Convert 1 011.000 110 101 110 000 9 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

1 011.000 110 101 110 000 9(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 1 011.
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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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.

1 011(10) =


11 1111 0011(2)


3. Convert to the binary (base 2) the fractional part: 0.000 110 101 110 000 9.

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) 0.000 110 101 110 000 9 × 2 = 0 + 0.000 220 202 220 001 8;
  • 2) 0.000 220 202 220 001 8 × 2 = 0 + 0.000 440 404 440 003 6;
  • 3) 0.000 440 404 440 003 6 × 2 = 0 + 0.000 880 808 880 007 2;
  • 4) 0.000 880 808 880 007 2 × 2 = 0 + 0.001 761 617 760 014 4;
  • 5) 0.001 761 617 760 014 4 × 2 = 0 + 0.003 523 235 520 028 8;
  • 6) 0.003 523 235 520 028 8 × 2 = 0 + 0.007 046 471 040 057 6;
  • 7) 0.007 046 471 040 057 6 × 2 = 0 + 0.014 092 942 080 115 2;
  • 8) 0.014 092 942 080 115 2 × 2 = 0 + 0.028 185 884 160 230 4;
  • 9) 0.028 185 884 160 230 4 × 2 = 0 + 0.056 371 768 320 460 8;
  • 10) 0.056 371 768 320 460 8 × 2 = 0 + 0.112 743 536 640 921 6;
  • 11) 0.112 743 536 640 921 6 × 2 = 0 + 0.225 487 073 281 843 2;
  • 12) 0.225 487 073 281 843 2 × 2 = 0 + 0.450 974 146 563 686 4;
  • 13) 0.450 974 146 563 686 4 × 2 = 0 + 0.901 948 293 127 372 8;
  • 14) 0.901 948 293 127 372 8 × 2 = 1 + 0.803 896 586 254 745 6;
  • 15) 0.803 896 586 254 745 6 × 2 = 1 + 0.607 793 172 509 491 2;
  • 16) 0.607 793 172 509 491 2 × 2 = 1 + 0.215 586 345 018 982 4;
  • 17) 0.215 586 345 018 982 4 × 2 = 0 + 0.431 172 690 037 964 8;
  • 18) 0.431 172 690 037 964 8 × 2 = 0 + 0.862 345 380 075 929 6;
  • 19) 0.862 345 380 075 929 6 × 2 = 1 + 0.724 690 760 151 859 2;
  • 20) 0.724 690 760 151 859 2 × 2 = 1 + 0.449 381 520 303 718 4;
  • 21) 0.449 381 520 303 718 4 × 2 = 0 + 0.898 763 040 607 436 8;
  • 22) 0.898 763 040 607 436 8 × 2 = 1 + 0.797 526 081 214 873 6;
  • 23) 0.797 526 081 214 873 6 × 2 = 1 + 0.595 052 162 429 747 2;
  • 24) 0.595 052 162 429 747 2 × 2 = 1 + 0.190 104 324 859 494 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...)


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:

0.000 110 101 110 000 9(10) =


0.0000 0000 0000 0111 0011 0111(2)


5. Positive number before normalization:

1 011.000 110 101 110 000 9(10) =


11 1111 0011.0000 0000 0000 0111 0011 0111(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the left so that only one non zero digit remains to the left of it:

1 011.000 110 101 110 000 9(10) =


11 1111 0011.0000 0000 0000 0111 0011 0111(2) =


11 1111 0011.0000 0000 0000 0111 0011 0111(2) × 20 =


1.1111 1001 1000 0000 0000 0011 1001 1011 1(2) × 29


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): 9


Mantissa (not normalized):
1.1111 1001 1000 0000 0000 0011 1001 1011 1


8. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


9 + 2(8-1) - 1 =


(9 + 127)(10) =


136(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;
  • 136 ÷ 2 = 68 + 0;
  • 68 ÷ 2 = 34 + 0;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


136(10) =


1000 1000(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, 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. 111 1100 1100 0000 0000 0001 11 0011 0111 =


111 1100 1100 0000 0000 0001


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) =
1000 1000


Mantissa (23 bits) =
111 1100 1100 0000 0000 0001


Number 1 011.000 110 101 110 000 9 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 1000 1000 - 111 1100 1100 0000 0000 0001

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 1

      30
    • 0

      29
    • 0

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
  • Mantissa (23 bits):

    • 1

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 0

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 1

      0

More operations of this kind:

1 011.000 110 101 110 000 8 = ? ... 1 011.000 110 101 110 001 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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