1.100 121 000 011 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.100 121 000 011 6(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
1.100 121 000 011 6(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: 0.100 121 000 011 6.

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) 0.100 121 000 011 6 × 2 = 0 + 0.200 242 000 023 2;
  • 2) 0.200 242 000 023 2 × 2 = 0 + 0.400 484 000 046 4;
  • 3) 0.400 484 000 046 4 × 2 = 0 + 0.800 968 000 092 8;
  • 4) 0.800 968 000 092 8 × 2 = 1 + 0.601 936 000 185 6;
  • 5) 0.601 936 000 185 6 × 2 = 1 + 0.203 872 000 371 2;
  • 6) 0.203 872 000 371 2 × 2 = 0 + 0.407 744 000 742 4;
  • 7) 0.407 744 000 742 4 × 2 = 0 + 0.815 488 001 484 8;
  • 8) 0.815 488 001 484 8 × 2 = 1 + 0.630 976 002 969 6;
  • 9) 0.630 976 002 969 6 × 2 = 1 + 0.261 952 005 939 2;
  • 10) 0.261 952 005 939 2 × 2 = 0 + 0.523 904 011 878 4;
  • 11) 0.523 904 011 878 4 × 2 = 1 + 0.047 808 023 756 8;
  • 12) 0.047 808 023 756 8 × 2 = 0 + 0.095 616 047 513 6;
  • 13) 0.095 616 047 513 6 × 2 = 0 + 0.191 232 095 027 2;
  • 14) 0.191 232 095 027 2 × 2 = 0 + 0.382 464 190 054 4;
  • 15) 0.382 464 190 054 4 × 2 = 0 + 0.764 928 380 108 8;
  • 16) 0.764 928 380 108 8 × 2 = 1 + 0.529 856 760 217 6;
  • 17) 0.529 856 760 217 6 × 2 = 1 + 0.059 713 520 435 2;
  • 18) 0.059 713 520 435 2 × 2 = 0 + 0.119 427 040 870 4;
  • 19) 0.119 427 040 870 4 × 2 = 0 + 0.238 854 081 740 8;
  • 20) 0.238 854 081 740 8 × 2 = 0 + 0.477 708 163 481 6;
  • 21) 0.477 708 163 481 6 × 2 = 0 + 0.955 416 326 963 2;
  • 22) 0.955 416 326 963 2 × 2 = 1 + 0.910 832 653 926 4;
  • 23) 0.910 832 653 926 4 × 2 = 1 + 0.821 665 307 852 8;
  • 24) 0.821 665 307 852 8 × 2 = 1 + 0.643 330 615 705 6;
  • 25) 0.643 330 615 705 6 × 2 = 1 + 0.286 661 231 411 2;
  • 26) 0.286 661 231 411 2 × 2 = 0 + 0.573 322 462 822 4;
  • 27) 0.573 322 462 822 4 × 2 = 1 + 0.146 644 925 644 8;
  • 28) 0.146 644 925 644 8 × 2 = 0 + 0.293 289 851 289 6;
  • 29) 0.293 289 851 289 6 × 2 = 0 + 0.586 579 702 579 2;
  • 30) 0.586 579 702 579 2 × 2 = 1 + 0.173 159 405 158 4;
  • 31) 0.173 159 405 158 4 × 2 = 0 + 0.346 318 810 316 8;
  • 32) 0.346 318 810 316 8 × 2 = 0 + 0.692 637 620 633 6;
  • 33) 0.692 637 620 633 6 × 2 = 1 + 0.385 275 241 267 2;
  • 34) 0.385 275 241 267 2 × 2 = 0 + 0.770 550 482 534 4;
  • 35) 0.770 550 482 534 4 × 2 = 1 + 0.541 100 965 068 8;
  • 36) 0.541 100 965 068 8 × 2 = 1 + 0.082 201 930 137 6;
  • 37) 0.082 201 930 137 6 × 2 = 0 + 0.164 403 860 275 2;
  • 38) 0.164 403 860 275 2 × 2 = 0 + 0.328 807 720 550 4;
  • 39) 0.328 807 720 550 4 × 2 = 0 + 0.657 615 441 100 8;
  • 40) 0.657 615 441 100 8 × 2 = 1 + 0.315 230 882 201 6;
  • 41) 0.315 230 882 201 6 × 2 = 0 + 0.630 461 764 403 2;
  • 42) 0.630 461 764 403 2 × 2 = 1 + 0.260 923 528 806 4;
  • 43) 0.260 923 528 806 4 × 2 = 0 + 0.521 847 057 612 8;
  • 44) 0.521 847 057 612 8 × 2 = 1 + 0.043 694 115 225 6;
  • 45) 0.043 694 115 225 6 × 2 = 0 + 0.087 388 230 451 2;
  • 46) 0.087 388 230 451 2 × 2 = 0 + 0.174 776 460 902 4;
  • 47) 0.174 776 460 902 4 × 2 = 0 + 0.349 552 921 804 8;
  • 48) 0.349 552 921 804 8 × 2 = 0 + 0.699 105 843 609 6;
  • 49) 0.699 105 843 609 6 × 2 = 1 + 0.398 211 687 219 2;
  • 50) 0.398 211 687 219 2 × 2 = 0 + 0.796 423 374 438 4;
  • 51) 0.796 423 374 438 4 × 2 = 1 + 0.592 846 748 876 8;
  • 52) 0.592 846 748 876 8 × 2 = 1 + 0.185 693 497 753 6;
  • 53) 0.185 693 497 753 6 × 2 = 0 + 0.371 386 995 507 2;

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:


0.100 121 000 011 6(10) =

0.0001 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011 0(2)

5. Positive number before normalization:

1.100 121 000 011 6(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011 0(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:


1.100 121 000 011 6(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011 0(2) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011 0(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 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011 0


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 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011 0 =

0001 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011


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 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011


Decimal number 1.100 121 000 011 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1001 1010 0001 1000 0111 1010 0100 1011 0001 0101 0000 1011

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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