1.100 121 000 016 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.100 121 000 016 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
1.100 121 000 016 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: 0.100 121 000 016 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) 0.100 121 000 016 2 × 2 = 0 + 0.200 242 000 032 4;
  • 2) 0.200 242 000 032 4 × 2 = 0 + 0.400 484 000 064 8;
  • 3) 0.400 484 000 064 8 × 2 = 0 + 0.800 968 000 129 6;
  • 4) 0.800 968 000 129 6 × 2 = 1 + 0.601 936 000 259 2;
  • 5) 0.601 936 000 259 2 × 2 = 1 + 0.203 872 000 518 4;
  • 6) 0.203 872 000 518 4 × 2 = 0 + 0.407 744 001 036 8;
  • 7) 0.407 744 001 036 8 × 2 = 0 + 0.815 488 002 073 6;
  • 8) 0.815 488 002 073 6 × 2 = 1 + 0.630 976 004 147 2;
  • 9) 0.630 976 004 147 2 × 2 = 1 + 0.261 952 008 294 4;
  • 10) 0.261 952 008 294 4 × 2 = 0 + 0.523 904 016 588 8;
  • 11) 0.523 904 016 588 8 × 2 = 1 + 0.047 808 033 177 6;
  • 12) 0.047 808 033 177 6 × 2 = 0 + 0.095 616 066 355 2;
  • 13) 0.095 616 066 355 2 × 2 = 0 + 0.191 232 132 710 4;
  • 14) 0.191 232 132 710 4 × 2 = 0 + 0.382 464 265 420 8;
  • 15) 0.382 464 265 420 8 × 2 = 0 + 0.764 928 530 841 6;
  • 16) 0.764 928 530 841 6 × 2 = 1 + 0.529 857 061 683 2;
  • 17) 0.529 857 061 683 2 × 2 = 1 + 0.059 714 123 366 4;
  • 18) 0.059 714 123 366 4 × 2 = 0 + 0.119 428 246 732 8;
  • 19) 0.119 428 246 732 8 × 2 = 0 + 0.238 856 493 465 6;
  • 20) 0.238 856 493 465 6 × 2 = 0 + 0.477 712 986 931 2;
  • 21) 0.477 712 986 931 2 × 2 = 0 + 0.955 425 973 862 4;
  • 22) 0.955 425 973 862 4 × 2 = 1 + 0.910 851 947 724 8;
  • 23) 0.910 851 947 724 8 × 2 = 1 + 0.821 703 895 449 6;
  • 24) 0.821 703 895 449 6 × 2 = 1 + 0.643 407 790 899 2;
  • 25) 0.643 407 790 899 2 × 2 = 1 + 0.286 815 581 798 4;
  • 26) 0.286 815 581 798 4 × 2 = 0 + 0.573 631 163 596 8;
  • 27) 0.573 631 163 596 8 × 2 = 1 + 0.147 262 327 193 6;
  • 28) 0.147 262 327 193 6 × 2 = 0 + 0.294 524 654 387 2;
  • 29) 0.294 524 654 387 2 × 2 = 0 + 0.589 049 308 774 4;
  • 30) 0.589 049 308 774 4 × 2 = 1 + 0.178 098 617 548 8;
  • 31) 0.178 098 617 548 8 × 2 = 0 + 0.356 197 235 097 6;
  • 32) 0.356 197 235 097 6 × 2 = 0 + 0.712 394 470 195 2;
  • 33) 0.712 394 470 195 2 × 2 = 1 + 0.424 788 940 390 4;
  • 34) 0.424 788 940 390 4 × 2 = 0 + 0.849 577 880 780 8;
  • 35) 0.849 577 880 780 8 × 2 = 1 + 0.699 155 761 561 6;
  • 36) 0.699 155 761 561 6 × 2 = 1 + 0.398 311 523 123 2;
  • 37) 0.398 311 523 123 2 × 2 = 0 + 0.796 623 046 246 4;
  • 38) 0.796 623 046 246 4 × 2 = 1 + 0.593 246 092 492 8;
  • 39) 0.593 246 092 492 8 × 2 = 1 + 0.186 492 184 985 6;
  • 40) 0.186 492 184 985 6 × 2 = 0 + 0.372 984 369 971 2;
  • 41) 0.372 984 369 971 2 × 2 = 0 + 0.745 968 739 942 4;
  • 42) 0.745 968 739 942 4 × 2 = 1 + 0.491 937 479 884 8;
  • 43) 0.491 937 479 884 8 × 2 = 0 + 0.983 874 959 769 6;
  • 44) 0.983 874 959 769 6 × 2 = 1 + 0.967 749 919 539 2;
  • 45) 0.967 749 919 539 2 × 2 = 1 + 0.935 499 839 078 4;
  • 46) 0.935 499 839 078 4 × 2 = 1 + 0.870 999 678 156 8;
  • 47) 0.870 999 678 156 8 × 2 = 1 + 0.741 999 356 313 6;
  • 48) 0.741 999 356 313 6 × 2 = 1 + 0.483 998 712 627 2;
  • 49) 0.483 998 712 627 2 × 2 = 0 + 0.967 997 425 254 4;
  • 50) 0.967 997 425 254 4 × 2 = 1 + 0.935 994 850 508 8;
  • 51) 0.935 994 850 508 8 × 2 = 1 + 0.871 989 701 017 6;
  • 52) 0.871 989 701 017 6 × 2 = 1 + 0.743 979 402 035 2;
  • 53) 0.743 979 402 035 2 × 2 = 1 + 0.487 958 804 070 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:


0.100 121 000 016 2(10) =

0.0001 1001 1010 0001 1000 0111 1010 0100 1011 0110 0101 1111 0111 1(2)

5. Positive number before normalization:

1.100 121 000 016 2(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 0110 0101 1111 0111 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:


1.100 121 000 016 2(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 0110 0101 1111 0111 1(2) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 0110 0101 1111 0111 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 1001 1010 0001 1000 0111 1010 0100 1011 0110 0101 1111 0111 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 1001 1010 0001 1000 0111 1010 0100 1011 0110 0101 1111 0111 1 =

0001 1001 1010 0001 1000 0111 1010 0100 1011 0110 0101 1111 0111


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 0110 0101 1111 0111


Decimal number 1.100 121 000 016 2 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 0110 0101 1111 0111

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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