 + 0.019 836 059 348 345 668 239 36;
  • 49) 0.019 836 059 348 345 668 239 36 × 2 = 0 + 0.039 672 118 696 691 336 478 72;
  • 50) 0.039 672 118 696 691 336 478 72 × 2 = 0 + 0.079 344 237 393 382 672 957 44;
  • 51) 0.079 344 237 393 382 672 957 44 × 2 = 0 + 0.158 688 474 786 765 345 914 88;
  • 52) 0.158 688 474 786 765 345 914 88 × 2 = 0 + 0.317 376 949 573 530 691 829 76;
  • 53) 0.317 376 949 573 530 691 829 76 × 2 = 0 + 0.634 753 899 147 061 383 659 52;
  • 54) 0.634 753 899 147 061 383 659 52 × 2 = 1 + 0.269 507 798 294 122 767 319 04;
  • 55) 0.269 507 798 294 122 767 319 04 × 2 = 0 + 0.539 015 596 588 245 534 638 08;
  • 56) 0.539 015 596 588 245 534 638 08 × 2 = 1 + 0.078 031 193 176 491 069 276 16;
  • 57) 0.078 031 193 176 491 069 276 16 × 2 = 0 + 0.156 062 386 352 982 138 552 32;
  • 58) 0.156 062 386 352 982 138 552 32 × 2 = 0 + 0.312 124 772 705 964 277 104 64;
  • 59) 0.312 124 772 705 964 277 104 64 × 2 = 0 + 0.624 249 545 411 928 554 209 28;
  • 60) 0.624 249 545 411 928 554 209 28 × 2 = 1 + 0.248 499 090 823 857 108 418 56;
  • 61) 0.248 499 090 823 857 108 418 56 × 2 = 0 + 0.496 998 181 647 714 216 837 12;
  • 62) 0.496 998 181 647 714 216 837 12 × 2 = 0 + 0.993 996 363 295 428 433 674 24;
  • 63) 0.993 996 363 295 428 433 674 24 × 2 = 1 + 0.987 992 726 590 856 867 348 48;
  • 64) 0.987 992 726 590 856 867 348 48 × 2 = 1 + 0.975 985 453 181 713 734 696 96;
  • 65) 0.975 985 453 181 713 734 696 96 × 2 = 1 + 0.951 970 906 363 427 469 393 92;
  • 66) 0.951 970 906 363 427 469 393 92 × 2 = 1 + 0.903 941 812 726 854 938 787 84;
  • 67) 0.903 941 812 726 854 938 787 84 × 2 = 1 + 0.807 883 625 453 709 877 575 68;
  • 68) 0.807 883 625 453 709 877 575 68 × 2 = 1 + 0.615 767 250 907 419 755 151 36;
  • 69) 0.615 767 250 907 419 755 151 36 × 2 = 1 + 0.231 534 501 814 839 510 302 72;
  • 70) 0.231 534 501 814 839 510 302 72 × 2 = 0 + 0.463 069 003 629 679 020 605 44;
  • 71) 0.463 069 003 629 679 020 605 44 × 2 = 0 + 0.926 138 007 259 358 041 210 88;
  • 72) 0.926 138 007 259 358 041 210 88 × 2 = 1 + 0.852 276 014 518 716 082 421 76;
  • 73) 0.852 276 014 518 716 082 421 76 × 2 = 1 + 0.704 552 029 037 432 164 843 52;
  • 74) 0.704 552 029 037 432 164 843 52 × 2 = 1 + 0.409 104 058 074 864 329 687 04;
  • 75) 0.409 104 058 074 864 329 687 04 × 2 = 0 + 0.818 208 116 149 728 659 374 08;
  • 76) 0.818 208 116 149 728 659 374 08 × 2 = 1 + 0.636 416 232 299 457 318 748 16;
  • 77) 0.636 416 232 299 457 318 748 16 × 2 = 1 + 0.272 832 464 598 914 637 496 32;
  • 78) 0.272 832 464 598 914 637 496 32 × 2 = 0 + 0.545 664 929 197 829 274 992 64;
  • 79) 0.545 664 929 197 829 274 992 64 × 2 = 1 + 0.091 329 858 395 658 549 985 28;
  • 80) 0.091 329 858 395 658 549 985 28 × 2 = 0 + 0.182 659 716 791 317 099 970 56;
  • 81) 0.182 659 716 791 317 099 970 56 × 2 = 0 + 0.365 319 433 582 634 199 941 12;
  • 82) 0.365 319 433 582 634 199 941 12 × 2 = 0 + 0.730 638 867 165 268 399 882 24;
  • 83) 0.730 638 867 165 268 399 882 24 × 2 = 1 + 0.461 277 734 330 536 799 764 48;
  • 84) 0.461 277 734 330 536 799 764 48 × 2 = 0 + 0.922 555 468 661 073 599 528 96;
  • 85) 0.922 555 468 661 073 599 528 96 × 2 = 1 + 0.845 110 937 322 147 199 057 92;
  • 86) 0.845 110 937 322 147 199 057 92 × 2 = 1 + 0.690 221 874 644 294 398 115 84;
  • 87) 0.690 221 874 644 294 398 115 84 × 2 = 1 + 0.380 443 749 288 588 796 231 68;
  • 88) 0.380 443 749 288 588 796 231 68 × 2 = 0 + 0.760 887 498 577 177 592 463 36;
  • 89) 0.760 887 498 577 177 592 463 36 × 2 = 1 + 0.521 774 997 154 355 184 926 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).


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:


0.000 000 000 010 068 461 037 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1001 1101 1010 0010 1110 1(2)

6. Positive number before normalization:

0.000 000 000 010 068 461 037 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1001 1101 1010 0010 1110 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 010 068 461 037 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1001 1101 1010 0010 1110 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1001 1101 1010 0010 1110 1(2) × 20 =

1.0110 0010 0100 0000 1010 0010 0111 1111 0011 1011 0100 0101 1101(2) × 2-37


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): -37

Mantissa (not normalized):
1.0110 0010 0100 0000 1010 0010 0111 1111 0011 1011 0100 0101 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-37 + 2(11-1) - 1 =

(-37 + 1 023)(10) =

986(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;
  • 986 ÷ 2 = 493 + 0;
  • 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;

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) =

986(10) =

011 1101 1010(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. 0110 0010 0100 0000 1010 0010 0111 1111 0011 1011 0100 0101 1101 =

0110 0010 0100 0000 1010 0010 0111 1111 0011 1011 0100 0101 1101


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 1010

Mantissa (52 bits) =
0110 0010 0100 0000 1010 0010 0111 1111 0011 1011 0100 0101 1101


Decimal number -0.000 000 000 010 068 461 037 56 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 0010 0100 0000 1010 0010 0111 1111 0011 1011 0100 0101 1101

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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