Decimal to 32 Bit IEEE 754 Binary: Convert Number 12 345 678 998 716 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 12 345 678 998 716(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

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;
  • 12 345 678 998 716 ÷ 2 = 6 172 839 499 358 + 0;
  • 6 172 839 499 358 ÷ 2 = 3 086 419 749 679 + 0;
  • 3 086 419 749 679 ÷ 2 = 1 543 209 874 839 + 1;
  • 1 543 209 874 839 ÷ 2 = 771 604 937 419 + 1;
  • 771 604 937 419 ÷ 2 = 385 802 468 709 + 1;
  • 385 802 468 709 ÷ 2 = 192 901 234 354 + 1;
  • 192 901 234 354 ÷ 2 = 96 450 617 177 + 0;
  • 96 450 617 177 ÷ 2 = 48 225 308 588 + 1;
  • 48 225 308 588 ÷ 2 = 24 112 654 294 + 0;
  • 24 112 654 294 ÷ 2 = 12 056 327 147 + 0;
  • 12 056 327 147 ÷ 2 = 6 028 163 573 + 1;
  • 6 028 163 573 ÷ 2 = 3 014 081 786 + 1;
  • 3 014 081 786 ÷ 2 = 1 507 040 893 + 0;
  • 1 507 040 893 ÷ 2 = 753 520 446 + 1;
  • 753 520 446 ÷ 2 = 376 760 223 + 0;
  • 376 760 223 ÷ 2 = 188 380 111 + 1;
  • 188 380 111 ÷ 2 = 94 190 055 + 1;
  • 94 190 055 ÷ 2 = 47 095 027 + 1;
  • 47 095 027 ÷ 2 = 23 547 513 + 1;
  • 23 547 513 ÷ 2 = 11 773 756 + 1;
  • 11 773 756 ÷ 2 = 5 886 878 + 0;
  • 5 886 878 ÷ 2 = 2 943 439 + 0;
  • 2 943 439 ÷ 2 = 1 471 719 + 1;
  • 1 471 719 ÷ 2 = 735 859 + 1;
  • 735 859 ÷ 2 = 367 929 + 1;
  • 367 929 ÷ 2 = 183 964 + 1;
  • 183 964 ÷ 2 = 91 982 + 0;
  • 91 982 ÷ 2 = 45 991 + 0;
  • 45 991 ÷ 2 = 22 995 + 1;
  • 22 995 ÷ 2 = 11 497 + 1;
  • 11 497 ÷ 2 = 5 748 + 1;
  • 5 748 ÷ 2 = 2 874 + 0;
  • 2 874 ÷ 2 = 1 437 + 0;
  • 1 437 ÷ 2 = 718 + 1;
  • 718 ÷ 2 = 359 + 0;
  • 359 ÷ 2 = 179 + 1;
  • 179 ÷ 2 = 89 + 1;
  • 89 ÷ 2 = 44 + 1;
  • 44 ÷ 2 = 22 + 0;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

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


12 345 678 998 716(10) =


1011 0011 1010 0111 0011 1100 1111 1010 1100 1011 1100(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the left, so that only one non zero digit remains to the left of it:


12 345 678 998 716(10) =


1011 0011 1010 0111 0011 1100 1111 1010 1100 1011 1100(2) =


1011 0011 1010 0111 0011 1100 1111 1010 1100 1011 1100(2) × 20 =


1.0110 0111 0100 1110 0111 1001 1111 0101 1001 0111 100(2) × 243


4. 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): 43


Mantissa (not normalized):
1.0110 0111 0100 1110 0111 1001 1111 0101 1001 0111 100


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


43 + 2(8-1) - 1 =


(43 + 127)(10) =


170(10)


6. 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;
  • 170 ÷ 2 = 85 + 0;
  • 85 ÷ 2 = 42 + 1;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


170(10) =


1010 1010(2)


8. 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 0011 1010 0111 0011 1100 1111 1010 1100 1011 1100 =


011 0011 1010 0111 0011 1100


9. 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) =
1010 1010


Mantissa (23 bits) =
011 0011 1010 0111 0011 1100


The base ten decimal number 12 345 678 998 716 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1010 1010 - 011 0011 1010 0111 0011 1100

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