-1.111 111 111 111 111 111 111 065 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1.111 111 111 111 111 111 111 065 4(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
-1.111 111 111 111 111 111 111 065 4(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. Start with the positive version of the number:

|-1.111 111 111 111 111 111 111 065 4| = 1.111 111 111 111 111 111 111 065 4


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.111 111 111 111 111 111 111 065 4.

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) 0.111 111 111 111 111 111 111 065 4 × 2 = 0 + 0.222 222 222 222 222 222 222 130 8;
  • 2) 0.222 222 222 222 222 222 222 130 8 × 2 = 0 + 0.444 444 444 444 444 444 444 261 6;
  • 3) 0.444 444 444 444 444 444 444 261 6 × 2 = 0 + 0.888 888 888 888 888 888 888 523 2;
  • 4) 0.888 888 888 888 888 888 888 523 2 × 2 = 1 + 0.777 777 777 777 777 777 777 046 4;
  • 5) 0.777 777 777 777 777 777 777 046 4 × 2 = 1 + 0.555 555 555 555 555 555 554 092 8;
  • 6) 0.555 555 555 555 555 555 554 092 8 × 2 = 1 + 0.111 111 111 111 111 111 108 185 6;
  • 7) 0.111 111 111 111 111 111 108 185 6 × 2 = 0 + 0.222 222 222 222 222 222 216 371 2;
  • 8) 0.222 222 222 222 222 222 216 371 2 × 2 = 0 + 0.444 444 444 444 444 444 432 742 4;
  • 9) 0.444 444 444 444 444 444 432 742 4 × 2 = 0 + 0.888 888 888 888 888 888 865 484 8;
  • 10) 0.888 888 888 888 888 888 865 484 8 × 2 = 1 + 0.777 777 777 777 777 777 730 969 6;
  • 11) 0.777 777 777 777 777 777 730 969 6 × 2 = 1 + 0.555 555 555 555 555 555 461 939 2;
  • 12) 0.555 555 555 555 555 555 461 939 2 × 2 = 1 + 0.111 111 111 111 111 110 923 878 4;
  • 13) 0.111 111 111 111 111 110 923 878 4 × 2 = 0 + 0.222 222 222 222 222 221 847 756 8;
  • 14) 0.222 222 222 222 222 221 847 756 8 × 2 = 0 + 0.444 444 444 444 444 443 695 513 6;
  • 15) 0.444 444 444 444 444 443 695 513 6 × 2 = 0 + 0.888 888 888 888 888 887 391 027 2;
  • 16) 0.888 888 888 888 888 887 391 027 2 × 2 = 1 + 0.777 777 777 777 777 774 782 054 4;
  • 17) 0.777 777 777 777 777 774 782 054 4 × 2 = 1 + 0.555 555 555 555 555 549 564 108 8;
  • 18) 0.555 555 555 555 555 549 564 108 8 × 2 = 1 + 0.111 111 111 111 111 099 128 217 6;
  • 19) 0.111 111 111 111 111 099 128 217 6 × 2 = 0 + 0.222 222 222 222 222 198 256 435 2;
  • 20) 0.222 222 222 222 222 198 256 435 2 × 2 = 0 + 0.444 444 444 444 444 396 512 870 4;
  • 21) 0.444 444 444 444 444 396 512 870 4 × 2 = 0 + 0.888 888 888 888 888 793 025 740 8;
  • 22) 0.888 888 888 888 888 793 025 740 8 × 2 = 1 + 0.777 777 777 777 777 586 051 481 6;
  • 23) 0.777 777 777 777 777 586 051 481 6 × 2 = 1 + 0.555 555 555 555 555 172 102 963 2;
  • 24) 0.555 555 555 555 555 172 102 963 2 × 2 = 1 + 0.111 111 111 111 110 344 205 926 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).


5. 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:


0.111 111 111 111 111 111 111 065 4(10) =

0.0001 1100 0111 0001 1100 0111(2)

6. Positive number before normalization:

1.111 111 111 111 111 111 111 065 4(10) =

1.0001 1100 0111 0001 1100 0111(2)

7. 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:


1.111 111 111 111 111 111 111 065 4(10) =

1.0001 1100 0111 0001 1100 0111(2) =

1.0001 1100 0111 0001 1100 0111(2) × 20


8. 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 1 (a negative number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.0001 1100 0111 0001 1100 0111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(10)


10. 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;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

127(10) =

0111 1111(2)


12. 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, 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. 000 1110 0011 1000 1110 0011 1 =

000 1110 0011 1000 1110 0011


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

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

Exponent (8 bits) =
0111 1111

Mantissa (23 bits) =
000 1110 0011 1000 1110 0011


Decimal number -1.111 111 111 111 111 111 111 065 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 1111 - 000 1110 0011 1000 1110 0011

Share

» Convert decimal -1.111 111 111 111 111 111 111 074 3 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Why Do Colleagues From Other Teams Have Better Results? Gemini Notebook YouTube Video of 11.31 minutes

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