1.039 720 770 837 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.039 720 770 837 2(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
1.039 720 770 837 2(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: 0.039 720 770 837 2.

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) 0.039 720 770 837 2 × 2 = 0 + 0.079 441 541 674 4;
  • 2) 0.079 441 541 674 4 × 2 = 0 + 0.158 883 083 348 8;
  • 3) 0.158 883 083 348 8 × 2 = 0 + 0.317 766 166 697 6;
  • 4) 0.317 766 166 697 6 × 2 = 0 + 0.635 532 333 395 2;
  • 5) 0.635 532 333 395 2 × 2 = 1 + 0.271 064 666 790 4;
  • 6) 0.271 064 666 790 4 × 2 = 0 + 0.542 129 333 580 8;
  • 7) 0.542 129 333 580 8 × 2 = 1 + 0.084 258 667 161 6;
  • 8) 0.084 258 667 161 6 × 2 = 0 + 0.168 517 334 323 2;
  • 9) 0.168 517 334 323 2 × 2 = 0 + 0.337 034 668 646 4;
  • 10) 0.337 034 668 646 4 × 2 = 0 + 0.674 069 337 292 8;
  • 11) 0.674 069 337 292 8 × 2 = 1 + 0.348 138 674 585 6;
  • 12) 0.348 138 674 585 6 × 2 = 0 + 0.696 277 349 171 2;
  • 13) 0.696 277 349 171 2 × 2 = 1 + 0.392 554 698 342 4;
  • 14) 0.392 554 698 342 4 × 2 = 0 + 0.785 109 396 684 8;
  • 15) 0.785 109 396 684 8 × 2 = 1 + 0.570 218 793 369 6;
  • 16) 0.570 218 793 369 6 × 2 = 1 + 0.140 437 586 739 2;
  • 17) 0.140 437 586 739 2 × 2 = 0 + 0.280 875 173 478 4;
  • 18) 0.280 875 173 478 4 × 2 = 0 + 0.561 750 346 956 8;
  • 19) 0.561 750 346 956 8 × 2 = 1 + 0.123 500 693 913 6;
  • 20) 0.123 500 693 913 6 × 2 = 0 + 0.247 001 387 827 2;
  • 21) 0.247 001 387 827 2 × 2 = 0 + 0.494 002 775 654 4;
  • 22) 0.494 002 775 654 4 × 2 = 0 + 0.988 005 551 308 8;
  • 23) 0.988 005 551 308 8 × 2 = 1 + 0.976 011 102 617 6;
  • 24) 0.976 011 102 617 6 × 2 = 1 + 0.952 022 205 235 2;
  • 25) 0.952 022 205 235 2 × 2 = 1 + 0.904 044 410 470 4;
  • 26) 0.904 044 410 470 4 × 2 = 1 + 0.808 088 820 940 8;
  • 27) 0.808 088 820 940 8 × 2 = 1 + 0.616 177 641 881 6;
  • 28) 0.616 177 641 881 6 × 2 = 1 + 0.232 355 283 763 2;
  • 29) 0.232 355 283 763 2 × 2 = 0 + 0.464 710 567 526 4;
  • 30) 0.464 710 567 526 4 × 2 = 0 + 0.929 421 135 052 8;
  • 31) 0.929 421 135 052 8 × 2 = 1 + 0.858 842 270 105 6;
  • 32) 0.858 842 270 105 6 × 2 = 1 + 0.717 684 540 211 2;
  • 33) 0.717 684 540 211 2 × 2 = 1 + 0.435 369 080 422 4;
  • 34) 0.435 369 080 422 4 × 2 = 0 + 0.870 738 160 844 8;
  • 35) 0.870 738 160 844 8 × 2 = 1 + 0.741 476 321 689 6;
  • 36) 0.741 476 321 689 6 × 2 = 1 + 0.482 952 643 379 2;
  • 37) 0.482 952 643 379 2 × 2 = 0 + 0.965 905 286 758 4;
  • 38) 0.965 905 286 758 4 × 2 = 1 + 0.931 810 573 516 8;
  • 39) 0.931 810 573 516 8 × 2 = 1 + 0.863 621 147 033 6;
  • 40) 0.863 621 147 033 6 × 2 = 1 + 0.727 242 294 067 2;
  • 41) 0.727 242 294 067 2 × 2 = 1 + 0.454 484 588 134 4;
  • 42) 0.454 484 588 134 4 × 2 = 0 + 0.908 969 176 268 8;
  • 43) 0.908 969 176 268 8 × 2 = 1 + 0.817 938 352 537 6;
  • 44) 0.817 938 352 537 6 × 2 = 1 + 0.635 876 705 075 2;
  • 45) 0.635 876 705 075 2 × 2 = 1 + 0.271 753 410 150 4;
  • 46) 0.271 753 410 150 4 × 2 = 0 + 0.543 506 820 300 8;
  • 47) 0.543 506 820 300 8 × 2 = 1 + 0.087 013 640 601 6;
  • 48) 0.087 013 640 601 6 × 2 = 0 + 0.174 027 281 203 2;
  • 49) 0.174 027 281 203 2 × 2 = 0 + 0.348 054 562 406 4;
  • 50) 0.348 054 562 406 4 × 2 = 0 + 0.696 109 124 812 8;
  • 51) 0.696 109 124 812 8 × 2 = 1 + 0.392 218 249 625 6;
  • 52) 0.392 218 249 625 6 × 2 = 0 + 0.784 436 499 251 2;
  • 53) 0.784 436 499 251 2 × 2 = 1 + 0.568 872 998 502 4;

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:


0.039 720 770 837 2(10) =

0.0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010 1(2)

5. Positive number before normalization:

1.039 720 770 837 2(10) =

1.0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010 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:


1.039 720 770 837 2(10) =

1.0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010 1(2) =

1.0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010 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):
1.0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010 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. 0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010 1 =

0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010


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) =
0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010


Decimal number 1.039 720 770 837 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0000 1010 0010 1011 0010 0011 1111 0011 1011 0111 1011 1010 0010

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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