-0.000 000 000 000 013 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 013 68(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 68(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 68| = 0.000 000 000 000 013 68


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 68.

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 68 × 2 = 0 + 0.000 000 000 000 027 36;
  • 2) 0.000 000 000 000 027 36 × 2 = 0 + 0.000 000 000 000 054 72;
  • 3) 0.000 000 000 000 054 72 × 2 = 0 + 0.000 000 000 000 109 44;
  • 4) 0.000 000 000 000 109 44 × 2 = 0 + 0.000 000 000 000 218 88;
  • 5) 0.000 000 000 000 218 88 × 2 = 0 + 0.000 000 000 000 437 76;
  • 6) 0.000 000 000 000 437 76 × 2 = 0 + 0.000 000 000 000 875 52;
  • 7) 0.000 000 000 000 875 52 × 2 = 0 + 0.000 000 000 001 751 04;
  • 8) 0.000 000 000 001 751 04 × 2 = 0 + 0.000 000 000 003 502 08;
  • 9) 0.000 000 000 003 502 08 × 2 = 0 + 0.000 000 000 007 004 16;
  • 10) 0.000 000 000 007 004 16 × 2 = 0 + 0.000 000 000 014 008 32;
  • 11) 0.000 000 000 014 008 32 × 2 = 0 + 0.000 000 000 028 016 64;
  • 12) 0.000 000 000 028 016 64 × 2 = 0 + 0.000 000 000 056 033 28;
  • 13) 0.000 000 000 056 033 28 × 2 = 0 + 0.000 000 000 112 066 56;
  • 14) 0.000 000 000 112 066 56 × 2 = 0 + 0.000 000 000 224 133 12;
  • 15) 0.000 000 000 224 133 12 × 2 = 0 + 0.000 000 000 448 266 24;
  • 16) 0.000 000 000 448 266 24 × 2 = 0 + 0.000 000 000 896 532 48;
  • 17) 0.000 000 000 896 532 48 × 2 = 0 + 0.000 000 001 793 064 96;
  • 18) 0.000 000 001 793 064 96 × 2 = 0 + 0.000 000 003 586 129 92;
  • 19) 0.000 000 003 586 129 92 × 2 = 0 + 0.000 000 007 172 259 84;
  • 20) 0.000 000 007 172 259 84 × 2 = 0 + 0.000 000 014 344 519 68;
  • 21) 0.000 000 014 344 519 68 × 2 = 0 + 0.000 000 028 689 039 36;
  • 22) 0.000 000 028 689 039 36 × 2 = 0 + 0.000 000 057 378 078 72;
  • 23) 0.000 000 057 378 078 72 × 2 = 0 + 0.000 000 114 756 157 44;
  • 24) 0.000 000 114 756 157 44 × 2 = 0 + 0.000 000 229 512 314 88;
  • 25) 0.000 000 229 512 314 88 × 2 = 0 + 0.000 000 459 024 629 76;
  • 26) 0.000 000 459 024 629 76 × 2 = 0 + 0.000 000 918 049 259 52;
  • 27) 0.000 000 918 049 259 52 × 2 = 0 + 0.000 001 836 098 519 04;
  • 28) 0.000 001 836 098 519 04 × 2 = 0 + 0.000 003 672 197 038 08;
  • 29) 0.000 003 672 197 038 08 × 2 = 0 + 0.000 007 344 394 076 16;
  • 30) 0.000 007 344 394 076 16 × 2 = 0 + 0.000 014 688 788 152 32;
  • 31) 0.000 014 688 788 152 32 × 2 = 0 + 0.000 029 377 576 304 64;
  • 32) 0.000 029 377 576 304 64 × 2 = 0 + 0.000 058 755 152 609 28;
  • 33) 0.000 058 755 152 609 28 × 2 = 0 + 0.000 117 510 305 218 56;
  • 34) 0.000 117 510 305 218 56 × 2 = 0 + 0.000 235 020 610 437 12;
  • 35) 0.000 235 020 610 437 12 × 2 = 0 + 0.000 470 041 220 874 24;
  • 36) 0.000 470 041 220 874 24 × 2 = 0 + 0.000 940 082 441 748 48;
  • 37) 0.000 940 082 441 748 48 × 2 = 0 + 0.001 880 164 883 496 96;
  • 38) 0.001 880 164 883 496 96 × 2 = 0 + 0.003 760 329 766 993 92;
  • 39) 0.003 760 329 766 993 92 × 2 = 0 + 0.007 520 659 533 987 84;
  • 40) 0.007 520 659 533 987 84 × 2 = 0 + 0.015 041 319 067 975 68;
  • 41) 0.015 041 319 067 975 68 × 2 = 0 + 0.030 082 638 135 951 36;
  • 42) 0.030 082 638 135 951 36 × 2 = 0 + 0.060 165 276 271 902 72;
  • 43) 0.060 165 276 271 902 72 × 2 = 0 + 0.120 330 552 543 805 44;
  • 44) 0.120 330 552 543 805 44 × 2 = 0 + 0.240 661 105 087 610 88;
  • 45) 0.240 661 105 087 610 88 × 2 = 0 + 0.481 322 210 175 221 76;
  • 46) 0.481 322 210 175 221 76 × 2 = 0 + 0.962 644 420 350 443 52;
  • 47) 0.962 644 420 350 443 52 × 2 = 1 + 0.925 288 840 700 887 04;
  • 48) 0.925 288 840 700 887 04 × 2 = 1 + 0.850 577 681 401 774 08;
  • 49) 0.850 577 681 401 774 08 × 2 = 1 + 0.701 155 362 803 548 16;
  • 50) 0.701 155 362 803 548 16 × 2 = 1 + 0.402 310 725 607 096 32;
  • 51) 0.402 310 725 607 096 32 × 2 = 0 + 0.804 621 451 214 192 64;
  • 52) 0.804 621 451 214 192 64 × 2 = 1 + 0.609 242 902 428 385 28;
  • 53) 0.609 242 902 428 385 28 × 2 = 1 + 0.218 485 804 856 770 56;
  • 54) 0.218 485 804 856 770 56 × 2 = 0 + 0.436 971 609 713 541 12;
  • 55) 0.436 971 609 713 541 12 × 2 = 0 + 0.873 943 219 427 082 24;
  • 56) 0.873 943 219 427 082 24 × 2 = 1 + 0.747 886 438 854 164 48;
  • 57) 0.747 886 438 854 164 48 × 2 = 1 + 0.495 772 877 708 328 96;
  • 58) 0.495 772 877 708 328 96 × 2 = 0 + 0.991 545 755 416 657 92;
  • 59) 0.991 545 755 416 657 92 × 2 = 1 + 0.983 091 510 833 315 84;
  • 60) 0.983 091 510 833 315 84 × 2 = 1 + 0.966 183 021 666 631 68;
  • 61) 0.966 183 021 666 631 68 × 2 = 1 + 0.932 366 043 333 263 36;
  • 62) 0.932 366 043 333 263 36 × 2 = 1 + 0.864 732 086 666 526 72;
  • 63) 0.864 732 086 666 526 72 × 2 = 1 + 0.729 464 173 333 053 44;
  • 64) 0.729 464 173 333 053 44 × 2 = 1 + 0.458 928 346 666 106 88;
  • 65) 0.458 928 346 666 106 88 × 2 = 0 + 0.917 856 693 332 213 76;
  • 66) 0.917 856 693 332 213 76 × 2 = 1 + 0.835 713 386 664 427 52;
  • 67) 0.835 713 386 664 427 52 × 2 = 1 + 0.671 426 773 328 855 04;
  • 68) 0.671 426 773 328 855 04 × 2 = 1 + 0.342 853 546 657 710 08;
  • 69) 0.342 853 546 657 710 08 × 2 = 0 + 0.685 707 093 315 420 16;
  • 70) 0.685 707 093 315 420 16 × 2 = 1 + 0.371 414 186 630 840 32;
  • 71) 0.371 414 186 630 840 32 × 2 = 0 + 0.742 828 373 261 680 64;
  • 72) 0.742 828 373 261 680 64 × 2 = 1 + 0.485 656 746 523 361 28;
  • 73) 0.485 656 746 523 361 28 × 2 = 0 + 0.971 313 493 046 722 56;
  • 74) 0.971 313 493 046 722 56 × 2 = 1 + 0.942 626 986 093 445 12;
  • 75) 0.942 626 986 093 445 12 × 2 = 1 + 0.885 253 972 186 890 24;
  • 76) 0.885 253 972 186 890 24 × 2 = 1 + 0.770 507 944 373 780 48;
  • 77) 0.770 507 944 373 780 48 × 2 = 1 + 0.541 015 888 747 560 96;
  • 78) 0.541 015 888 747 560 96 × 2 = 1 + 0.082 031 777 495 121 92;
  • 79) 0.082 031 777 495 121 92 × 2 = 0 + 0.164 063 554 990 243 84;
  • 80) 0.164 063 554 990 243 84 × 2 = 0 + 0.328 127 109 980 487 68;
  • 81) 0.328 127 109 980 487 68 × 2 = 0 + 0.656 254 219 960 975 36;
  • 82) 0.656 254 219 960 975 36 × 2 = 1 + 0.312 508 439 921 950 72;
  • 83) 0.312 508 439 921 950 72 × 2 = 0 + 0.625 016 879 843 901 44;
  • 84) 0.625 016 879 843 901 44 × 2 = 1 + 0.250 033 759 687 802 88;
  • 85) 0.250 033 759 687 802 88 × 2 = 0 + 0.500 067 519 375 605 76;
  • 86) 0.500 067 519 375 605 76 × 2 = 1 + 0.000 135 038 751 211 52;
  • 87) 0.000 135 038 751 211 52 × 2 = 0 + 0.000 270 077 502 423 04;
  • 88) 0.000 270 077 502 423 04 × 2 = 0 + 0.000 540 155 004 846 08;
  • 89) 0.000 540 155 004 846 08 × 2 = 0 + 0.001 080 310 009 692 16;
  • 90) 0.001 080 310 009 692 16 × 2 = 0 + 0.002 160 620 019 384 32;
  • 91) 0.002 160 620 019 384 32 × 2 = 0 + 0.004 321 240 038 768 64;
  • 92) 0.004 321 240 038 768 64 × 2 = 0 + 0.008 642 480 077 537 28;
  • 93) 0.008 642 480 077 537 28 × 2 = 0 + 0.017 284 960 155 074 56;
  • 94) 0.017 284 960 155 074 56 × 2 = 0 + 0.034 569 920 310 149 12;
  • 95) 0.034 569 920 310 149 12 × 2 = 0 + 0.069 139 840 620 298 24;
  • 96) 0.069 139 840 620 298 24 × 2 = 0 + 0.138 279 681 240 596 48;
  • 97) 0.138 279 681 240 596 48 × 2 = 0 + 0.276 559 362 481 192 96;
  • 98) 0.276 559 362 481 192 96 × 2 = 0 + 0.553 118 724 962 385 92;
  • 99) 0.553 118 724 962 385 92 × 2 = 1 + 0.106 237 449 924 771 84;

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 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 1001 1011 1111 0111 0101 0111 1100 0101 0100 0000 0000 001(2)

6. Positive number before normalization:

0.000 000 000 000 013 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 1001 1011 1111 0111 0101 0111 1100 0101 0100 0000 0000 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 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 1001 1011 1111 0111 0101 0111 1100 0101 0100 0000 0000 001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 1001 1011 1111 0111 0101 0111 1100 0101 0100 0000 0000 001(2) × 20 =

1.1110 1100 1101 1111 1011 1010 1011 1110 0010 1010 0000 0000 0001(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):
1.1110 1100 1101 1111 1011 1010 1011 1110 0010 1010 0000 0000 0001


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. 1110 1100 1101 1111 1011 1010 1011 1110 0010 1010 0000 0000 0001 =

1110 1100 1101 1111 1011 1010 1011 1110 0010 1010 0000 0000 0001


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) =
1110 1100 1101 1111 1011 1010 1011 1110 0010 1010 0000 0000 0001


Decimal number -0.000 000 000 000 013 68 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0000 - 1110 1100 1101 1111 1011 1010 1011 1110 0010 1010 0000 0000 0001

» Convert decimal -0.000 000 000 000 013 3 to 64 bit double precision IEEE 754 binary floating point representation standard

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» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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