-202 212 342 143.123 123 91 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -202 212 342 143.123 123 91(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
-202 212 342 143.123 123 91(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:

|-202 212 342 143.123 123 91| = 202 212 342 143.123 123 91


2. First, convert to binary (in base 2) the integer part: 202 212 342 143.
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;
  • 202 212 342 143 ÷ 2 = 101 106 171 071 + 1;
  • 101 106 171 071 ÷ 2 = 50 553 085 535 + 1;
  • 50 553 085 535 ÷ 2 = 25 276 542 767 + 1;
  • 25 276 542 767 ÷ 2 = 12 638 271 383 + 1;
  • 12 638 271 383 ÷ 2 = 6 319 135 691 + 1;
  • 6 319 135 691 ÷ 2 = 3 159 567 845 + 1;
  • 3 159 567 845 ÷ 2 = 1 579 783 922 + 1;
  • 1 579 783 922 ÷ 2 = 789 891 961 + 0;
  • 789 891 961 ÷ 2 = 394 945 980 + 1;
  • 394 945 980 ÷ 2 = 197 472 990 + 0;
  • 197 472 990 ÷ 2 = 98 736 495 + 0;
  • 98 736 495 ÷ 2 = 49 368 247 + 1;
  • 49 368 247 ÷ 2 = 24 684 123 + 1;
  • 24 684 123 ÷ 2 = 12 342 061 + 1;
  • 12 342 061 ÷ 2 = 6 171 030 + 1;
  • 6 171 030 ÷ 2 = 3 085 515 + 0;
  • 3 085 515 ÷ 2 = 1 542 757 + 1;
  • 1 542 757 ÷ 2 = 771 378 + 1;
  • 771 378 ÷ 2 = 385 689 + 0;
  • 385 689 ÷ 2 = 192 844 + 1;
  • 192 844 ÷ 2 = 96 422 + 0;
  • 96 422 ÷ 2 = 48 211 + 0;
  • 48 211 ÷ 2 = 24 105 + 1;
  • 24 105 ÷ 2 = 12 052 + 1;
  • 12 052 ÷ 2 = 6 026 + 0;
  • 6 026 ÷ 2 = 3 013 + 0;
  • 3 013 ÷ 2 = 1 506 + 1;
  • 1 506 ÷ 2 = 753 + 0;
  • 753 ÷ 2 = 376 + 1;
  • 376 ÷ 2 = 188 + 0;
  • 188 ÷ 2 = 94 + 0;
  • 94 ÷ 2 = 47 + 0;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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.

202 212 342 143(10) =

10 1111 0001 0100 1100 1011 0111 1001 0111 1111(2)


4. Convert to binary (base 2) the fractional part: 0.123 123 91.

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) 0.123 123 91 × 2 = 0 + 0.246 247 82;
  • 2) 0.246 247 82 × 2 = 0 + 0.492 495 64;
  • 3) 0.492 495 64 × 2 = 0 + 0.984 991 28;
  • 4) 0.984 991 28 × 2 = 1 + 0.969 982 56;
  • 5) 0.969 982 56 × 2 = 1 + 0.939 965 12;
  • 6) 0.939 965 12 × 2 = 1 + 0.879 930 24;
  • 7) 0.879 930 24 × 2 = 1 + 0.759 860 48;
  • 8) 0.759 860 48 × 2 = 1 + 0.519 720 96;
  • 9) 0.519 720 96 × 2 = 1 + 0.039 441 92;
  • 10) 0.039 441 92 × 2 = 0 + 0.078 883 84;
  • 11) 0.078 883 84 × 2 = 0 + 0.157 767 68;
  • 12) 0.157 767 68 × 2 = 0 + 0.315 535 36;
  • 13) 0.315 535 36 × 2 = 0 + 0.631 070 72;
  • 14) 0.631 070 72 × 2 = 1 + 0.262 141 44;
  • 15) 0.262 141 44 × 2 = 0 + 0.524 282 88;
  • 16) 0.524 282 88 × 2 = 1 + 0.048 565 76;
  • 17) 0.048 565 76 × 2 = 0 + 0.097 131 52;
  • 18) 0.097 131 52 × 2 = 0 + 0.194 263 04;
  • 19) 0.194 263 04 × 2 = 0 + 0.388 526 08;
  • 20) 0.388 526 08 × 2 = 0 + 0.777 052 16;
  • 21) 0.777 052 16 × 2 = 1 + 0.554 104 32;
  • 22) 0.554 104 32 × 2 = 1 + 0.108 208 64;
  • 23) 0.108 208 64 × 2 = 0 + 0.216 417 28;
  • 24) 0.216 417 28 × 2 = 0 + 0.432 834 56;

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:


0.123 123 91(10) =

0.0001 1111 1000 0101 0000 1100(2)

6. Positive number before normalization:

202 212 342 143.123 123 91(10) =

10 1111 0001 0100 1100 1011 0111 1001 0111 1111.0001 1111 1000 0101 0000 1100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the left, so that only one non zero digit remains to the left of it:


202 212 342 143.123 123 91(10) =

10 1111 0001 0100 1100 1011 0111 1001 0111 1111.0001 1111 1000 0101 0000 1100(2) =

10 1111 0001 0100 1100 1011 0111 1001 0111 1111.0001 1111 1000 0101 0000 1100(2) × 20 =

1.0111 1000 1010 0110 0101 1011 1100 1011 1111 1000 1111 1100 0010 1000 0110 0(2) × 237


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): 37

Mantissa (not normalized):
1.0111 1000 1010 0110 0101 1011 1100 1011 1111 1000 1111 1100 0010 1000 0110 0


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

37 + 2(8-1) - 1 =

(37 + 127)(10) =

164(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;
  • 164 ÷ 2 = 82 + 0;
  • 82 ÷ 2 = 41 + 0;
  • 41 ÷ 2 = 20 + 1;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

164(10) =

1010 0100(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. 011 1100 0101 0011 0010 1101 11 1001 0111 1111 0001 1111 1000 0101 0000 1100 =

011 1100 0101 0011 0010 1101


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) =
1010 0100

Mantissa (23 bits) =
011 1100 0101 0011 0010 1101


Decimal number -202 212 342 143.123 123 91 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 1010 0100 - 011 1100 0101 0011 0010 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