0.000 000 110 009 32 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 009 32(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
0.000 000 110 009 32(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 110 009 32.

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) 0.000 000 110 009 32 × 2 = 0 + 0.000 000 220 018 64;
  • 2) 0.000 000 220 018 64 × 2 = 0 + 0.000 000 440 037 28;
  • 3) 0.000 000 440 037 28 × 2 = 0 + 0.000 000 880 074 56;
  • 4) 0.000 000 880 074 56 × 2 = 0 + 0.000 001 760 149 12;
  • 5) 0.000 001 760 149 12 × 2 = 0 + 0.000 003 520 298 24;
  • 6) 0.000 003 520 298 24 × 2 = 0 + 0.000 007 040 596 48;
  • 7) 0.000 007 040 596 48 × 2 = 0 + 0.000 014 081 192 96;
  • 8) 0.000 014 081 192 96 × 2 = 0 + 0.000 028 162 385 92;
  • 9) 0.000 028 162 385 92 × 2 = 0 + 0.000 056 324 771 84;
  • 10) 0.000 056 324 771 84 × 2 = 0 + 0.000 112 649 543 68;
  • 11) 0.000 112 649 543 68 × 2 = 0 + 0.000 225 299 087 36;
  • 12) 0.000 225 299 087 36 × 2 = 0 + 0.000 450 598 174 72;
  • 13) 0.000 450 598 174 72 × 2 = 0 + 0.000 901 196 349 44;
  • 14) 0.000 901 196 349 44 × 2 = 0 + 0.001 802 392 698 88;
  • 15) 0.001 802 392 698 88 × 2 = 0 + 0.003 604 785 397 76;
  • 16) 0.003 604 785 397 76 × 2 = 0 + 0.007 209 570 795 52;
  • 17) 0.007 209 570 795 52 × 2 = 0 + 0.014 419 141 591 04;
  • 18) 0.014 419 141 591 04 × 2 = 0 + 0.028 838 283 182 08;
  • 19) 0.028 838 283 182 08 × 2 = 0 + 0.057 676 566 364 16;
  • 20) 0.057 676 566 364 16 × 2 = 0 + 0.115 353 132 728 32;
  • 21) 0.115 353 132 728 32 × 2 = 0 + 0.230 706 265 456 64;
  • 22) 0.230 706 265 456 64 × 2 = 0 + 0.461 412 530 913 28;
  • 23) 0.461 412 530 913 28 × 2 = 0 + 0.922 825 061 826 56;
  • 24) 0.922 825 061 826 56 × 2 = 1 + 0.845 650 123 653 12;
  • 25) 0.845 650 123 653 12 × 2 = 1 + 0.691 300 247 306 24;
  • 26) 0.691 300 247 306 24 × 2 = 1 + 0.382 600 494 612 48;
  • 27) 0.382 600 494 612 48 × 2 = 0 + 0.765 200 989 224 96;
  • 28) 0.765 200 989 224 96 × 2 = 1 + 0.530 401 978 449 92;
  • 29) 0.530 401 978 449 92 × 2 = 1 + 0.060 803 956 899 84;
  • 30) 0.060 803 956 899 84 × 2 = 0 + 0.121 607 913 799 68;
  • 31) 0.121 607 913 799 68 × 2 = 0 + 0.243 215 827 599 36;
  • 32) 0.243 215 827 599 36 × 2 = 0 + 0.486 431 655 198 72;
  • 33) 0.486 431 655 198 72 × 2 = 0 + 0.972 863 310 397 44;
  • 34) 0.972 863 310 397 44 × 2 = 1 + 0.945 726 620 794 88;
  • 35) 0.945 726 620 794 88 × 2 = 1 + 0.891 453 241 589 76;
  • 36) 0.891 453 241 589 76 × 2 = 1 + 0.782 906 483 179 52;
  • 37) 0.782 906 483 179 52 × 2 = 1 + 0.565 812 966 359 04;
  • 38) 0.565 812 966 359 04 × 2 = 1 + 0.131 625 932 718 08;
  • 39) 0.131 625 932 718 08 × 2 = 0 + 0.263 251 865 436 16;
  • 40) 0.263 251 865 436 16 × 2 = 0 + 0.526 503 730 872 32;
  • 41) 0.526 503 730 872 32 × 2 = 1 + 0.053 007 461 744 64;
  • 42) 0.053 007 461 744 64 × 2 = 0 + 0.106 014 923 489 28;
  • 43) 0.106 014 923 489 28 × 2 = 0 + 0.212 029 846 978 56;
  • 44) 0.212 029 846 978 56 × 2 = 0 + 0.424 059 693 957 12;
  • 45) 0.424 059 693 957 12 × 2 = 0 + 0.848 119 387 914 24;
  • 46) 0.848 119 387 914 24 × 2 = 1 + 0.696 238 775 828 48;
  • 47) 0.696 238 775 828 48 × 2 = 1 + 0.392 477 551 656 96;

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:


0.000 000 110 009 32(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 1000 011(2)

5. Positive number before normalization:

0.000 000 110 009 32(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 1000 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 009 32(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 1000 011(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 1000 011(2) × 20 =

1.1101 1000 0111 1100 1000 011(2) × 2-24


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): -24

Mantissa (not normalized):
1.1101 1000 0111 1100 1000 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

103(10) =

0110 0111(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 1100 0011 1110 0100 0011 =

110 1100 0011 1110 0100 0011


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) =
0110 0111

Mantissa (23 bits) =
110 1100 0011 1110 0100 0011


Decimal number 0.000 000 110 009 32 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 0100 0011

Share

» Convert decimal 0.000 000 110 010 3 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: Your Greatest Competitor, Your Most Powerful Ally. Gemini Notebook YouTube Video of 7.54 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