1.111 111 111 111 111 111 111 636 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.111 111 111 111 111 111 111 636(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
1.111 111 111 111 111 111 111 636(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: 0.111 111 111 111 111 111 111 636.

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) 0.111 111 111 111 111 111 111 636 × 2 = 0 + 0.222 222 222 222 222 222 223 272;
  • 2) 0.222 222 222 222 222 222 223 272 × 2 = 0 + 0.444 444 444 444 444 444 446 544;
  • 3) 0.444 444 444 444 444 444 446 544 × 2 = 0 + 0.888 888 888 888 888 888 893 088;
  • 4) 0.888 888 888 888 888 888 893 088 × 2 = 1 + 0.777 777 777 777 777 777 786 176;
  • 5) 0.777 777 777 777 777 777 786 176 × 2 = 1 + 0.555 555 555 555 555 555 572 352;
  • 6) 0.555 555 555 555 555 555 572 352 × 2 = 1 + 0.111 111 111 111 111 111 144 704;
  • 7) 0.111 111 111 111 111 111 144 704 × 2 = 0 + 0.222 222 222 222 222 222 289 408;
  • 8) 0.222 222 222 222 222 222 289 408 × 2 = 0 + 0.444 444 444 444 444 444 578 816;
  • 9) 0.444 444 444 444 444 444 578 816 × 2 = 0 + 0.888 888 888 888 888 889 157 632;
  • 10) 0.888 888 888 888 888 889 157 632 × 2 = 1 + 0.777 777 777 777 777 778 315 264;
  • 11) 0.777 777 777 777 777 778 315 264 × 2 = 1 + 0.555 555 555 555 555 556 630 528;
  • 12) 0.555 555 555 555 555 556 630 528 × 2 = 1 + 0.111 111 111 111 111 113 261 056;
  • 13) 0.111 111 111 111 111 113 261 056 × 2 = 0 + 0.222 222 222 222 222 226 522 112;
  • 14) 0.222 222 222 222 222 226 522 112 × 2 = 0 + 0.444 444 444 444 444 453 044 224;
  • 15) 0.444 444 444 444 444 453 044 224 × 2 = 0 + 0.888 888 888 888 888 906 088 448;
  • 16) 0.888 888 888 888 888 906 088 448 × 2 = 1 + 0.777 777 777 777 777 812 176 896;
  • 17) 0.777 777 777 777 777 812 176 896 × 2 = 1 + 0.555 555 555 555 555 624 353 792;
  • 18) 0.555 555 555 555 555 624 353 792 × 2 = 1 + 0.111 111 111 111 111 248 707 584;
  • 19) 0.111 111 111 111 111 248 707 584 × 2 = 0 + 0.222 222 222 222 222 497 415 168;
  • 20) 0.222 222 222 222 222 497 415 168 × 2 = 0 + 0.444 444 444 444 444 994 830 336;
  • 21) 0.444 444 444 444 444 994 830 336 × 2 = 0 + 0.888 888 888 888 889 989 660 672;
  • 22) 0.888 888 888 888 889 989 660 672 × 2 = 1 + 0.777 777 777 777 779 979 321 344;
  • 23) 0.777 777 777 777 779 979 321 344 × 2 = 1 + 0.555 555 555 555 559 958 642 688;
  • 24) 0.555 555 555 555 559 958 642 688 × 2 = 1 + 0.111 111 111 111 119 917 285 376;
  • 25) 0.111 111 111 111 119 917 285 376 × 2 = 0 + 0.222 222 222 222 239 834 570 752;
  • 26) 0.222 222 222 222 239 834 570 752 × 2 = 0 + 0.444 444 444 444 479 669 141 504;
  • 27) 0.444 444 444 444 479 669 141 504 × 2 = 0 + 0.888 888 888 888 959 338 283 008;
  • 28) 0.888 888 888 888 959 338 283 008 × 2 = 1 + 0.777 777 777 777 918 676 566 016;
  • 29) 0.777 777 777 777 918 676 566 016 × 2 = 1 + 0.555 555 555 555 837 353 132 032;
  • 30) 0.555 555 555 555 837 353 132 032 × 2 = 1 + 0.111 111 111 111 674 706 264 064;
  • 31) 0.111 111 111 111 674 706 264 064 × 2 = 0 + 0.222 222 222 223 349 412 528 128;
  • 32) 0.222 222 222 223 349 412 528 128 × 2 = 0 + 0.444 444 444 446 698 825 056 256;
  • 33) 0.444 444 444 446 698 825 056 256 × 2 = 0 + 0.888 888 888 893 397 650 112 512;
  • 34) 0.888 888 888 893 397 650 112 512 × 2 = 1 + 0.777 777 777 786 795 300 225 024;
  • 35) 0.777 777 777 786 795 300 225 024 × 2 = 1 + 0.555 555 555 573 590 600 450 048;
  • 36) 0.555 555 555 573 590 600 450 048 × 2 = 1 + 0.111 111 111 147 181 200 900 096;
  • 37) 0.111 111 111 147 181 200 900 096 × 2 = 0 + 0.222 222 222 294 362 401 800 192;
  • 38) 0.222 222 222 294 362 401 800 192 × 2 = 0 + 0.444 444 444 588 724 803 600 384;
  • 39) 0.444 444 444 588 724 803 600 384 × 2 = 0 + 0.888 888 889 177 449 607 200 768;
  • 40) 0.888 888 889 177 449 607 200 768 × 2 = 1 + 0.777 777 778 354 899 214 401 536;
  • 41) 0.777 777 778 354 899 214 401 536 × 2 = 1 + 0.555 555 556 709 798 428 803 072;
  • 42) 0.555 555 556 709 798 428 803 072 × 2 = 1 + 0.111 111 113 419 596 857 606 144;
  • 43) 0.111 111 113 419 596 857 606 144 × 2 = 0 + 0.222 222 226 839 193 715 212 288;
  • 44) 0.222 222 226 839 193 715 212 288 × 2 = 0 + 0.444 444 453 678 387 430 424 576;
  • 45) 0.444 444 453 678 387 430 424 576 × 2 = 0 + 0.888 888 907 356 774 860 849 152;
  • 46) 0.888 888 907 356 774 860 849 152 × 2 = 1 + 0.777 777 814 713 549 721 698 304;
  • 47) 0.777 777 814 713 549 721 698 304 × 2 = 1 + 0.555 555 629 427 099 443 396 608;
  • 48) 0.555 555 629 427 099 443 396 608 × 2 = 1 + 0.111 111 258 854 198 886 793 216;
  • 49) 0.111 111 258 854 198 886 793 216 × 2 = 0 + 0.222 222 517 708 397 773 586 432;
  • 50) 0.222 222 517 708 397 773 586 432 × 2 = 0 + 0.444 445 035 416 795 547 172 864;
  • 51) 0.444 445 035 416 795 547 172 864 × 2 = 0 + 0.888 890 070 833 591 094 345 728;
  • 52) 0.888 890 070 833 591 094 345 728 × 2 = 1 + 0.777 780 141 667 182 188 691 456;
  • 53) 0.777 780 141 667 182 188 691 456 × 2 = 1 + 0.555 560 283 334 364 377 382 912;

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:


0.111 111 111 111 111 111 111 636(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2)

5. Positive number before normalization:

1.111 111 111 111 111 111 111 636(10) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 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:


1.111 111 111 111 111 111 111 636(10) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 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):
1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 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. 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1 =

0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


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) =
0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


Decimal number 1.111 111 111 111 111 111 111 636 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001

Share

» Convert decimal 1.111 111 111 111 111 111 111 696 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 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 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