0.000 000 109 988 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 988 5(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
0.000 000 109 988 5(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: 0.000 000 109 988 5.

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) 0.000 000 109 988 5 × 2 = 0 + 0.000 000 219 977;
  • 2) 0.000 000 219 977 × 2 = 0 + 0.000 000 439 954;
  • 3) 0.000 000 439 954 × 2 = 0 + 0.000 000 879 908;
  • 4) 0.000 000 879 908 × 2 = 0 + 0.000 001 759 816;
  • 5) 0.000 001 759 816 × 2 = 0 + 0.000 003 519 632;
  • 6) 0.000 003 519 632 × 2 = 0 + 0.000 007 039 264;
  • 7) 0.000 007 039 264 × 2 = 0 + 0.000 014 078 528;
  • 8) 0.000 014 078 528 × 2 = 0 + 0.000 028 157 056;
  • 9) 0.000 028 157 056 × 2 = 0 + 0.000 056 314 112;
  • 10) 0.000 056 314 112 × 2 = 0 + 0.000 112 628 224;
  • 11) 0.000 112 628 224 × 2 = 0 + 0.000 225 256 448;
  • 12) 0.000 225 256 448 × 2 = 0 + 0.000 450 512 896;
  • 13) 0.000 450 512 896 × 2 = 0 + 0.000 901 025 792;
  • 14) 0.000 901 025 792 × 2 = 0 + 0.001 802 051 584;
  • 15) 0.001 802 051 584 × 2 = 0 + 0.003 604 103 168;
  • 16) 0.003 604 103 168 × 2 = 0 + 0.007 208 206 336;
  • 17) 0.007 208 206 336 × 2 = 0 + 0.014 416 412 672;
  • 18) 0.014 416 412 672 × 2 = 0 + 0.028 832 825 344;
  • 19) 0.028 832 825 344 × 2 = 0 + 0.057 665 650 688;
  • 20) 0.057 665 650 688 × 2 = 0 + 0.115 331 301 376;
  • 21) 0.115 331 301 376 × 2 = 0 + 0.230 662 602 752;
  • 22) 0.230 662 602 752 × 2 = 0 + 0.461 325 205 504;
  • 23) 0.461 325 205 504 × 2 = 0 + 0.922 650 411 008;
  • 24) 0.922 650 411 008 × 2 = 1 + 0.845 300 822 016;
  • 25) 0.845 300 822 016 × 2 = 1 + 0.690 601 644 032;
  • 26) 0.690 601 644 032 × 2 = 1 + 0.381 203 288 064;
  • 27) 0.381 203 288 064 × 2 = 0 + 0.762 406 576 128;
  • 28) 0.762 406 576 128 × 2 = 1 + 0.524 813 152 256;
  • 29) 0.524 813 152 256 × 2 = 1 + 0.049 626 304 512;
  • 30) 0.049 626 304 512 × 2 = 0 + 0.099 252 609 024;
  • 31) 0.099 252 609 024 × 2 = 0 + 0.198 505 218 048;
  • 32) 0.198 505 218 048 × 2 = 0 + 0.397 010 436 096;
  • 33) 0.397 010 436 096 × 2 = 0 + 0.794 020 872 192;
  • 34) 0.794 020 872 192 × 2 = 1 + 0.588 041 744 384;
  • 35) 0.588 041 744 384 × 2 = 1 + 0.176 083 488 768;
  • 36) 0.176 083 488 768 × 2 = 0 + 0.352 166 977 536;
  • 37) 0.352 166 977 536 × 2 = 0 + 0.704 333 955 072;
  • 38) 0.704 333 955 072 × 2 = 1 + 0.408 667 910 144;
  • 39) 0.408 667 910 144 × 2 = 0 + 0.817 335 820 288;
  • 40) 0.817 335 820 288 × 2 = 1 + 0.634 671 640 576;
  • 41) 0.634 671 640 576 × 2 = 1 + 0.269 343 281 152;
  • 42) 0.269 343 281 152 × 2 = 0 + 0.538 686 562 304;
  • 43) 0.538 686 562 304 × 2 = 1 + 0.077 373 124 608;
  • 44) 0.077 373 124 608 × 2 = 0 + 0.154 746 249 216;
  • 45) 0.154 746 249 216 × 2 = 0 + 0.309 492 498 432;
  • 46) 0.309 492 498 432 × 2 = 0 + 0.618 984 996 864;
  • 47) 0.618 984 996 864 × 2 = 1 + 0.237 969 993 728;

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:


0.000 000 109 988 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0101 1010 001(2)

5. Positive number before normalization:

0.000 000 109 988 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0101 1010 001(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:


0.000 000 109 988 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0101 1010 001(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0101 1010 001(2) × 20 =

1.1101 1000 0110 0101 1010 001(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 0110 0101 1010 001


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 0010 1101 0001 =

110 1100 0011 0010 1101 0001


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 0010 1101 0001


Decimal number 0.000 000 109 988 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 0010 1101 0001

Share

» Convert decimal 0.000 000 109 993 2 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: Reputation: Bridge Between Company's Actions & Market Value. Gemini Notebook YouTube Video of 6.55 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