1.100 000 009 79 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.100 000 009 79(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
1.100 000 009 79(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: 0.100 000 009 79.

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) 0.100 000 009 79 × 2 = 0 + 0.200 000 019 58;
  • 2) 0.200 000 019 58 × 2 = 0 + 0.400 000 039 16;
  • 3) 0.400 000 039 16 × 2 = 0 + 0.800 000 078 32;
  • 4) 0.800 000 078 32 × 2 = 1 + 0.600 000 156 64;
  • 5) 0.600 000 156 64 × 2 = 1 + 0.200 000 313 28;
  • 6) 0.200 000 313 28 × 2 = 0 + 0.400 000 626 56;
  • 7) 0.400 000 626 56 × 2 = 0 + 0.800 001 253 12;
  • 8) 0.800 001 253 12 × 2 = 1 + 0.600 002 506 24;
  • 9) 0.600 002 506 24 × 2 = 1 + 0.200 005 012 48;
  • 10) 0.200 005 012 48 × 2 = 0 + 0.400 010 024 96;
  • 11) 0.400 010 024 96 × 2 = 0 + 0.800 020 049 92;
  • 12) 0.800 020 049 92 × 2 = 1 + 0.600 040 099 84;
  • 13) 0.600 040 099 84 × 2 = 1 + 0.200 080 199 68;
  • 14) 0.200 080 199 68 × 2 = 0 + 0.400 160 399 36;
  • 15) 0.400 160 399 36 × 2 = 0 + 0.800 320 798 72;
  • 16) 0.800 320 798 72 × 2 = 1 + 0.600 641 597 44;
  • 17) 0.600 641 597 44 × 2 = 1 + 0.201 283 194 88;
  • 18) 0.201 283 194 88 × 2 = 0 + 0.402 566 389 76;
  • 19) 0.402 566 389 76 × 2 = 0 + 0.805 132 779 52;
  • 20) 0.805 132 779 52 × 2 = 1 + 0.610 265 559 04;
  • 21) 0.610 265 559 04 × 2 = 1 + 0.220 531 118 08;
  • 22) 0.220 531 118 08 × 2 = 0 + 0.441 062 236 16;
  • 23) 0.441 062 236 16 × 2 = 0 + 0.882 124 472 32;
  • 24) 0.882 124 472 32 × 2 = 1 + 0.764 248 944 64;
  • 25) 0.764 248 944 64 × 2 = 1 + 0.528 497 889 28;
  • 26) 0.528 497 889 28 × 2 = 1 + 0.056 995 778 56;
  • 27) 0.056 995 778 56 × 2 = 0 + 0.113 991 557 12;
  • 28) 0.113 991 557 12 × 2 = 0 + 0.227 983 114 24;
  • 29) 0.227 983 114 24 × 2 = 0 + 0.455 966 228 48;
  • 30) 0.455 966 228 48 × 2 = 0 + 0.911 932 456 96;
  • 31) 0.911 932 456 96 × 2 = 1 + 0.823 864 913 92;
  • 32) 0.823 864 913 92 × 2 = 1 + 0.647 729 827 84;
  • 33) 0.647 729 827 84 × 2 = 1 + 0.295 459 655 68;
  • 34) 0.295 459 655 68 × 2 = 0 + 0.590 919 311 36;
  • 35) 0.590 919 311 36 × 2 = 1 + 0.181 838 622 72;
  • 36) 0.181 838 622 72 × 2 = 0 + 0.363 677 245 44;
  • 37) 0.363 677 245 44 × 2 = 0 + 0.727 354 490 88;
  • 38) 0.727 354 490 88 × 2 = 1 + 0.454 708 981 76;
  • 39) 0.454 708 981 76 × 2 = 0 + 0.909 417 963 52;
  • 40) 0.909 417 963 52 × 2 = 1 + 0.818 835 927 04;
  • 41) 0.818 835 927 04 × 2 = 1 + 0.637 671 854 08;
  • 42) 0.637 671 854 08 × 2 = 1 + 0.275 343 708 16;
  • 43) 0.275 343 708 16 × 2 = 0 + 0.550 687 416 32;
  • 44) 0.550 687 416 32 × 2 = 1 + 0.101 374 832 64;
  • 45) 0.101 374 832 64 × 2 = 0 + 0.202 749 665 28;
  • 46) 0.202 749 665 28 × 2 = 0 + 0.405 499 330 56;
  • 47) 0.405 499 330 56 × 2 = 0 + 0.810 998 661 12;
  • 48) 0.810 998 661 12 × 2 = 1 + 0.621 997 322 24;
  • 49) 0.621 997 322 24 × 2 = 1 + 0.243 994 644 48;
  • 50) 0.243 994 644 48 × 2 = 0 + 0.487 989 288 96;
  • 51) 0.487 989 288 96 × 2 = 0 + 0.975 978 577 92;
  • 52) 0.975 978 577 92 × 2 = 1 + 0.951 957 155 84;
  • 53) 0.951 957 155 84 × 2 = 1 + 0.903 914 311 68;

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:


0.100 000 009 79(10) =

0.0001 1001 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001 1(2)

5. Positive number before normalization:

1.100 000 009 79(10) =

1.0001 1001 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001 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:


1.100 000 009 79(10) =

1.0001 1001 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001 1(2) =

1.0001 1001 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001 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 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001 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 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001 1 =

0001 1001 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001


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 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001


Decimal number 1.100 000 009 79 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1001 1001 1001 1001 1001 1100 0011 1010 0101 1101 0001 1001

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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