11 001.010 110 001 101 010 124 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 11 001.010 110 001 101 010 124(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
11 001.010 110 001 101 010 124(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: 11 001.
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;
  • 11 001 ÷ 2 = 5 500 + 1;
  • 5 500 ÷ 2 = 2 750 + 0;
  • 2 750 ÷ 2 = 1 375 + 0;
  • 1 375 ÷ 2 = 687 + 1;
  • 687 ÷ 2 = 343 + 1;
  • 343 ÷ 2 = 171 + 1;
  • 171 ÷ 2 = 85 + 1;
  • 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;

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.

11 001(10) =

10 1010 1111 1001(2)


3. Convert to binary (base 2) the fractional part: 0.010 110 001 101 010 124.

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) 0.010 110 001 101 010 124 × 2 = 0 + 0.020 220 002 202 020 248;
  • 2) 0.020 220 002 202 020 248 × 2 = 0 + 0.040 440 004 404 040 496;
  • 3) 0.040 440 004 404 040 496 × 2 = 0 + 0.080 880 008 808 080 992;
  • 4) 0.080 880 008 808 080 992 × 2 = 0 + 0.161 760 017 616 161 984;
  • 5) 0.161 760 017 616 161 984 × 2 = 0 + 0.323 520 035 232 323 968;
  • 6) 0.323 520 035 232 323 968 × 2 = 0 + 0.647 040 070 464 647 936;
  • 7) 0.647 040 070 464 647 936 × 2 = 1 + 0.294 080 140 929 295 872;
  • 8) 0.294 080 140 929 295 872 × 2 = 0 + 0.588 160 281 858 591 744;
  • 9) 0.588 160 281 858 591 744 × 2 = 1 + 0.176 320 563 717 183 488;
  • 10) 0.176 320 563 717 183 488 × 2 = 0 + 0.352 641 127 434 366 976;
  • 11) 0.352 641 127 434 366 976 × 2 = 0 + 0.705 282 254 868 733 952;
  • 12) 0.705 282 254 868 733 952 × 2 = 1 + 0.410 564 509 737 467 904;
  • 13) 0.410 564 509 737 467 904 × 2 = 0 + 0.821 129 019 474 935 808;
  • 14) 0.821 129 019 474 935 808 × 2 = 1 + 0.642 258 038 949 871 616;
  • 15) 0.642 258 038 949 871 616 × 2 = 1 + 0.284 516 077 899 743 232;
  • 16) 0.284 516 077 899 743 232 × 2 = 0 + 0.569 032 155 799 486 464;
  • 17) 0.569 032 155 799 486 464 × 2 = 1 + 0.138 064 311 598 972 928;
  • 18) 0.138 064 311 598 972 928 × 2 = 0 + 0.276 128 623 197 945 856;
  • 19) 0.276 128 623 197 945 856 × 2 = 0 + 0.552 257 246 395 891 712;
  • 20) 0.552 257 246 395 891 712 × 2 = 1 + 0.104 514 492 791 783 424;
  • 21) 0.104 514 492 791 783 424 × 2 = 0 + 0.209 028 985 583 566 848;
  • 22) 0.209 028 985 583 566 848 × 2 = 0 + 0.418 057 971 167 133 696;
  • 23) 0.418 057 971 167 133 696 × 2 = 0 + 0.836 115 942 334 267 392;
  • 24) 0.836 115 942 334 267 392 × 2 = 1 + 0.672 231 884 668 534 784;

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:


0.010 110 001 101 010 124(10) =

0.0000 0010 1001 0110 1001 0001(2)

5. Positive number before normalization:

11 001.010 110 001 101 010 124(10) =

10 1010 1111 1001.0000 0010 1001 0110 1001 0001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the left, so that only one non zero digit remains to the left of it:


11 001.010 110 001 101 010 124(10) =

10 1010 1111 1001.0000 0010 1001 0110 1001 0001(2) =

10 1010 1111 1001.0000 0010 1001 0110 1001 0001(2) × 20 =

1.0101 0111 1100 1000 0001 0100 1011 0100 1000 1(2) × 213


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): 13

Mantissa (not normalized):
1.0101 0111 1100 1000 0001 0100 1011 0100 1000 1


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

13 + 2(8-1) - 1 =

(13 + 127)(10) =

140(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;
  • 140 ÷ 2 = 70 + 0;
  • 70 ÷ 2 = 35 + 0;
  • 35 ÷ 2 = 17 + 1;
  • 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) =

140(10) =

1000 1100(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 1011 1110 0100 0000 1010 01 0110 1001 0001 =

010 1011 1110 0100 0000 1010


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 1100

Mantissa (23 bits) =
010 1011 1110 0100 0000 1010


Decimal number 11 001.010 110 001 101 010 124 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 1100 - 010 1011 1110 0100 0000 1010

Share

» Convert decimal 11 001.010 110 001 101 010 163 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Show, Don’t Tell, the Strategic Power of Action. Gemini Notebook YouTube Video of 9.21 minutes

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