111 100 110.000 010 012 79 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 111 100 110.000 010 012 79(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
111 100 110.000 010 012 79(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: 111 100 110.
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;
  • 111 100 110 ÷ 2 = 55 550 055 + 0;
  • 55 550 055 ÷ 2 = 27 775 027 + 1;
  • 27 775 027 ÷ 2 = 13 887 513 + 1;
  • 13 887 513 ÷ 2 = 6 943 756 + 1;
  • 6 943 756 ÷ 2 = 3 471 878 + 0;
  • 3 471 878 ÷ 2 = 1 735 939 + 0;
  • 1 735 939 ÷ 2 = 867 969 + 1;
  • 867 969 ÷ 2 = 433 984 + 1;
  • 433 984 ÷ 2 = 216 992 + 0;
  • 216 992 ÷ 2 = 108 496 + 0;
  • 108 496 ÷ 2 = 54 248 + 0;
  • 54 248 ÷ 2 = 27 124 + 0;
  • 27 124 ÷ 2 = 13 562 + 0;
  • 13 562 ÷ 2 = 6 781 + 0;
  • 6 781 ÷ 2 = 3 390 + 1;
  • 3 390 ÷ 2 = 1 695 + 0;
  • 1 695 ÷ 2 = 847 + 1;
  • 847 ÷ 2 = 423 + 1;
  • 423 ÷ 2 = 211 + 1;
  • 211 ÷ 2 = 105 + 1;
  • 105 ÷ 2 = 52 + 1;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 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.

111 100 110(10) =

110 1001 1111 0100 0000 1100 1110(2)


3. Convert to binary (base 2) the fractional part: 0.000 010 012 79.

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) 0.000 010 012 79 × 2 = 0 + 0.000 020 025 58;
  • 2) 0.000 020 025 58 × 2 = 0 + 0.000 040 051 16;
  • 3) 0.000 040 051 16 × 2 = 0 + 0.000 080 102 32;
  • 4) 0.000 080 102 32 × 2 = 0 + 0.000 160 204 64;
  • 5) 0.000 160 204 64 × 2 = 0 + 0.000 320 409 28;
  • 6) 0.000 320 409 28 × 2 = 0 + 0.000 640 818 56;
  • 7) 0.000 640 818 56 × 2 = 0 + 0.001 281 637 12;
  • 8) 0.001 281 637 12 × 2 = 0 + 0.002 563 274 24;
  • 9) 0.002 563 274 24 × 2 = 0 + 0.005 126 548 48;
  • 10) 0.005 126 548 48 × 2 = 0 + 0.010 253 096 96;
  • 11) 0.010 253 096 96 × 2 = 0 + 0.020 506 193 92;
  • 12) 0.020 506 193 92 × 2 = 0 + 0.041 012 387 84;
  • 13) 0.041 012 387 84 × 2 = 0 + 0.082 024 775 68;
  • 14) 0.082 024 775 68 × 2 = 0 + 0.164 049 551 36;
  • 15) 0.164 049 551 36 × 2 = 0 + 0.328 099 102 72;
  • 16) 0.328 099 102 72 × 2 = 0 + 0.656 198 205 44;
  • 17) 0.656 198 205 44 × 2 = 1 + 0.312 396 410 88;
  • 18) 0.312 396 410 88 × 2 = 0 + 0.624 792 821 76;
  • 19) 0.624 792 821 76 × 2 = 1 + 0.249 585 643 52;
  • 20) 0.249 585 643 52 × 2 = 0 + 0.499 171 287 04;
  • 21) 0.499 171 287 04 × 2 = 0 + 0.998 342 574 08;
  • 22) 0.998 342 574 08 × 2 = 1 + 0.996 685 148 16;
  • 23) 0.996 685 148 16 × 2 = 1 + 0.993 370 296 32;
  • 24) 0.993 370 296 32 × 2 = 1 + 0.986 740 592 64;

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:


0.000 010 012 79(10) =

0.0000 0000 0000 0000 1010 0111(2)

5. Positive number before normalization:

111 100 110.000 010 012 79(10) =

110 1001 1111 0100 0000 1100 1110.0000 0000 0000 0000 1010 0111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the left, so that only one non zero digit remains to the left of it:


111 100 110.000 010 012 79(10) =

110 1001 1111 0100 0000 1100 1110.0000 0000 0000 0000 1010 0111(2) =

110 1001 1111 0100 0000 1100 1110.0000 0000 0000 0000 1010 0111(2) × 20 =

1.1010 0111 1101 0000 0011 0011 1000 0000 0000 0000 0010 1001 11(2) × 226


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): 26

Mantissa (not normalized):
1.1010 0111 1101 0000 0011 0011 1000 0000 0000 0000 0010 1001 11


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

26 + 2(8-1) - 1 =

(26 + 127)(10) =

153(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;
  • 153 ÷ 2 = 76 + 1;
  • 76 ÷ 2 = 38 + 0;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 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) =

153(10) =

1001 1001(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. 101 0011 1110 1000 0001 1001 110 0000 0000 0000 0000 1010 0111 =

101 0011 1110 1000 0001 1001


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) =
1001 1001

Mantissa (23 bits) =
101 0011 1110 1000 0001 1001


Decimal number 111 100 110.000 010 012 79 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1001 1001 - 101 0011 1110 1000 0001 1001

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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