1.792 505 054 354 527 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.792 505 054 354 527 9(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
1.792 505 054 354 527 9(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: 0.792 505 054 354 527 9.

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) 0.792 505 054 354 527 9 × 2 = 1 + 0.585 010 108 709 055 8;
  • 2) 0.585 010 108 709 055 8 × 2 = 1 + 0.170 020 217 418 111 6;
  • 3) 0.170 020 217 418 111 6 × 2 = 0 + 0.340 040 434 836 223 2;
  • 4) 0.340 040 434 836 223 2 × 2 = 0 + 0.680 080 869 672 446 4;
  • 5) 0.680 080 869 672 446 4 × 2 = 1 + 0.360 161 739 344 892 8;
  • 6) 0.360 161 739 344 892 8 × 2 = 0 + 0.720 323 478 689 785 6;
  • 7) 0.720 323 478 689 785 6 × 2 = 1 + 0.440 646 957 379 571 2;
  • 8) 0.440 646 957 379 571 2 × 2 = 0 + 0.881 293 914 759 142 4;
  • 9) 0.881 293 914 759 142 4 × 2 = 1 + 0.762 587 829 518 284 8;
  • 10) 0.762 587 829 518 284 8 × 2 = 1 + 0.525 175 659 036 569 6;
  • 11) 0.525 175 659 036 569 6 × 2 = 1 + 0.050 351 318 073 139 2;
  • 12) 0.050 351 318 073 139 2 × 2 = 0 + 0.100 702 636 146 278 4;
  • 13) 0.100 702 636 146 278 4 × 2 = 0 + 0.201 405 272 292 556 8;
  • 14) 0.201 405 272 292 556 8 × 2 = 0 + 0.402 810 544 585 113 6;
  • 15) 0.402 810 544 585 113 6 × 2 = 0 + 0.805 621 089 170 227 2;
  • 16) 0.805 621 089 170 227 2 × 2 = 1 + 0.611 242 178 340 454 4;
  • 17) 0.611 242 178 340 454 4 × 2 = 1 + 0.222 484 356 680 908 8;
  • 18) 0.222 484 356 680 908 8 × 2 = 0 + 0.444 968 713 361 817 6;
  • 19) 0.444 968 713 361 817 6 × 2 = 0 + 0.889 937 426 723 635 2;
  • 20) 0.889 937 426 723 635 2 × 2 = 1 + 0.779 874 853 447 270 4;
  • 21) 0.779 874 853 447 270 4 × 2 = 1 + 0.559 749 706 894 540 8;
  • 22) 0.559 749 706 894 540 8 × 2 = 1 + 0.119 499 413 789 081 6;
  • 23) 0.119 499 413 789 081 6 × 2 = 0 + 0.238 998 827 578 163 2;
  • 24) 0.238 998 827 578 163 2 × 2 = 0 + 0.477 997 655 156 326 4;
  • 25) 0.477 997 655 156 326 4 × 2 = 0 + 0.955 995 310 312 652 8;
  • 26) 0.955 995 310 312 652 8 × 2 = 1 + 0.911 990 620 625 305 6;
  • 27) 0.911 990 620 625 305 6 × 2 = 1 + 0.823 981 241 250 611 2;
  • 28) 0.823 981 241 250 611 2 × 2 = 1 + 0.647 962 482 501 222 4;
  • 29) 0.647 962 482 501 222 4 × 2 = 1 + 0.295 924 965 002 444 8;
  • 30) 0.295 924 965 002 444 8 × 2 = 0 + 0.591 849 930 004 889 6;
  • 31) 0.591 849 930 004 889 6 × 2 = 1 + 0.183 699 860 009 779 2;
  • 32) 0.183 699 860 009 779 2 × 2 = 0 + 0.367 399 720 019 558 4;
  • 33) 0.367 399 720 019 558 4 × 2 = 0 + 0.734 799 440 039 116 8;
  • 34) 0.734 799 440 039 116 8 × 2 = 1 + 0.469 598 880 078 233 6;
  • 35) 0.469 598 880 078 233 6 × 2 = 0 + 0.939 197 760 156 467 2;
  • 36) 0.939 197 760 156 467 2 × 2 = 1 + 0.878 395 520 312 934 4;
  • 37) 0.878 395 520 312 934 4 × 2 = 1 + 0.756 791 040 625 868 8;
  • 38) 0.756 791 040 625 868 8 × 2 = 1 + 0.513 582 081 251 737 6;
  • 39) 0.513 582 081 251 737 6 × 2 = 1 + 0.027 164 162 503 475 2;
  • 40) 0.027 164 162 503 475 2 × 2 = 0 + 0.054 328 325 006 950 4;
  • 41) 0.054 328 325 006 950 4 × 2 = 0 + 0.108 656 650 013 900 8;
  • 42) 0.108 656 650 013 900 8 × 2 = 0 + 0.217 313 300 027 801 6;
  • 43) 0.217 313 300 027 801 6 × 2 = 0 + 0.434 626 600 055 603 2;
  • 44) 0.434 626 600 055 603 2 × 2 = 0 + 0.869 253 200 111 206 4;
  • 45) 0.869 253 200 111 206 4 × 2 = 1 + 0.738 506 400 222 412 8;
  • 46) 0.738 506 400 222 412 8 × 2 = 1 + 0.477 012 800 444 825 6;
  • 47) 0.477 012 800 444 825 6 × 2 = 0 + 0.954 025 600 889 651 2;
  • 48) 0.954 025 600 889 651 2 × 2 = 1 + 0.908 051 201 779 302 4;
  • 49) 0.908 051 201 779 302 4 × 2 = 1 + 0.816 102 403 558 604 8;
  • 50) 0.816 102 403 558 604 8 × 2 = 1 + 0.632 204 807 117 209 6;
  • 51) 0.632 204 807 117 209 6 × 2 = 1 + 0.264 409 614 234 419 2;
  • 52) 0.264 409 614 234 419 2 × 2 = 0 + 0.528 819 228 468 838 4;
  • 53) 0.528 819 228 468 838 4 × 2 = 1 + 0.057 638 456 937 676 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:


0.792 505 054 354 527 9(10) =

0.1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110 1(2)

5. Positive number before normalization:

1.792 505 054 354 527 9(10) =

1.1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110 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:


1.792 505 054 354 527 9(10) =

1.1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110 1(2) =

1.1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110 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):
1.1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110 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. 1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110 1 =

1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110


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) =
1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110


Decimal number 1.792 505 054 354 527 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1100 1010 1110 0001 1001 1100 0111 1010 0101 1110 0000 1101 1110

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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