-0.000 000 000 000 013 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 013 1(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 1(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 1| = 0.000 000 000 000 013 1


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 1.

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 1 × 2 = 0 + 0.000 000 000 000 026 2;
  • 2) 0.000 000 000 000 026 2 × 2 = 0 + 0.000 000 000 000 052 4;
  • 3) 0.000 000 000 000 052 4 × 2 = 0 + 0.000 000 000 000 104 8;
  • 4) 0.000 000 000 000 104 8 × 2 = 0 + 0.000 000 000 000 209 6;
  • 5) 0.000 000 000 000 209 6 × 2 = 0 + 0.000 000 000 000 419 2;
  • 6) 0.000 000 000 000 419 2 × 2 = 0 + 0.000 000 000 000 838 4;
  • 7) 0.000 000 000 000 838 4 × 2 = 0 + 0.000 000 000 001 676 8;
  • 8) 0.000 000 000 001 676 8 × 2 = 0 + 0.000 000 000 003 353 6;
  • 9) 0.000 000 000 003 353 6 × 2 = 0 + 0.000 000 000 006 707 2;
  • 10) 0.000 000 000 006 707 2 × 2 = 0 + 0.000 000 000 013 414 4;
  • 11) 0.000 000 000 013 414 4 × 2 = 0 + 0.000 000 000 026 828 8;
  • 12) 0.000 000 000 026 828 8 × 2 = 0 + 0.000 000 000 053 657 6;
  • 13) 0.000 000 000 053 657 6 × 2 = 0 + 0.000 000 000 107 315 2;
  • 14) 0.000 000 000 107 315 2 × 2 = 0 + 0.000 000 000 214 630 4;
  • 15) 0.000 000 000 214 630 4 × 2 = 0 + 0.000 000 000 429 260 8;
  • 16) 0.000 000 000 429 260 8 × 2 = 0 + 0.000 000 000 858 521 6;
  • 17) 0.000 000 000 858 521 6 × 2 = 0 + 0.000 000 001 717 043 2;
  • 18) 0.000 000 001 717 043 2 × 2 = 0 + 0.000 000 003 434 086 4;
  • 19) 0.000 000 003 434 086 4 × 2 = 0 + 0.000 000 006 868 172 8;
  • 20) 0.000 000 006 868 172 8 × 2 = 0 + 0.000 000 013 736 345 6;
  • 21) 0.000 000 013 736 345 6 × 2 = 0 + 0.000 000 027 472 691 2;
  • 22) 0.000 000 027 472 691 2 × 2 = 0 + 0.000 000 054 945 382 4;
  • 23) 0.000 000 054 945 382 4 × 2 = 0 + 0.000 000 109 890 764 8;
  • 24) 0.000 000 109 890 764 8 × 2 = 0 + 0.000 000 219 781 529 6;
  • 25) 0.000 000 219 781 529 6 × 2 = 0 + 0.000 000 439 563 059 2;
  • 26) 0.000 000 439 563 059 2 × 2 = 0 + 0.000 000 879 126 118 4;
  • 27) 0.000 000 879 126 118 4 × 2 = 0 + 0.000 001 758 252 236 8;
  • 28) 0.000 001 758 252 236 8 × 2 = 0 + 0.000 003 516 504 473 6;
  • 29) 0.000 003 516 504 473 6 × 2 = 0 + 0.000 007 033 008 947 2;
  • 30) 0.000 007 033 008 947 2 × 2 = 0 + 0.000 014 066 017 894 4;
  • 31) 0.000 014 066 017 894 4 × 2 = 0 + 0.000 028 132 035 788 8;
  • 32) 0.000 028 132 035 788 8 × 2 = 0 + 0.000 056 264 071 577 6;
  • 33) 0.000 056 264 071 577 6 × 2 = 0 + 0.000 112 528 143 155 2;
  • 34) 0.000 112 528 143 155 2 × 2 = 0 + 0.000 225 056 286 310 4;
  • 35) 0.000 225 056 286 310 4 × 2 = 0 + 0.000 450 112 572 620 8;
  • 36) 0.000 450 112 572 620 8 × 2 = 0 + 0.000 900 225 145 241 6;
  • 37) 0.000 900 225 145 241 6 × 2 = 0 + 0.001 800 450 290 483 2;
  • 38) 0.001 800 450 290 483 2 × 2 = 0 + 0.003 600 900 580 966 4;
  • 39) 0.003 600 900 580 966 4 × 2 = 0 + 0.007 201 801 161 932 8;
  • 40) 0.007 201 801 161 932 8 × 2 = 0 + 0.014 403 602 323 865 6;
  • 41) 0.014 403 602 323 865 6 × 2 = 0 + 0.028 807 204 647 731 2;
  • 42) 0.028 807 204 647 731 2 × 2 = 0 + 0.057 614 409 295 462 4;
  • 43) 0.057 614 409 295 462 4 × 2 = 0 + 0.115 228 818 590 924 8;
  • 44) 0.115 228 818 590 924 8 × 2 = 0 + 0.230 457 637 181 849 6;
  • 45) 0.230 457 637 181 849 6 × 2 = 0 + 0.460 915 274 363 699 2;
  • 46) 0.460 915 274 363 699 2 × 2 = 0 + 0.921 830 548 727 398 4;
  • 47) 0.921 830 548 727 398 4 × 2 = 1 + 0.843 661 097 454 796 8;
  • 48) 0.843 661 097 454 796 8 × 2 = 1 + 0.687 322 194 909 593 6;
  • 49) 0.687 322 194 909 593 6 × 2 = 1 + 0.374 644 389 819 187 2;
  • 50) 0.374 644 389 819 187 2 × 2 = 0 + 0.749 288 779 638 374 4;
  • 51) 0.749 288 779 638 374 4 × 2 = 1 + 0.498 577 559 276 748 8;
  • 52) 0.498 577 559 276 748 8 × 2 = 0 + 0.997 155 118 553 497 6;
  • 53) 0.997 155 118 553 497 6 × 2 = 1 + 0.994 310 237 106 995 2;
  • 54) 0.994 310 237 106 995 2 × 2 = 1 + 0.988 620 474 213 990 4;
  • 55) 0.988 620 474 213 990 4 × 2 = 1 + 0.977 240 948 427 980 8;
  • 56) 0.977 240 948 427 980 8 × 2 = 1 + 0.954 481 896 855 961 6;
  • 57) 0.954 481 896 855 961 6 × 2 = 1 + 0.908 963 793 711 923 2;
  • 58) 0.908 963 793 711 923 2 × 2 = 1 + 0.817 927 587 423 846 4;
  • 59) 0.817 927 587 423 846 4 × 2 = 1 + 0.635 855 174 847 692 8;
  • 60) 0.635 855 174 847 692 8 × 2 = 1 + 0.271 710 349 695 385 6;
  • 61) 0.271 710 349 695 385 6 × 2 = 0 + 0.543 420 699 390 771 2;
  • 62) 0.543 420 699 390 771 2 × 2 = 1 + 0.086 841 398 781 542 4;
  • 63) 0.086 841 398 781 542 4 × 2 = 0 + 0.173 682 797 563 084 8;
  • 64) 0.173 682 797 563 084 8 × 2 = 0 + 0.347 365 595 126 169 6;
  • 65) 0.347 365 595 126 169 6 × 2 = 0 + 0.694 731 190 252 339 2;
  • 66) 0.694 731 190 252 339 2 × 2 = 1 + 0.389 462 380 504 678 4;
  • 67) 0.389 462 380 504 678 4 × 2 = 0 + 0.778 924 761 009 356 8;
  • 68) 0.778 924 761 009 356 8 × 2 = 1 + 0.557 849 522 018 713 6;
  • 69) 0.557 849 522 018 713 6 × 2 = 1 + 0.115 699 044 037 427 2;
  • 70) 0.115 699 044 037 427 2 × 2 = 0 + 0.231 398 088 074 854 4;
  • 71) 0.231 398 088 074 854 4 × 2 = 0 + 0.462 796 176 149 708 8;
  • 72) 0.462 796 176 149 708 8 × 2 = 0 + 0.925 592 352 299 417 6;
  • 73) 0.925 592 352 299 417 6 × 2 = 1 + 0.851 184 704 598 835 2;
  • 74) 0.851 184 704 598 835 2 × 2 = 1 + 0.702 369 409 197 670 4;
  • 75) 0.702 369 409 197 670 4 × 2 = 1 + 0.404 738 818 395 340 8;
  • 76) 0.404 738 818 395 340 8 × 2 = 0 + 0.809 477 636 790 681 6;
  • 77) 0.809 477 636 790 681 6 × 2 = 1 + 0.618 955 273 581 363 2;
  • 78) 0.618 955 273 581 363 2 × 2 = 1 + 0.237 910 547 162 726 4;
  • 79) 0.237 910 547 162 726 4 × 2 = 0 + 0.475 821 094 325 452 8;
  • 80) 0.475 821 094 325 452 8 × 2 = 0 + 0.951 642 188 650 905 6;
  • 81) 0.951 642 188 650 905 6 × 2 = 1 + 0.903 284 377 301 811 2;
  • 82) 0.903 284 377 301 811 2 × 2 = 1 + 0.806 568 754 603 622 4;
  • 83) 0.806 568 754 603 622 4 × 2 = 1 + 0.613 137 509 207 244 8;
  • 84) 0.613 137 509 207 244 8 × 2 = 1 + 0.226 275 018 414 489 6;
  • 85) 0.226 275 018 414 489 6 × 2 = 0 + 0.452 550 036 828 979 2;
  • 86) 0.452 550 036 828 979 2 × 2 = 0 + 0.905 100 073 657 958 4;
  • 87) 0.905 100 073 657 958 4 × 2 = 1 + 0.810 200 147 315 916 8;
  • 88) 0.810 200 147 315 916 8 × 2 = 1 + 0.620 400 294 631 833 6;
  • 89) 0.620 400 294 631 833 6 × 2 = 1 + 0.240 800 589 263 667 2;
  • 90) 0.240 800 589 263 667 2 × 2 = 0 + 0.481 601 178 527 334 4;
  • 91) 0.481 601 178 527 334 4 × 2 = 0 + 0.963 202 357 054 668 8;
  • 92) 0.963 202 357 054 668 8 × 2 = 1 + 0.926 404 714 109 337 6;
  • 93) 0.926 404 714 109 337 6 × 2 = 1 + 0.852 809 428 218 675 2;
  • 94) 0.852 809 428 218 675 2 × 2 = 1 + 0.705 618 856 437 350 4;
  • 95) 0.705 618 856 437 350 4 × 2 = 1 + 0.411 237 712 874 700 8;
  • 96) 0.411 237 712 874 700 8 × 2 = 0 + 0.822 475 425 749 401 6;
  • 97) 0.822 475 425 749 401 6 × 2 = 1 + 0.644 950 851 498 803 2;
  • 98) 0.644 950 851 498 803 2 × 2 = 1 + 0.289 901 702 997 606 4;
  • 99) 0.289 901 702 997 606 4 × 2 = 0 + 0.579 803 405 995 212 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:


0.000 000 000 000 013 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1010 1111 1111 0100 0101 1000 1110 1100 1111 0011 1001 1110 110(2)

6. Positive number before normalization:

0.000 000 000 000 013 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1010 1111 1111 0100 0101 1000 1110 1100 1111 0011 1001 1110 110(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 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1010 1111 1111 0100 0101 1000 1110 1100 1111 0011 1001 1110 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1010 1111 1111 0100 0101 1000 1110 1100 1111 0011 1001 1110 110(2) × 20 =

1.1101 0111 1111 1010 0010 1100 0111 0110 0111 1001 1100 1111 0110(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):
1.1101 0111 1111 1010 0010 1100 0111 0110 0111 1001 1100 1111 0110


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. 1101 0111 1111 1010 0010 1100 0111 0110 0111 1001 1100 1111 0110 =

1101 0111 1111 1010 0010 1100 0111 0110 0111 1001 1100 1111 0110


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) =
1101 0111 1111 1010 0010 1100 0111 0110 0111 1001 1100 1111 0110


Decimal number -0.000 000 000 000 013 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0000 - 1101 0111 1111 1010 0010 1100 0111 0110 0111 1001 1100 1111 0110

» Convert decimal -0.000 000 000 000 011 2 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