-0.000 000 000 000 176 556 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 35(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
-0.000 000 000 000 176 556 35(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:

|-0.000 000 000 000 176 556 35| = 0.000 000 000 000 176 556 35


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: 0.000 000 000 000 176 556 35.

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) 0.000 000 000 000 176 556 35 × 2 = 0 + 0.000 000 000 000 353 112 7;
  • 2) 0.000 000 000 000 353 112 7 × 2 = 0 + 0.000 000 000 000 706 225 4;
  • 3) 0.000 000 000 000 706 225 4 × 2 = 0 + 0.000 000 000 001 412 450 8;
  • 4) 0.000 000 000 001 412 450 8 × 2 = 0 + 0.000 000 000 002 824 901 6;
  • 5) 0.000 000 000 002 824 901 6 × 2 = 0 + 0.000 000 000 005 649 803 2;
  • 6) 0.000 000 000 005 649 803 2 × 2 = 0 + 0.000 000 000 011 299 606 4;
  • 7) 0.000 000 000 011 299 606 4 × 2 = 0 + 0.000 000 000 022 599 212 8;
  • 8) 0.000 000 000 022 599 212 8 × 2 = 0 + 0.000 000 000 045 198 425 6;
  • 9) 0.000 000 000 045 198 425 6 × 2 = 0 + 0.000 000 000 090 396 851 2;
  • 10) 0.000 000 000 090 396 851 2 × 2 = 0 + 0.000 000 000 180 793 702 4;
  • 11) 0.000 000 000 180 793 702 4 × 2 = 0 + 0.000 000 000 361 587 404 8;
  • 12) 0.000 000 000 361 587 404 8 × 2 = 0 + 0.000 000 000 723 174 809 6;
  • 13) 0.000 000 000 723 174 809 6 × 2 = 0 + 0.000 000 001 446 349 619 2;
  • 14) 0.000 000 001 446 349 619 2 × 2 = 0 + 0.000 000 002 892 699 238 4;
  • 15) 0.000 000 002 892 699 238 4 × 2 = 0 + 0.000 000 005 785 398 476 8;
  • 16) 0.000 000 005 785 398 476 8 × 2 = 0 + 0.000 000 011 570 796 953 6;
  • 17) 0.000 000 011 570 796 953 6 × 2 = 0 + 0.000 000 023 141 593 907 2;
  • 18) 0.000 000 023 141 593 907 2 × 2 = 0 + 0.000 000 046 283 187 814 4;
  • 19) 0.000 000 046 283 187 814 4 × 2 = 0 + 0.000 000 092 566 375 628 8;
  • 20) 0.000 000 092 566 375 628 8 × 2 = 0 + 0.000 000 185 132 751 257 6;
  • 21) 0.000 000 185 132 751 257 6 × 2 = 0 + 0.000 000 370 265 502 515 2;
  • 22) 0.000 000 370 265 502 515 2 × 2 = 0 + 0.000 000 740 531 005 030 4;
  • 23) 0.000 000 740 531 005 030 4 × 2 = 0 + 0.000 001 481 062 010 060 8;
  • 24) 0.000 001 481 062 010 060 8 × 2 = 0 + 0.000 002 962 124 020 121 6;
  • 25) 0.000 002 962 124 020 121 6 × 2 = 0 + 0.000 005 924 248 040 243 2;
  • 26) 0.000 005 924 248 040 243 2 × 2 = 0 + 0.000 011 848 496 080 486 4;
  • 27) 0.000 011 848 496 080 486 4 × 2 = 0 + 0.000 023 696 992 160 972 8;
  • 28) 0.000 023 696 992 160 972 8 × 2 = 0 + 0.000 047 393 984 321 945 6;
  • 29) 0.000 047 393 984 321 945 6 × 2 = 0 + 0.000 094 787 968 643 891 2;
  • 30) 0.000 094 787 968 643 891 2 × 2 = 0 + 0.000 189 575 937 287 782 4;
  • 31) 0.000 189 575 937 287 782 4 × 2 = 0 + 0.000 379 151 874 575 564 8;
  • 32) 0.000 379 151 874 575 564 8 × 2 = 0 + 0.000 758 303 749 151 129 6;
  • 33) 0.000 758 303 749 151 129 6 × 2 = 0 + 0.001 516 607 498 302 259 2;
  • 34) 0.001 516 607 498 302 259 2 × 2 = 0 + 0.003 033 214 996 604 518 4;
  • 35) 0.003 033 214 996 604 518 4 × 2 = 0 + 0.006 066 429 993 209 036 8;
  • 36) 0.006 066 429 993 209 036 8 × 2 = 0 + 0.012 132 859 986 418 073 6;
  • 37) 0.012 132 859 986 418 073 6 × 2 = 0 + 0.024 265 719 972 836 147 2;
  • 38) 0.024 265 719 972 836 147 2 × 2 = 0 + 0.048 531 439 945 672 294 4;
  • 39) 0.048 531 439 945 672 294 4 × 2 = 0 + 0.097 062 879 891 344 588 8;
  • 40) 0.097 062 879 891 344 588 8 × 2 = 0 + 0.194 125 759 782 689 177 6;
  • 41) 0.194 125 759 782 689 177 6 × 2 = 0 + 0.388 251 519 565 378 355 2;
  • 42) 0.388 251 519 565 378 355 2 × 2 = 0 + 0.776 503 039 130 756 710 4;
  • 43) 0.776 503 039 130 756 710 4 × 2 = 1 + 0.553 006 078 261 513 420 8;
  • 44) 0.553 006 078 261 513 420 8 × 2 = 1 + 0.106 012 156 523 026 841 6;
  • 45) 0.106 012 156 523 026 841 6 × 2 = 0 + 0.212 024 313 046 053 683 2;
  • 46) 0.212 024 313 046 053 683 2 × 2 = 0 + 0.424 048 626 092 107 366 4;
  • 47) 0.424 048 626 092 107 366 4 × 2 = 0 + 0.848 097 252 184 214 732 8;
  • 48) 0.848 097 252 184 214 732 8 × 2 = 1 + 0.696 194 504 368 429 465 6;
  • 49) 0.696 194 504 368 429 465 6 × 2 = 1 + 0.392 389 008 736 858 931 2;
  • 50) 0.392 389 008 736 858 931 2 × 2 = 0 + 0.784 778 017 473 717 862 4;
  • 51) 0.784 778 017 473 717 862 4 × 2 = 1 + 0.569 556 034 947 435 724 8;
  • 52) 0.569 556 034 947 435 724 8 × 2 = 1 + 0.139 112 069 894 871 449 6;
  • 53) 0.139 112 069 894 871 449 6 × 2 = 0 + 0.278 224 139 789 742 899 2;
  • 54) 0.278 224 139 789 742 899 2 × 2 = 0 + 0.556 448 279 579 485 798 4;
  • 55) 0.556 448 279 579 485 798 4 × 2 = 1 + 0.112 896 559 158 971 596 8;
  • 56) 0.112 896 559 158 971 596 8 × 2 = 0 + 0.225 793 118 317 943 193 6;
  • 57) 0.225 793 118 317 943 193 6 × 2 = 0 + 0.451 586 236 635 886 387 2;
  • 58) 0.451 586 236 635 886 387 2 × 2 = 0 + 0.903 172 473 271 772 774 4;
  • 59) 0.903 172 473 271 772 774 4 × 2 = 1 + 0.806 344 946 543 545 548 8;
  • 60) 0.806 344 946 543 545 548 8 × 2 = 1 + 0.612 689 893 087 091 097 6;
  • 61) 0.612 689 893 087 091 097 6 × 2 = 1 + 0.225 379 786 174 182 195 2;
  • 62) 0.225 379 786 174 182 195 2 × 2 = 0 + 0.450 759 572 348 364 390 4;
  • 63) 0.450 759 572 348 364 390 4 × 2 = 0 + 0.901 519 144 696 728 780 8;
  • 64) 0.901 519 144 696 728 780 8 × 2 = 1 + 0.803 038 289 393 457 561 6;
  • 65) 0.803 038 289 393 457 561 6 × 2 = 1 + 0.606 076 578 786 915 123 2;
  • 66) 0.606 076 578 786 915 123 2 × 2 = 1 + 0.212 153 157 573 830 246 4;
  • 67) 0.212 153 157 573 830 246 4 × 2 = 0 + 0.424 306 315 147 660 492 8;
  • 68) 0.424 306 315 147 660 492 8 × 2 = 0 + 0.848 612 630 295 320 985 6;
  • 69) 0.848 612 630 295 320 985 6 × 2 = 1 + 0.697 225 260 590 641 971 2;
  • 70) 0.697 225 260 590 641 971 2 × 2 = 1 + 0.394 450 521 181 283 942 4;
  • 71) 0.394 450 521 181 283 942 4 × 2 = 0 + 0.788 901 042 362 567 884 8;
  • 72) 0.788 901 042 362 567 884 8 × 2 = 1 + 0.577 802 084 725 135 769 6;
  • 73) 0.577 802 084 725 135 769 6 × 2 = 1 + 0.155 604 169 450 271 539 2;
  • 74) 0.155 604 169 450 271 539 2 × 2 = 0 + 0.311 208 338 900 543 078 4;
  • 75) 0.311 208 338 900 543 078 4 × 2 = 0 + 0.622 416 677 801 086 156 8;
  • 76) 0.622 416 677 801 086 156 8 × 2 = 1 + 0.244 833 355 602 172 313 6;
  • 77) 0.244 833 355 602 172 313 6 × 2 = 0 + 0.489 666 711 204 344 627 2;
  • 78) 0.489 666 711 204 344 627 2 × 2 = 0 + 0.979 333 422 408 689 254 4;
  • 79) 0.979 333 422 408 689 254 4 × 2 = 1 + 0.958 666 844 817 378 508 8;
  • 80) 0.958 666 844 817 378 508 8 × 2 = 1 + 0.917 333 689 634 757 017 6;
  • 81) 0.917 333 689 634 757 017 6 × 2 = 1 + 0.834 667 379 269 514 035 2;
  • 82) 0.834 667 379 269 514 035 2 × 2 = 1 + 0.669 334 758 539 028 070 4;
  • 83) 0.669 334 758 539 028 070 4 × 2 = 1 + 0.338 669 517 078 056 140 8;
  • 84) 0.338 669 517 078 056 140 8 × 2 = 0 + 0.677 339 034 156 112 281 6;
  • 85) 0.677 339 034 156 112 281 6 × 2 = 1 + 0.354 678 068 312 224 563 2;
  • 86) 0.354 678 068 312 224 563 2 × 2 = 0 + 0.709 356 136 624 449 126 4;
  • 87) 0.709 356 136 624 449 126 4 × 2 = 1 + 0.418 712 273 248 898 252 8;
  • 88) 0.418 712 273 248 898 252 8 × 2 = 0 + 0.837 424 546 497 796 505 6;
  • 89) 0.837 424 546 497 796 505 6 × 2 = 1 + 0.674 849 092 995 593 011 2;
  • 90) 0.674 849 092 995 593 011 2 × 2 = 1 + 0.349 698 185 991 186 022 4;
  • 91) 0.349 698 185 991 186 022 4 × 2 = 0 + 0.699 396 371 982 372 044 8;
  • 92) 0.699 396 371 982 372 044 8 × 2 = 1 + 0.398 792 743 964 744 089 6;
  • 93) 0.398 792 743 964 744 089 6 × 2 = 0 + 0.797 585 487 929 488 179 2;
  • 94) 0.797 585 487 929 488 179 2 × 2 = 1 + 0.595 170 975 858 976 358 4;
  • 95) 0.595 170 975 858 976 358 4 × 2 = 1 + 0.190 341 951 717 952 716 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:


0.000 000 000 000 176 556 35(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1001 1100 1101 1001 0011 1110 1010 1101 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 35(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1001 1100 1101 1001 0011 1110 1010 1101 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 556 35(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1001 1100 1101 1001 0011 1110 1010 1101 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1001 1100 1101 1001 0011 1110 1010 1101 011(2) × 20 =

1.1000 1101 1001 0001 1100 1110 0110 1100 1001 1111 0101 0110 1011(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0001 1100 1110 0110 1100 1001 1111 0101 0110 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 1001 0001 1100 1110 0110 1100 1001 1111 0101 0110 1011 =

1000 1101 1001 0001 1100 1110 0110 1100 1001 1111 0101 0110 1011


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 0100

Mantissa (52 bits) =
1000 1101 1001 0001 1100 1110 0110 1100 1001 1111 0101 0110 1011


Decimal number -0.000 000 000 000 176 556 35 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1100 1110 0110 1100 1001 1111 0101 0110 1011

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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