0.000 000 005 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 005 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
0.000 000 005 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: 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: 0.000 000 005 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) 0.000 000 005 4 × 2 = 0 + 0.000 000 010 8;
  • 2) 0.000 000 010 8 × 2 = 0 + 0.000 000 021 6;
  • 3) 0.000 000 021 6 × 2 = 0 + 0.000 000 043 2;
  • 4) 0.000 000 043 2 × 2 = 0 + 0.000 000 086 4;
  • 5) 0.000 000 086 4 × 2 = 0 + 0.000 000 172 8;
  • 6) 0.000 000 172 8 × 2 = 0 + 0.000 000 345 6;
  • 7) 0.000 000 345 6 × 2 = 0 + 0.000 000 691 2;
  • 8) 0.000 000 691 2 × 2 = 0 + 0.000 001 382 4;
  • 9) 0.000 001 382 4 × 2 = 0 + 0.000 002 764 8;
  • 10) 0.000 002 764 8 × 2 = 0 + 0.000 005 529 6;
  • 11) 0.000 005 529 6 × 2 = 0 + 0.000 011 059 2;
  • 12) 0.000 011 059 2 × 2 = 0 + 0.000 022 118 4;
  • 13) 0.000 022 118 4 × 2 = 0 + 0.000 044 236 8;
  • 14) 0.000 044 236 8 × 2 = 0 + 0.000 088 473 6;
  • 15) 0.000 088 473 6 × 2 = 0 + 0.000 176 947 2;
  • 16) 0.000 176 947 2 × 2 = 0 + 0.000 353 894 4;
  • 17) 0.000 353 894 4 × 2 = 0 + 0.000 707 788 8;
  • 18) 0.000 707 788 8 × 2 = 0 + 0.001 415 577 6;
  • 19) 0.001 415 577 6 × 2 = 0 + 0.002 831 155 2;
  • 20) 0.002 831 155 2 × 2 = 0 + 0.005 662 310 4;
  • 21) 0.005 662 310 4 × 2 = 0 + 0.011 324 620 8;
  • 22) 0.011 324 620 8 × 2 = 0 + 0.022 649 241 6;
  • 23) 0.022 649 241 6 × 2 = 0 + 0.045 298 483 2;
  • 24) 0.045 298 483 2 × 2 = 0 + 0.090 596 966 4;
  • 25) 0.090 596 966 4 × 2 = 0 + 0.181 193 932 8;
  • 26) 0.181 193 932 8 × 2 = 0 + 0.362 387 865 6;
  • 27) 0.362 387 865 6 × 2 = 0 + 0.724 775 731 2;
  • 28) 0.724 775 731 2 × 2 = 1 + 0.449 551 462 4;
  • 29) 0.449 551 462 4 × 2 = 0 + 0.899 102 924 8;
  • 30) 0.899 102 924 8 × 2 = 1 + 0.798 205 849 6;
  • 31) 0.798 205 849 6 × 2 = 1 + 0.596 411 699 2;
  • 32) 0.596 411 699 2 × 2 = 1 + 0.192 823 398 4;
  • 33) 0.192 823 398 4 × 2 = 0 + 0.385 646 796 8;
  • 34) 0.385 646 796 8 × 2 = 0 + 0.771 293 593 6;
  • 35) 0.771 293 593 6 × 2 = 1 + 0.542 587 187 2;
  • 36) 0.542 587 187 2 × 2 = 1 + 0.085 174 374 4;
  • 37) 0.085 174 374 4 × 2 = 0 + 0.170 348 748 8;
  • 38) 0.170 348 748 8 × 2 = 0 + 0.340 697 497 6;
  • 39) 0.340 697 497 6 × 2 = 0 + 0.681 394 995 2;
  • 40) 0.681 394 995 2 × 2 = 1 + 0.362 789 990 4;
  • 41) 0.362 789 990 4 × 2 = 0 + 0.725 579 980 8;
  • 42) 0.725 579 980 8 × 2 = 1 + 0.451 159 961 6;
  • 43) 0.451 159 961 6 × 2 = 0 + 0.902 319 923 2;
  • 44) 0.902 319 923 2 × 2 = 1 + 0.804 639 846 4;
  • 45) 0.804 639 846 4 × 2 = 1 + 0.609 279 692 8;
  • 46) 0.609 279 692 8 × 2 = 1 + 0.218 559 385 6;
  • 47) 0.218 559 385 6 × 2 = 0 + 0.437 118 771 2;
  • 48) 0.437 118 771 2 × 2 = 0 + 0.874 237 542 4;
  • 49) 0.874 237 542 4 × 2 = 1 + 0.748 475 084 8;
  • 50) 0.748 475 084 8 × 2 = 1 + 0.496 950 169 6;
  • 51) 0.496 950 169 6 × 2 = 0 + 0.993 900 339 2;

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:


0.000 000 005 4(10) =

0.0000 0000 0000 0000 0000 0000 0001 0111 0011 0001 0101 1100 110(2)

5. Positive number before normalization:

0.000 000 005 4(10) =

0.0000 0000 0000 0000 0000 0000 0001 0111 0011 0001 0101 1100 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 005 4(10) =

0.0000 0000 0000 0000 0000 0000 0001 0111 0011 0001 0101 1100 110(2) =

0.0000 0000 0000 0000 0000 0000 0001 0111 0011 0001 0101 1100 110(2) × 20 =

1.0111 0011 0001 0101 1100 110(2) × 2-28


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): -28

Mantissa (not normalized):
1.0111 0011 0001 0101 1100 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

99(10) =

0110 0011(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. 011 1001 1000 1010 1110 0110 =

011 1001 1000 1010 1110 0110


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 0011

Mantissa (23 bits) =
011 1001 1000 1010 1110 0110


Decimal number 0.000 000 005 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0011 - 011 1001 1000 1010 1110 0110

Share

» Convert decimal 0.000 000 013 9 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 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM 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