32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -0.711 111 111 115 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -0.711 111 111 115(10) converted and written in 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:

|-0.711 111 111 115| = 0.711 111 111 115

2. 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;

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.


0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.711 111 111 115.

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) 0.711 111 111 115 × 2 = 1 + 0.422 222 222 23;
  • 2) 0.422 222 222 23 × 2 = 0 + 0.844 444 444 46;
  • 3) 0.844 444 444 46 × 2 = 1 + 0.688 888 888 92;
  • 4) 0.688 888 888 92 × 2 = 1 + 0.377 777 777 84;
  • 5) 0.377 777 777 84 × 2 = 0 + 0.755 555 555 68;
  • 6) 0.755 555 555 68 × 2 = 1 + 0.511 111 111 36;
  • 7) 0.511 111 111 36 × 2 = 1 + 0.022 222 222 72;
  • 8) 0.022 222 222 72 × 2 = 0 + 0.044 444 445 44;
  • 9) 0.044 444 445 44 × 2 = 0 + 0.088 888 890 88;
  • 10) 0.088 888 890 88 × 2 = 0 + 0.177 777 781 76;
  • 11) 0.177 777 781 76 × 2 = 0 + 0.355 555 563 52;
  • 12) 0.355 555 563 52 × 2 = 0 + 0.711 111 127 04;
  • 13) 0.711 111 127 04 × 2 = 1 + 0.422 222 254 08;
  • 14) 0.422 222 254 08 × 2 = 0 + 0.844 444 508 16;
  • 15) 0.844 444 508 16 × 2 = 1 + 0.688 889 016 32;
  • 16) 0.688 889 016 32 × 2 = 1 + 0.377 778 032 64;
  • 17) 0.377 778 032 64 × 2 = 0 + 0.755 556 065 28;
  • 18) 0.755 556 065 28 × 2 = 1 + 0.511 112 130 56;
  • 19) 0.511 112 130 56 × 2 = 1 + 0.022 224 261 12;
  • 20) 0.022 224 261 12 × 2 = 0 + 0.044 448 522 24;
  • 21) 0.044 448 522 24 × 2 = 0 + 0.088 897 044 48;
  • 22) 0.088 897 044 48 × 2 = 0 + 0.177 794 088 96;
  • 23) 0.177 794 088 96 × 2 = 0 + 0.355 588 177 92;
  • 24) 0.355 588 177 92 × 2 = 0 + 0.711 176 355 84;

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...)


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:


0.711 111 111 115(10) =

0.1011 0110 0000 1011 0110 0000(2)


6. Positive number before normalization:

0.711 111 111 115(10) =

0.1011 0110 0000 1011 0110 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.711 111 111 115(10) =

0.1011 0110 0000 1011 0110 0000(2) =

0.1011 0110 0000 1011 0110 0000(2) × 20 =

1.0110 1100 0001 0110 1100 000(2) × 2-1


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): -1

Mantissa (not normalized):
1.0110 1100 0001 0110 1100 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 127)(10) =

126(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;
  • 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;

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) =

126(10) =

0111 1110(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 011 0110 0000 1011 0110 0000 =

011 0110 0000 1011 0110 0000


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 1110

Mantissa (23 bits) =
011 0110 0000 1011 0110 0000


The base ten decimal number -0.711 111 111 115 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0111 1110 - 011 0110 0000 1011 0110 0000

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number -0.711 111 111 116 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number -0.711 111 111 114 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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

Number 283 695 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:42 UTC (GMT)
Number 0.8 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
Number 3 879 716 556 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
Number 2 604 738 988 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
Number 1 201 174 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
Number 72 008 087 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
Number 1 110 101 011 110 101 010 101 010 100 968 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
Number 1 264 779 328 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
Number 1 195 319 289 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
Number 0.070 32 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Sep 14 01:41 UTC (GMT)
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