1.100 121 000 017 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.100 121 000 017 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
1.100 121 000 017 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: 0.100 121 000 017 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) 0.100 121 000 017 2 × 2 = 0 + 0.200 242 000 034 4;
  • 2) 0.200 242 000 034 4 × 2 = 0 + 0.400 484 000 068 8;
  • 3) 0.400 484 000 068 8 × 2 = 0 + 0.800 968 000 137 6;
  • 4) 0.800 968 000 137 6 × 2 = 1 + 0.601 936 000 275 2;
  • 5) 0.601 936 000 275 2 × 2 = 1 + 0.203 872 000 550 4;
  • 6) 0.203 872 000 550 4 × 2 = 0 + 0.407 744 001 100 8;
  • 7) 0.407 744 001 100 8 × 2 = 0 + 0.815 488 002 201 6;
  • 8) 0.815 488 002 201 6 × 2 = 1 + 0.630 976 004 403 2;
  • 9) 0.630 976 004 403 2 × 2 = 1 + 0.261 952 008 806 4;
  • 10) 0.261 952 008 806 4 × 2 = 0 + 0.523 904 017 612 8;
  • 11) 0.523 904 017 612 8 × 2 = 1 + 0.047 808 035 225 6;
  • 12) 0.047 808 035 225 6 × 2 = 0 + 0.095 616 070 451 2;
  • 13) 0.095 616 070 451 2 × 2 = 0 + 0.191 232 140 902 4;
  • 14) 0.191 232 140 902 4 × 2 = 0 + 0.382 464 281 804 8;
  • 15) 0.382 464 281 804 8 × 2 = 0 + 0.764 928 563 609 6;
  • 16) 0.764 928 563 609 6 × 2 = 1 + 0.529 857 127 219 2;
  • 17) 0.529 857 127 219 2 × 2 = 1 + 0.059 714 254 438 4;
  • 18) 0.059 714 254 438 4 × 2 = 0 + 0.119 428 508 876 8;
  • 19) 0.119 428 508 876 8 × 2 = 0 + 0.238 857 017 753 6;
  • 20) 0.238 857 017 753 6 × 2 = 0 + 0.477 714 035 507 2;
  • 21) 0.477 714 035 507 2 × 2 = 0 + 0.955 428 071 014 4;
  • 22) 0.955 428 071 014 4 × 2 = 1 + 0.910 856 142 028 8;
  • 23) 0.910 856 142 028 8 × 2 = 1 + 0.821 712 284 057 6;
  • 24) 0.821 712 284 057 6 × 2 = 1 + 0.643 424 568 115 2;
  • 25) 0.643 424 568 115 2 × 2 = 1 + 0.286 849 136 230 4;
  • 26) 0.286 849 136 230 4 × 2 = 0 + 0.573 698 272 460 8;
  • 27) 0.573 698 272 460 8 × 2 = 1 + 0.147 396 544 921 6;
  • 28) 0.147 396 544 921 6 × 2 = 0 + 0.294 793 089 843 2;
  • 29) 0.294 793 089 843 2 × 2 = 0 + 0.589 586 179 686 4;
  • 30) 0.589 586 179 686 4 × 2 = 1 + 0.179 172 359 372 8;
  • 31) 0.179 172 359 372 8 × 2 = 0 + 0.358 344 718 745 6;
  • 32) 0.358 344 718 745 6 × 2 = 0 + 0.716 689 437 491 2;
  • 33) 0.716 689 437 491 2 × 2 = 1 + 0.433 378 874 982 4;
  • 34) 0.433 378 874 982 4 × 2 = 0 + 0.866 757 749 964 8;
  • 35) 0.866 757 749 964 8 × 2 = 1 + 0.733 515 499 929 6;
  • 36) 0.733 515 499 929 6 × 2 = 1 + 0.467 030 999 859 2;
  • 37) 0.467 030 999 859 2 × 2 = 0 + 0.934 061 999 718 4;
  • 38) 0.934 061 999 718 4 × 2 = 1 + 0.868 123 999 436 8;
  • 39) 0.868 123 999 436 8 × 2 = 1 + 0.736 247 998 873 6;
  • 40) 0.736 247 998 873 6 × 2 = 1 + 0.472 495 997 747 2;
  • 41) 0.472 495 997 747 2 × 2 = 0 + 0.944 991 995 494 4;
  • 42) 0.944 991 995 494 4 × 2 = 1 + 0.889 983 990 988 8;
  • 43) 0.889 983 990 988 8 × 2 = 1 + 0.779 967 981 977 6;
  • 44) 0.779 967 981 977 6 × 2 = 1 + 0.559 935 963 955 2;
  • 45) 0.559 935 963 955 2 × 2 = 1 + 0.119 871 927 910 4;
  • 46) 0.119 871 927 910 4 × 2 = 0 + 0.239 743 855 820 8;
  • 47) 0.239 743 855 820 8 × 2 = 0 + 0.479 487 711 641 6;
  • 48) 0.479 487 711 641 6 × 2 = 0 + 0.958 975 423 283 2;
  • 49) 0.958 975 423 283 2 × 2 = 1 + 0.917 950 846 566 4;
  • 50) 0.917 950 846 566 4 × 2 = 1 + 0.835 901 693 132 8;
  • 51) 0.835 901 693 132 8 × 2 = 1 + 0.671 803 386 265 6;
  • 52) 0.671 803 386 265 6 × 2 = 1 + 0.343 606 772 531 2;
  • 53) 0.343 606 772 531 2 × 2 = 0 + 0.687 213 545 062 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:


0.100 121 000 017 2(10) =

0.0001 1001 1010 0001 1000 0111 1010 0100 1011 0111 0111 1000 1111 0(2)

5. Positive number before normalization:

1.100 121 000 017 2(10) =

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


1.100 121 000 017 2(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 0111 0111 1000 1111 0(2) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 0111 0111 1000 1111 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 0111 0111 1000 1111 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 0111 0111 1000 1111 0 =

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


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 0111 0111 1000 1111


Decimal number 1.100 121 000 017 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 0111 0111 1000 1111

Share

» Convert decimal 1.100 121 000 021 7 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 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions & Market Value. Gemini Notebook YouTube Video of 6.55 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