1.100 121 000 019 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.100 121 000 019 7(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
1.100 121 000 019 7(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: 0.100 121 000 019 7.

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) 0.100 121 000 019 7 × 2 = 0 + 0.200 242 000 039 4;
  • 2) 0.200 242 000 039 4 × 2 = 0 + 0.400 484 000 078 8;
  • 3) 0.400 484 000 078 8 × 2 = 0 + 0.800 968 000 157 6;
  • 4) 0.800 968 000 157 6 × 2 = 1 + 0.601 936 000 315 2;
  • 5) 0.601 936 000 315 2 × 2 = 1 + 0.203 872 000 630 4;
  • 6) 0.203 872 000 630 4 × 2 = 0 + 0.407 744 001 260 8;
  • 7) 0.407 744 001 260 8 × 2 = 0 + 0.815 488 002 521 6;
  • 8) 0.815 488 002 521 6 × 2 = 1 + 0.630 976 005 043 2;
  • 9) 0.630 976 005 043 2 × 2 = 1 + 0.261 952 010 086 4;
  • 10) 0.261 952 010 086 4 × 2 = 0 + 0.523 904 020 172 8;
  • 11) 0.523 904 020 172 8 × 2 = 1 + 0.047 808 040 345 6;
  • 12) 0.047 808 040 345 6 × 2 = 0 + 0.095 616 080 691 2;
  • 13) 0.095 616 080 691 2 × 2 = 0 + 0.191 232 161 382 4;
  • 14) 0.191 232 161 382 4 × 2 = 0 + 0.382 464 322 764 8;
  • 15) 0.382 464 322 764 8 × 2 = 0 + 0.764 928 645 529 6;
  • 16) 0.764 928 645 529 6 × 2 = 1 + 0.529 857 291 059 2;
  • 17) 0.529 857 291 059 2 × 2 = 1 + 0.059 714 582 118 4;
  • 18) 0.059 714 582 118 4 × 2 = 0 + 0.119 429 164 236 8;
  • 19) 0.119 429 164 236 8 × 2 = 0 + 0.238 858 328 473 6;
  • 20) 0.238 858 328 473 6 × 2 = 0 + 0.477 716 656 947 2;
  • 21) 0.477 716 656 947 2 × 2 = 0 + 0.955 433 313 894 4;
  • 22) 0.955 433 313 894 4 × 2 = 1 + 0.910 866 627 788 8;
  • 23) 0.910 866 627 788 8 × 2 = 1 + 0.821 733 255 577 6;
  • 24) 0.821 733 255 577 6 × 2 = 1 + 0.643 466 511 155 2;
  • 25) 0.643 466 511 155 2 × 2 = 1 + 0.286 933 022 310 4;
  • 26) 0.286 933 022 310 4 × 2 = 0 + 0.573 866 044 620 8;
  • 27) 0.573 866 044 620 8 × 2 = 1 + 0.147 732 089 241 6;
  • 28) 0.147 732 089 241 6 × 2 = 0 + 0.295 464 178 483 2;
  • 29) 0.295 464 178 483 2 × 2 = 0 + 0.590 928 356 966 4;
  • 30) 0.590 928 356 966 4 × 2 = 1 + 0.181 856 713 932 8;
  • 31) 0.181 856 713 932 8 × 2 = 0 + 0.363 713 427 865 6;
  • 32) 0.363 713 427 865 6 × 2 = 0 + 0.727 426 855 731 2;
  • 33) 0.727 426 855 731 2 × 2 = 1 + 0.454 853 711 462 4;
  • 34) 0.454 853 711 462 4 × 2 = 0 + 0.909 707 422 924 8;
  • 35) 0.909 707 422 924 8 × 2 = 1 + 0.819 414 845 849 6;
  • 36) 0.819 414 845 849 6 × 2 = 1 + 0.638 829 691 699 2;
  • 37) 0.638 829 691 699 2 × 2 = 1 + 0.277 659 383 398 4;
  • 38) 0.277 659 383 398 4 × 2 = 0 + 0.555 318 766 796 8;
  • 39) 0.555 318 766 796 8 × 2 = 1 + 0.110 637 533 593 6;
  • 40) 0.110 637 533 593 6 × 2 = 0 + 0.221 275 067 187 2;
  • 41) 0.221 275 067 187 2 × 2 = 0 + 0.442 550 134 374 4;
  • 42) 0.442 550 134 374 4 × 2 = 0 + 0.885 100 268 748 8;
  • 43) 0.885 100 268 748 8 × 2 = 1 + 0.770 200 537 497 6;
  • 44) 0.770 200 537 497 6 × 2 = 1 + 0.540 401 074 995 2;
  • 45) 0.540 401 074 995 2 × 2 = 1 + 0.080 802 149 990 4;
  • 46) 0.080 802 149 990 4 × 2 = 0 + 0.161 604 299 980 8;
  • 47) 0.161 604 299 980 8 × 2 = 0 + 0.323 208 599 961 6;
  • 48) 0.323 208 599 961 6 × 2 = 0 + 0.646 417 199 923 2;
  • 49) 0.646 417 199 923 2 × 2 = 1 + 0.292 834 399 846 4;
  • 50) 0.292 834 399 846 4 × 2 = 0 + 0.585 668 799 692 8;
  • 51) 0.585 668 799 692 8 × 2 = 1 + 0.171 337 599 385 6;
  • 52) 0.171 337 599 385 6 × 2 = 0 + 0.342 675 198 771 2;
  • 53) 0.342 675 198 771 2 × 2 = 0 + 0.685 350 397 542 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:


0.100 121 000 019 7(10) =

0.0001 1001 1010 0001 1000 0111 1010 0100 1011 1010 0011 1000 1010 0(2)

5. Positive number before normalization:

1.100 121 000 019 7(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 1010 0011 1000 1010 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:


1.100 121 000 019 7(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 1010 0011 1000 1010 0(2) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1011 1010 0011 1000 1010 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 1010 0011 1000 1010 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 1010 0011 1000 1010 0 =

0001 1001 1010 0001 1000 0111 1010 0100 1011 1010 0011 1000 1010


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 1010 0011 1000 1010


Decimal number 1.100 121 000 019 7 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 1010 0011 1000 1010

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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