0.000 006 14 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 006 14(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
0.000 006 14(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: 0.000 006 14.

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) 0.000 006 14 × 2 = 0 + 0.000 012 28;
  • 2) 0.000 012 28 × 2 = 0 + 0.000 024 56;
  • 3) 0.000 024 56 × 2 = 0 + 0.000 049 12;
  • 4) 0.000 049 12 × 2 = 0 + 0.000 098 24;
  • 5) 0.000 098 24 × 2 = 0 + 0.000 196 48;
  • 6) 0.000 196 48 × 2 = 0 + 0.000 392 96;
  • 7) 0.000 392 96 × 2 = 0 + 0.000 785 92;
  • 8) 0.000 785 92 × 2 = 0 + 0.001 571 84;
  • 9) 0.001 571 84 × 2 = 0 + 0.003 143 68;
  • 10) 0.003 143 68 × 2 = 0 + 0.006 287 36;
  • 11) 0.006 287 36 × 2 = 0 + 0.012 574 72;
  • 12) 0.012 574 72 × 2 = 0 + 0.025 149 44;
  • 13) 0.025 149 44 × 2 = 0 + 0.050 298 88;
  • 14) 0.050 298 88 × 2 = 0 + 0.100 597 76;
  • 15) 0.100 597 76 × 2 = 0 + 0.201 195 52;
  • 16) 0.201 195 52 × 2 = 0 + 0.402 391 04;
  • 17) 0.402 391 04 × 2 = 0 + 0.804 782 08;
  • 18) 0.804 782 08 × 2 = 1 + 0.609 564 16;
  • 19) 0.609 564 16 × 2 = 1 + 0.219 128 32;
  • 20) 0.219 128 32 × 2 = 0 + 0.438 256 64;
  • 21) 0.438 256 64 × 2 = 0 + 0.876 513 28;
  • 22) 0.876 513 28 × 2 = 1 + 0.753 026 56;
  • 23) 0.753 026 56 × 2 = 1 + 0.506 053 12;
  • 24) 0.506 053 12 × 2 = 1 + 0.012 106 24;
  • 25) 0.012 106 24 × 2 = 0 + 0.024 212 48;
  • 26) 0.024 212 48 × 2 = 0 + 0.048 424 96;
  • 27) 0.048 424 96 × 2 = 0 + 0.096 849 92;
  • 28) 0.096 849 92 × 2 = 0 + 0.193 699 84;
  • 29) 0.193 699 84 × 2 = 0 + 0.387 399 68;
  • 30) 0.387 399 68 × 2 = 0 + 0.774 799 36;
  • 31) 0.774 799 36 × 2 = 1 + 0.549 598 72;
  • 32) 0.549 598 72 × 2 = 1 + 0.099 197 44;
  • 33) 0.099 197 44 × 2 = 0 + 0.198 394 88;
  • 34) 0.198 394 88 × 2 = 0 + 0.396 789 76;
  • 35) 0.396 789 76 × 2 = 0 + 0.793 579 52;
  • 36) 0.793 579 52 × 2 = 1 + 0.587 159 04;
  • 37) 0.587 159 04 × 2 = 1 + 0.174 318 08;
  • 38) 0.174 318 08 × 2 = 0 + 0.348 636 16;
  • 39) 0.348 636 16 × 2 = 0 + 0.697 272 32;
  • 40) 0.697 272 32 × 2 = 1 + 0.394 544 64;
  • 41) 0.394 544 64 × 2 = 0 + 0.789 089 28;

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:


0.000 006 14(10) =

0.0000 0000 0000 0000 0110 0111 0000 0011 0001 1001 0(2)

5. Positive number before normalization:

0.000 006 14(10) =

0.0000 0000 0000 0000 0110 0111 0000 0011 0001 1001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right, so that only one non zero digit remains to the left of it:


0.000 006 14(10) =

0.0000 0000 0000 0000 0110 0111 0000 0011 0001 1001 0(2) =

0.0000 0000 0000 0000 0110 0111 0000 0011 0001 1001 0(2) × 20 =

1.1001 1100 0000 1100 0110 010(2) × 2-18


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): -18

Mantissa (not normalized):
1.1001 1100 0000 1100 0110 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-18 + 2(8-1) - 1 =

(-18 + 127)(10) =

109(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;
  • 109 ÷ 2 = 54 + 1;
  • 54 ÷ 2 = 27 + 0;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

109(10) =

0110 1101(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. 100 1110 0000 0110 0011 0010 =

100 1110 0000 0110 0011 0010


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 1101

Mantissa (23 bits) =
100 1110 0000 0110 0011 0010


Decimal number 0.000 006 14 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1101 - 100 1110 0000 0110 0011 0010

Share

» Convert decimal 0.000 006 05 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: Your greatest rival, your most powerful ally. NotebookLM 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