-0.000 000 000 000 013 82 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 013 82(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
-0.000 000 000 000 013 82(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:

|-0.000 000 000 000 013 82| = 0.000 000 000 000 013 82


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: 0.000 000 000 000 013 82.

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) 0.000 000 000 000 013 82 × 2 = 0 + 0.000 000 000 000 027 64;
  • 2) 0.000 000 000 000 027 64 × 2 = 0 + 0.000 000 000 000 055 28;
  • 3) 0.000 000 000 000 055 28 × 2 = 0 + 0.000 000 000 000 110 56;
  • 4) 0.000 000 000 000 110 56 × 2 = 0 + 0.000 000 000 000 221 12;
  • 5) 0.000 000 000 000 221 12 × 2 = 0 + 0.000 000 000 000 442 24;
  • 6) 0.000 000 000 000 442 24 × 2 = 0 + 0.000 000 000 000 884 48;
  • 7) 0.000 000 000 000 884 48 × 2 = 0 + 0.000 000 000 001 768 96;
  • 8) 0.000 000 000 001 768 96 × 2 = 0 + 0.000 000 000 003 537 92;
  • 9) 0.000 000 000 003 537 92 × 2 = 0 + 0.000 000 000 007 075 84;
  • 10) 0.000 000 000 007 075 84 × 2 = 0 + 0.000 000 000 014 151 68;
  • 11) 0.000 000 000 014 151 68 × 2 = 0 + 0.000 000 000 028 303 36;
  • 12) 0.000 000 000 028 303 36 × 2 = 0 + 0.000 000 000 056 606 72;
  • 13) 0.000 000 000 056 606 72 × 2 = 0 + 0.000 000 000 113 213 44;
  • 14) 0.000 000 000 113 213 44 × 2 = 0 + 0.000 000 000 226 426 88;
  • 15) 0.000 000 000 226 426 88 × 2 = 0 + 0.000 000 000 452 853 76;
  • 16) 0.000 000 000 452 853 76 × 2 = 0 + 0.000 000 000 905 707 52;
  • 17) 0.000 000 000 905 707 52 × 2 = 0 + 0.000 000 001 811 415 04;
  • 18) 0.000 000 001 811 415 04 × 2 = 0 + 0.000 000 003 622 830 08;
  • 19) 0.000 000 003 622 830 08 × 2 = 0 + 0.000 000 007 245 660 16;
  • 20) 0.000 000 007 245 660 16 × 2 = 0 + 0.000 000 014 491 320 32;
  • 21) 0.000 000 014 491 320 32 × 2 = 0 + 0.000 000 028 982 640 64;
  • 22) 0.000 000 028 982 640 64 × 2 = 0 + 0.000 000 057 965 281 28;
  • 23) 0.000 000 057 965 281 28 × 2 = 0 + 0.000 000 115 930 562 56;
  • 24) 0.000 000 115 930 562 56 × 2 = 0 + 0.000 000 231 861 125 12;
  • 25) 0.000 000 231 861 125 12 × 2 = 0 + 0.000 000 463 722 250 24;
  • 26) 0.000 000 463 722 250 24 × 2 = 0 + 0.000 000 927 444 500 48;
  • 27) 0.000 000 927 444 500 48 × 2 = 0 + 0.000 001 854 889 000 96;
  • 28) 0.000 001 854 889 000 96 × 2 = 0 + 0.000 003 709 778 001 92;
  • 29) 0.000 003 709 778 001 92 × 2 = 0 + 0.000 007 419 556 003 84;
  • 30) 0.000 007 419 556 003 84 × 2 = 0 + 0.000 014 839 112 007 68;
  • 31) 0.000 014 839 112 007 68 × 2 = 0 + 0.000 029 678 224 015 36;
  • 32) 0.000 029 678 224 015 36 × 2 = 0 + 0.000 059 356 448 030 72;
  • 33) 0.000 059 356 448 030 72 × 2 = 0 + 0.000 118 712 896 061 44;
  • 34) 0.000 118 712 896 061 44 × 2 = 0 + 0.000 237 425 792 122 88;
  • 35) 0.000 237 425 792 122 88 × 2 = 0 + 0.000 474 851 584 245 76;
  • 36) 0.000 474 851 584 245 76 × 2 = 0 + 0.000 949 703 168 491 52;
  • 37) 0.000 949 703 168 491 52 × 2 = 0 + 0.001 899 406 336 983 04;
  • 38) 0.001 899 406 336 983 04 × 2 = 0 + 0.003 798 812 673 966 08;
  • 39) 0.003 798 812 673 966 08 × 2 = 0 + 0.007 597 625 347 932 16;
  • 40) 0.007 597 625 347 932 16 × 2 = 0 + 0.015 195 250 695 864 32;
  • 41) 0.015 195 250 695 864 32 × 2 = 0 + 0.030 390 501 391 728 64;
  • 42) 0.030 390 501 391 728 64 × 2 = 0 + 0.060 781 002 783 457 28;
  • 43) 0.060 781 002 783 457 28 × 2 = 0 + 0.121 562 005 566 914 56;
  • 44) 0.121 562 005 566 914 56 × 2 = 0 + 0.243 124 011 133 829 12;
  • 45) 0.243 124 011 133 829 12 × 2 = 0 + 0.486 248 022 267 658 24;
  • 46) 0.486 248 022 267 658 24 × 2 = 0 + 0.972 496 044 535 316 48;
  • 47) 0.972 496 044 535 316 48 × 2 = 1 + 0.944 992 089 070 632 96;
  • 48) 0.944 992 089 070 632 96 × 2 = 1 + 0.889 984 178 141 265 92;
  • 49) 0.889 984 178 141 265 92 × 2 = 1 + 0.779 968 356 282 531 84;
  • 50) 0.779 968 356 282 531 84 × 2 = 1 + 0.559 936 712 565 063 68;
  • 51) 0.559 936 712 565 063 68 × 2 = 1 + 0.119 873 425 130 127 36;
  • 52) 0.119 873 425 130 127 36 × 2 = 0 + 0.239 746 850 260 254 72;
  • 53) 0.239 746 850 260 254 72 × 2 = 0 + 0.479 493 700 520 509 44;
  • 54) 0.479 493 700 520 509 44 × 2 = 0 + 0.958 987 401 041 018 88;
  • 55) 0.958 987 401 041 018 88 × 2 = 1 + 0.917 974 802 082 037 76;
  • 56) 0.917 974 802 082 037 76 × 2 = 1 + 0.835 949 604 164 075 52;
  • 57) 0.835 949 604 164 075 52 × 2 = 1 + 0.671 899 208 328 151 04;
  • 58) 0.671 899 208 328 151 04 × 2 = 1 + 0.343 798 416 656 302 08;
  • 59) 0.343 798 416 656 302 08 × 2 = 0 + 0.687 596 833 312 604 16;
  • 60) 0.687 596 833 312 604 16 × 2 = 1 + 0.375 193 666 625 208 32;
  • 61) 0.375 193 666 625 208 32 × 2 = 0 + 0.750 387 333 250 416 64;
  • 62) 0.750 387 333 250 416 64 × 2 = 1 + 0.500 774 666 500 833 28;
  • 63) 0.500 774 666 500 833 28 × 2 = 1 + 0.001 549 333 001 666 56;
  • 64) 0.001 549 333 001 666 56 × 2 = 0 + 0.003 098 666 003 333 12;
  • 65) 0.003 098 666 003 333 12 × 2 = 0 + 0.006 197 332 006 666 24;
  • 66) 0.006 197 332 006 666 24 × 2 = 0 + 0.012 394 664 013 332 48;
  • 67) 0.012 394 664 013 332 48 × 2 = 0 + 0.024 789 328 026 664 96;
  • 68) 0.024 789 328 026 664 96 × 2 = 0 + 0.049 578 656 053 329 92;
  • 69) 0.049 578 656 053 329 92 × 2 = 0 + 0.099 157 312 106 659 84;
  • 70) 0.099 157 312 106 659 84 × 2 = 0 + 0.198 314 624 213 319 68;
  • 71) 0.198 314 624 213 319 68 × 2 = 0 + 0.396 629 248 426 639 36;
  • 72) 0.396 629 248 426 639 36 × 2 = 0 + 0.793 258 496 853 278 72;
  • 73) 0.793 258 496 853 278 72 × 2 = 1 + 0.586 516 993 706 557 44;
  • 74) 0.586 516 993 706 557 44 × 2 = 1 + 0.173 033 987 413 114 88;
  • 75) 0.173 033 987 413 114 88 × 2 = 0 + 0.346 067 974 826 229 76;
  • 76) 0.346 067 974 826 229 76 × 2 = 0 + 0.692 135 949 652 459 52;
  • 77) 0.692 135 949 652 459 52 × 2 = 1 + 0.384 271 899 304 919 04;
  • 78) 0.384 271 899 304 919 04 × 2 = 0 + 0.768 543 798 609 838 08;
  • 79) 0.768 543 798 609 838 08 × 2 = 1 + 0.537 087 597 219 676 16;
  • 80) 0.537 087 597 219 676 16 × 2 = 1 + 0.074 175 194 439 352 32;
  • 81) 0.074 175 194 439 352 32 × 2 = 0 + 0.148 350 388 878 704 64;
  • 82) 0.148 350 388 878 704 64 × 2 = 0 + 0.296 700 777 757 409 28;
  • 83) 0.296 700 777 757 409 28 × 2 = 0 + 0.593 401 555 514 818 56;
  • 84) 0.593 401 555 514 818 56 × 2 = 1 + 0.186 803 111 029 637 12;
  • 85) 0.186 803 111 029 637 12 × 2 = 0 + 0.373 606 222 059 274 24;
  • 86) 0.373 606 222 059 274 24 × 2 = 0 + 0.747 212 444 118 548 48;
  • 87) 0.747 212 444 118 548 48 × 2 = 1 + 0.494 424 888 237 096 96;
  • 88) 0.494 424 888 237 096 96 × 2 = 0 + 0.988 849 776 474 193 92;
  • 89) 0.988 849 776 474 193 92 × 2 = 1 + 0.977 699 552 948 387 84;
  • 90) 0.977 699 552 948 387 84 × 2 = 1 + 0.955 399 105 896 775 68;
  • 91) 0.955 399 105 896 775 68 × 2 = 1 + 0.910 798 211 793 551 36;
  • 92) 0.910 798 211 793 551 36 × 2 = 1 + 0.821 596 423 587 102 72;
  • 93) 0.821 596 423 587 102 72 × 2 = 1 + 0.643 192 847 174 205 44;
  • 94) 0.643 192 847 174 205 44 × 2 = 1 + 0.286 385 694 348 410 88;
  • 95) 0.286 385 694 348 410 88 × 2 = 0 + 0.572 771 388 696 821 76;
  • 96) 0.572 771 388 696 821 76 × 2 = 1 + 0.145 542 777 393 643 52;
  • 97) 0.145 542 777 393 643 52 × 2 = 0 + 0.291 085 554 787 287 04;
  • 98) 0.291 085 554 787 287 04 × 2 = 0 + 0.582 171 109 574 574 08;
  • 99) 0.582 171 109 574 574 08 × 2 = 1 + 0.164 342 219 149 148 16;

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:


0.000 000 000 000 013 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1110 0011 1101 0110 0000 0000 1100 1011 0001 0010 1111 1101 001(2)

6. Positive number before normalization:

0.000 000 000 000 013 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1110 0011 1101 0110 0000 0000 1100 1011 0001 0010 1111 1101 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 47 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 013 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1110 0011 1101 0110 0000 0000 1100 1011 0001 0010 1111 1101 001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1110 0011 1101 0110 0000 0000 1100 1011 0001 0010 1111 1101 001(2) × 20 =

1.1111 0001 1110 1011 0000 0000 0110 0101 1000 1001 0111 1110 1001(2) × 2-47


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): -47

Mantissa (not normalized):
1.1111 0001 1110 1011 0000 0000 0110 0101 1000 1001 0111 1110 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-47 + 2(11-1) - 1 =

(-47 + 1 023)(10) =

976(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;
  • 976 ÷ 2 = 488 + 0;
  • 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) =

976(10) =

011 1101 0000(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. 1111 0001 1110 1011 0000 0000 0110 0101 1000 1001 0111 1110 1001 =

1111 0001 1110 1011 0000 0000 0110 0101 1000 1001 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 0000

Mantissa (52 bits) =
1111 0001 1110 1011 0000 0000 0110 0101 1000 1001 0111 1110 1001


Decimal number -0.000 000 000 000 013 82 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0000 - 1111 0001 1110 1011 0000 0000 0110 0101 1000 1001 0111 1110 1001

» Convert decimal -0.000 000 000 000 013 1 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