0.000 000 110 008 82 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 008 82(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 110 008 82(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 110 008 82.

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) 0.000 000 110 008 82 × 2 = 0 + 0.000 000 220 017 64;
  • 2) 0.000 000 220 017 64 × 2 = 0 + 0.000 000 440 035 28;
  • 3) 0.000 000 440 035 28 × 2 = 0 + 0.000 000 880 070 56;
  • 4) 0.000 000 880 070 56 × 2 = 0 + 0.000 001 760 141 12;
  • 5) 0.000 001 760 141 12 × 2 = 0 + 0.000 003 520 282 24;
  • 6) 0.000 003 520 282 24 × 2 = 0 + 0.000 007 040 564 48;
  • 7) 0.000 007 040 564 48 × 2 = 0 + 0.000 014 081 128 96;
  • 8) 0.000 014 081 128 96 × 2 = 0 + 0.000 028 162 257 92;
  • 9) 0.000 028 162 257 92 × 2 = 0 + 0.000 056 324 515 84;
  • 10) 0.000 056 324 515 84 × 2 = 0 + 0.000 112 649 031 68;
  • 11) 0.000 112 649 031 68 × 2 = 0 + 0.000 225 298 063 36;
  • 12) 0.000 225 298 063 36 × 2 = 0 + 0.000 450 596 126 72;
  • 13) 0.000 450 596 126 72 × 2 = 0 + 0.000 901 192 253 44;
  • 14) 0.000 901 192 253 44 × 2 = 0 + 0.001 802 384 506 88;
  • 15) 0.001 802 384 506 88 × 2 = 0 + 0.003 604 769 013 76;
  • 16) 0.003 604 769 013 76 × 2 = 0 + 0.007 209 538 027 52;
  • 17) 0.007 209 538 027 52 × 2 = 0 + 0.014 419 076 055 04;
  • 18) 0.014 419 076 055 04 × 2 = 0 + 0.028 838 152 110 08;
  • 19) 0.028 838 152 110 08 × 2 = 0 + 0.057 676 304 220 16;
  • 20) 0.057 676 304 220 16 × 2 = 0 + 0.115 352 608 440 32;
  • 21) 0.115 352 608 440 32 × 2 = 0 + 0.230 705 216 880 64;
  • 22) 0.230 705 216 880 64 × 2 = 0 + 0.461 410 433 761 28;
  • 23) 0.461 410 433 761 28 × 2 = 0 + 0.922 820 867 522 56;
  • 24) 0.922 820 867 522 56 × 2 = 1 + 0.845 641 735 045 12;
  • 25) 0.845 641 735 045 12 × 2 = 1 + 0.691 283 470 090 24;
  • 26) 0.691 283 470 090 24 × 2 = 1 + 0.382 566 940 180 48;
  • 27) 0.382 566 940 180 48 × 2 = 0 + 0.765 133 880 360 96;
  • 28) 0.765 133 880 360 96 × 2 = 1 + 0.530 267 760 721 92;
  • 29) 0.530 267 760 721 92 × 2 = 1 + 0.060 535 521 443 84;
  • 30) 0.060 535 521 443 84 × 2 = 0 + 0.121 071 042 887 68;
  • 31) 0.121 071 042 887 68 × 2 = 0 + 0.242 142 085 775 36;
  • 32) 0.242 142 085 775 36 × 2 = 0 + 0.484 284 171 550 72;
  • 33) 0.484 284 171 550 72 × 2 = 0 + 0.968 568 343 101 44;
  • 34) 0.968 568 343 101 44 × 2 = 1 + 0.937 136 686 202 88;
  • 35) 0.937 136 686 202 88 × 2 = 1 + 0.874 273 372 405 76;
  • 36) 0.874 273 372 405 76 × 2 = 1 + 0.748 546 744 811 52;
  • 37) 0.748 546 744 811 52 × 2 = 1 + 0.497 093 489 623 04;
  • 38) 0.497 093 489 623 04 × 2 = 0 + 0.994 186 979 246 08;
  • 39) 0.994 186 979 246 08 × 2 = 1 + 0.988 373 958 492 16;
  • 40) 0.988 373 958 492 16 × 2 = 1 + 0.976 747 916 984 32;
  • 41) 0.976 747 916 984 32 × 2 = 1 + 0.953 495 833 968 64;
  • 42) 0.953 495 833 968 64 × 2 = 1 + 0.906 991 667 937 28;
  • 43) 0.906 991 667 937 28 × 2 = 1 + 0.813 983 335 874 56;
  • 44) 0.813 983 335 874 56 × 2 = 1 + 0.627 966 671 749 12;
  • 45) 0.627 966 671 749 12 × 2 = 1 + 0.255 933 343 498 24;
  • 46) 0.255 933 343 498 24 × 2 = 0 + 0.511 866 686 996 48;
  • 47) 0.511 866 686 996 48 × 2 = 1 + 0.023 733 373 992 96;

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:


0.000 000 110 008 82(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1111 101(2)

5. Positive number before normalization:

0.000 000 110 008 82(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1111 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 008 82(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1111 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1111 101(2) × 20 =

1.1101 1000 0111 1011 1111 101(2) × 2-24


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1000 0111 1011 1111 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

103(10) =

0110 0111(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 1100 0011 1101 1111 1101 =

110 1100 0011 1101 1111 1101


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0111

Mantissa (23 bits) =
110 1100 0011 1101 1111 1101


Decimal number 0.000 000 110 008 82 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1101 1111 1101

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111