5 252.113 122 032 546 105 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5 252.113 122 032 546 105 4(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
5 252.113 122 032 546 105 4(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: 5 252.
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;
  • 5 252 ÷ 2 = 2 626 + 0;
  • 2 626 ÷ 2 = 1 313 + 0;
  • 1 313 ÷ 2 = 656 + 1;
  • 656 ÷ 2 = 328 + 0;
  • 328 ÷ 2 = 164 + 0;
  • 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;

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.

5 252(10) =

1 0100 1000 0100(2)


3. Convert to binary (base 2) the fractional part: 0.113 122 032 546 105 4.

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) 0.113 122 032 546 105 4 × 2 = 0 + 0.226 244 065 092 210 8;
  • 2) 0.226 244 065 092 210 8 × 2 = 0 + 0.452 488 130 184 421 6;
  • 3) 0.452 488 130 184 421 6 × 2 = 0 + 0.904 976 260 368 843 2;
  • 4) 0.904 976 260 368 843 2 × 2 = 1 + 0.809 952 520 737 686 4;
  • 5) 0.809 952 520 737 686 4 × 2 = 1 + 0.619 905 041 475 372 8;
  • 6) 0.619 905 041 475 372 8 × 2 = 1 + 0.239 810 082 950 745 6;
  • 7) 0.239 810 082 950 745 6 × 2 = 0 + 0.479 620 165 901 491 2;
  • 8) 0.479 620 165 901 491 2 × 2 = 0 + 0.959 240 331 802 982 4;
  • 9) 0.959 240 331 802 982 4 × 2 = 1 + 0.918 480 663 605 964 8;
  • 10) 0.918 480 663 605 964 8 × 2 = 1 + 0.836 961 327 211 929 6;
  • 11) 0.836 961 327 211 929 6 × 2 = 1 + 0.673 922 654 423 859 2;
  • 12) 0.673 922 654 423 859 2 × 2 = 1 + 0.347 845 308 847 718 4;
  • 13) 0.347 845 308 847 718 4 × 2 = 0 + 0.695 690 617 695 436 8;
  • 14) 0.695 690 617 695 436 8 × 2 = 1 + 0.391 381 235 390 873 6;
  • 15) 0.391 381 235 390 873 6 × 2 = 0 + 0.782 762 470 781 747 2;
  • 16) 0.782 762 470 781 747 2 × 2 = 1 + 0.565 524 941 563 494 4;
  • 17) 0.565 524 941 563 494 4 × 2 = 1 + 0.131 049 883 126 988 8;
  • 18) 0.131 049 883 126 988 8 × 2 = 0 + 0.262 099 766 253 977 6;
  • 19) 0.262 099 766 253 977 6 × 2 = 0 + 0.524 199 532 507 955 2;
  • 20) 0.524 199 532 507 955 2 × 2 = 1 + 0.048 399 065 015 910 4;
  • 21) 0.048 399 065 015 910 4 × 2 = 0 + 0.096 798 130 031 820 8;
  • 22) 0.096 798 130 031 820 8 × 2 = 0 + 0.193 596 260 063 641 6;
  • 23) 0.193 596 260 063 641 6 × 2 = 0 + 0.387 192 520 127 283 2;
  • 24) 0.387 192 520 127 283 2 × 2 = 0 + 0.774 385 040 254 566 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 - 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:


0.113 122 032 546 105 4(10) =

0.0001 1100 1111 0101 1001 0000(2)

5. Positive number before normalization:

5 252.113 122 032 546 105 4(10) =

1 0100 1000 0100.0001 1100 1111 0101 1001 0000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the left, so that only one non zero digit remains to the left of it:


5 252.113 122 032 546 105 4(10) =

1 0100 1000 0100.0001 1100 1111 0101 1001 0000(2) =

1 0100 1000 0100.0001 1100 1111 0101 1001 0000(2) × 20 =

1.0100 1000 0100 0001 1100 1111 0101 1001 0000(2) × 212


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): 12

Mantissa (not normalized):
1.0100 1000 0100 0001 1100 1111 0101 1001 0000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

12 + 2(8-1) - 1 =

(12 + 127)(10) =

139(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;
  • 139 ÷ 2 = 69 + 1;
  • 69 ÷ 2 = 34 + 1;
  • 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) =

139(10) =

1000 1011(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. 010 0100 0010 0000 1110 0111 1 0101 1001 0000 =

010 0100 0010 0000 1110 0111


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 1011

Mantissa (23 bits) =
010 0100 0010 0000 1110 0111


Decimal number 5 252.113 122 032 546 105 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 1011 - 010 0100 0010 0000 1110 0111

» Convert decimal 5 252.113 122 032 546 110 5 to 32 bit single precision IEEE 754 binary floating point representation standard

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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