1.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15(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
1.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15.

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) 0.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15 × 2 = 0 + 0.666 666 666 666 666 518 636 930 049 979 127 943 515 776 3;
  • 2) 0.666 666 666 666 666 518 636 930 049 979 127 943 515 776 3 × 2 = 1 + 0.333 333 333 333 333 037 273 860 099 958 255 887 031 552 6;
  • 3) 0.333 333 333 333 333 037 273 860 099 958 255 887 031 552 6 × 2 = 0 + 0.666 666 666 666 666 074 547 720 199 916 511 774 063 105 2;
  • 4) 0.666 666 666 666 666 074 547 720 199 916 511 774 063 105 2 × 2 = 1 + 0.333 333 333 333 332 149 095 440 399 833 023 548 126 210 4;
  • 5) 0.333 333 333 333 332 149 095 440 399 833 023 548 126 210 4 × 2 = 0 + 0.666 666 666 666 664 298 190 880 799 666 047 096 252 420 8;
  • 6) 0.666 666 666 666 664 298 190 880 799 666 047 096 252 420 8 × 2 = 1 + 0.333 333 333 333 328 596 381 761 599 332 094 192 504 841 6;
  • 7) 0.333 333 333 333 328 596 381 761 599 332 094 192 504 841 6 × 2 = 0 + 0.666 666 666 666 657 192 763 523 198 664 188 385 009 683 2;
  • 8) 0.666 666 666 666 657 192 763 523 198 664 188 385 009 683 2 × 2 = 1 + 0.333 333 333 333 314 385 527 046 397 328 376 770 019 366 4;
  • 9) 0.333 333 333 333 314 385 527 046 397 328 376 770 019 366 4 × 2 = 0 + 0.666 666 666 666 628 771 054 092 794 656 753 540 038 732 8;
  • 10) 0.666 666 666 666 628 771 054 092 794 656 753 540 038 732 8 × 2 = 1 + 0.333 333 333 333 257 542 108 185 589 313 507 080 077 465 6;
  • 11) 0.333 333 333 333 257 542 108 185 589 313 507 080 077 465 6 × 2 = 0 + 0.666 666 666 666 515 084 216 371 178 627 014 160 154 931 2;
  • 12) 0.666 666 666 666 515 084 216 371 178 627 014 160 154 931 2 × 2 = 1 + 0.333 333 333 333 030 168 432 742 357 254 028 320 309 862 4;
  • 13) 0.333 333 333 333 030 168 432 742 357 254 028 320 309 862 4 × 2 = 0 + 0.666 666 666 666 060 336 865 484 714 508 056 640 619 724 8;
  • 14) 0.666 666 666 666 060 336 865 484 714 508 056 640 619 724 8 × 2 = 1 + 0.333 333 333 332 120 673 730 969 429 016 113 281 239 449 6;
  • 15) 0.333 333 333 332 120 673 730 969 429 016 113 281 239 449 6 × 2 = 0 + 0.666 666 666 664 241 347 461 938 858 032 226 562 478 899 2;
  • 16) 0.666 666 666 664 241 347 461 938 858 032 226 562 478 899 2 × 2 = 1 + 0.333 333 333 328 482 694 923 877 716 064 453 124 957 798 4;
  • 17) 0.333 333 333 328 482 694 923 877 716 064 453 124 957 798 4 × 2 = 0 + 0.666 666 666 656 965 389 847 755 432 128 906 249 915 596 8;
  • 18) 0.666 666 666 656 965 389 847 755 432 128 906 249 915 596 8 × 2 = 1 + 0.333 333 333 313 930 779 695 510 864 257 812 499 831 193 6;
  • 19) 0.333 333 333 313 930 779 695 510 864 257 812 499 831 193 6 × 2 = 0 + 0.666 666 666 627 861 559 391 021 728 515 624 999 662 387 2;
  • 20) 0.666 666 666 627 861 559 391 021 728 515 624 999 662 387 2 × 2 = 1 + 0.333 333 333 255 723 118 782 043 457 031 249 999 324 774 4;
  • 21) 0.333 333 333 255 723 118 782 043 457 031 249 999 324 774 4 × 2 = 0 + 0.666 666 666 511 446 237 564 086 914 062 499 998 649 548 8;
  • 22) 0.666 666 666 511 446 237 564 086 914 062 499 998 649 548 8 × 2 = 1 + 0.333 333 333 022 892 475 128 173 828 124 999 997 299 097 6;
  • 23) 0.333 333 333 022 892 475 128 173 828 124 999 997 299 097 6 × 2 = 0 + 0.666 666 666 045 784 950 256 347 656 249 999 994 598 195 2;
  • 24) 0.666 666 666 045 784 950 256 347 656 249 999 994 598 195 2 × 2 = 1 + 0.333 333 332 091 569 900 512 695 312 499 999 989 196 390 4;
  • 25) 0.333 333 332 091 569 900 512 695 312 499 999 989 196 390 4 × 2 = 0 + 0.666 666 664 183 139 801 025 390 624 999 999 978 392 780 8;
  • 26) 0.666 666 664 183 139 801 025 390 624 999 999 978 392 780 8 × 2 = 1 + 0.333 333 328 366 279 602 050 781 249 999 999 956 785 561 6;
  • 27) 0.333 333 328 366 279 602 050 781 249 999 999 956 785 561 6 × 2 = 0 + 0.666 666 656 732 559 204 101 562 499 999 999 913 571 123 2;
  • 28) 0.666 666 656 732 559 204 101 562 499 999 999 913 571 123 2 × 2 = 1 + 0.333 333 313 465 118 408 203 124 999 999 999 827 142 246 4;
  • 29) 0.333 333 313 465 118 408 203 124 999 999 999 827 142 246 4 × 2 = 0 + 0.666 666 626 930 236 816 406 249 999 999 999 654 284 492 8;
  • 30) 0.666 666 626 930 236 816 406 249 999 999 999 654 284 492 8 × 2 = 1 + 0.333 333 253 860 473 632 812 499 999 999 999 308 568 985 6;
  • 31) 0.333 333 253 860 473 632 812 499 999 999 999 308 568 985 6 × 2 = 0 + 0.666 666 507 720 947 265 624 999 999 999 998 617 137 971 2;
  • 32) 0.666 666 507 720 947 265 624 999 999 999 998 617 137 971 2 × 2 = 1 + 0.333 333 015 441 894 531 249 999 999 999 997 234 275 942 4;
  • 33) 0.333 333 015 441 894 531 249 999 999 999 997 234 275 942 4 × 2 = 0 + 0.666 666 030 883 789 062 499 999 999 999 994 468 551 884 8;
  • 34) 0.666 666 030 883 789 062 499 999 999 999 994 468 551 884 8 × 2 = 1 + 0.333 332 061 767 578 124 999 999 999 999 988 937 103 769 6;
  • 35) 0.333 332 061 767 578 124 999 999 999 999 988 937 103 769 6 × 2 = 0 + 0.666 664 123 535 156 249 999 999 999 999 977 874 207 539 2;
  • 36) 0.666 664 123 535 156 249 999 999 999 999 977 874 207 539 2 × 2 = 1 + 0.333 328 247 070 312 499 999 999 999 999 955 748 415 078 4;
  • 37) 0.333 328 247 070 312 499 999 999 999 999 955 748 415 078 4 × 2 = 0 + 0.666 656 494 140 624 999 999 999 999 999 911 496 830 156 8;
  • 38) 0.666 656 494 140 624 999 999 999 999 999 911 496 830 156 8 × 2 = 1 + 0.333 312 988 281 249 999 999 999 999 999 822 993 660 313 6;
  • 39) 0.333 312 988 281 249 999 999 999 999 999 822 993 660 313 6 × 2 = 0 + 0.666 625 976 562 499 999 999 999 999 999 645 987 320 627 2;
  • 40) 0.666 625 976 562 499 999 999 999 999 999 645 987 320 627 2 × 2 = 1 + 0.333 251 953 124 999 999 999 999 999 999 291 974 641 254 4;
  • 41) 0.333 251 953 124 999 999 999 999 999 999 291 974 641 254 4 × 2 = 0 + 0.666 503 906 249 999 999 999 999 999 998 583 949 282 508 8;
  • 42) 0.666 503 906 249 999 999 999 999 999 998 583 949 282 508 8 × 2 = 1 + 0.333 007 812 499 999 999 999 999 999 997 167 898 565 017 6;
  • 43) 0.333 007 812 499 999 999 999 999 999 997 167 898 565 017 6 × 2 = 0 + 0.666 015 624 999 999 999 999 999 999 994 335 797 130 035 2;
  • 44) 0.666 015 624 999 999 999 999 999 999 994 335 797 130 035 2 × 2 = 1 + 0.332 031 249 999 999 999 999 999 999 988 671 594 260 070 4;
  • 45) 0.332 031 249 999 999 999 999 999 999 988 671 594 260 070 4 × 2 = 0 + 0.664 062 499 999 999 999 999 999 999 977 343 188 520 140 8;
  • 46) 0.664 062 499 999 999 999 999 999 999 977 343 188 520 140 8 × 2 = 1 + 0.328 124 999 999 999 999 999 999 999 954 686 377 040 281 6;
  • 47) 0.328 124 999 999 999 999 999 999 999 954 686 377 040 281 6 × 2 = 0 + 0.656 249 999 999 999 999 999 999 999 909 372 754 080 563 2;
  • 48) 0.656 249 999 999 999 999 999 999 999 909 372 754 080 563 2 × 2 = 1 + 0.312 499 999 999 999 999 999 999 999 818 745 508 161 126 4;
  • 49) 0.312 499 999 999 999 999 999 999 999 818 745 508 161 126 4 × 2 = 0 + 0.624 999 999 999 999 999 999 999 999 637 491 016 322 252 8;
  • 50) 0.624 999 999 999 999 999 999 999 999 637 491 016 322 252 8 × 2 = 1 + 0.249 999 999 999 999 999 999 999 999 274 982 032 644 505 6;
  • 51) 0.249 999 999 999 999 999 999 999 999 274 982 032 644 505 6 × 2 = 0 + 0.499 999 999 999 999 999 999 999 998 549 964 065 289 011 2;
  • 52) 0.499 999 999 999 999 999 999 999 998 549 964 065 289 011 2 × 2 = 0 + 0.999 999 999 999 999 999 999 999 997 099 928 130 578 022 4;
  • 53) 0.999 999 999 999 999 999 999 999 997 099 928 130 578 022 4 × 2 = 1 + 0.999 999 999 999 999 999 999 999 994 199 856 261 156 044 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).


4. 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:


0.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 1(2)

5. Positive number before normalization:

1.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15(10) =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15(10) =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 1(2) =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 1(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(10)


9. 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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1023(10) =

011 1111 1111(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 1 =

0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1111

Mantissa (52 bits) =
0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100


Decimal number 1.333 333 333 333 333 259 318 465 024 989 563 971 757 888 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100

Share

» Convert decimal 1.333 333 333 333 333 259 318 465 024 989 563 971 757 888 12 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: My Manager Only Says “We” When Referring to “My Work”. Gemini Notebook YouTube Video of 11.13 minutes

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