0.000 000 69 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 69(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 69(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 69.

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 69 × 2 = 0 + 0.000 001 38;
  • 2) 0.000 001 38 × 2 = 0 + 0.000 002 76;
  • 3) 0.000 002 76 × 2 = 0 + 0.000 005 52;
  • 4) 0.000 005 52 × 2 = 0 + 0.000 011 04;
  • 5) 0.000 011 04 × 2 = 0 + 0.000 022 08;
  • 6) 0.000 022 08 × 2 = 0 + 0.000 044 16;
  • 7) 0.000 044 16 × 2 = 0 + 0.000 088 32;
  • 8) 0.000 088 32 × 2 = 0 + 0.000 176 64;
  • 9) 0.000 176 64 × 2 = 0 + 0.000 353 28;
  • 10) 0.000 353 28 × 2 = 0 + 0.000 706 56;
  • 11) 0.000 706 56 × 2 = 0 + 0.001 413 12;
  • 12) 0.001 413 12 × 2 = 0 + 0.002 826 24;
  • 13) 0.002 826 24 × 2 = 0 + 0.005 652 48;
  • 14) 0.005 652 48 × 2 = 0 + 0.011 304 96;
  • 15) 0.011 304 96 × 2 = 0 + 0.022 609 92;
  • 16) 0.022 609 92 × 2 = 0 + 0.045 219 84;
  • 17) 0.045 219 84 × 2 = 0 + 0.090 439 68;
  • 18) 0.090 439 68 × 2 = 0 + 0.180 879 36;
  • 19) 0.180 879 36 × 2 = 0 + 0.361 758 72;
  • 20) 0.361 758 72 × 2 = 0 + 0.723 517 44;
  • 21) 0.723 517 44 × 2 = 1 + 0.447 034 88;
  • 22) 0.447 034 88 × 2 = 0 + 0.894 069 76;
  • 23) 0.894 069 76 × 2 = 1 + 0.788 139 52;
  • 24) 0.788 139 52 × 2 = 1 + 0.576 279 04;
  • 25) 0.576 279 04 × 2 = 1 + 0.152 558 08;
  • 26) 0.152 558 08 × 2 = 0 + 0.305 116 16;
  • 27) 0.305 116 16 × 2 = 0 + 0.610 232 32;
  • 28) 0.610 232 32 × 2 = 1 + 0.220 464 64;
  • 29) 0.220 464 64 × 2 = 0 + 0.440 929 28;
  • 30) 0.440 929 28 × 2 = 0 + 0.881 858 56;
  • 31) 0.881 858 56 × 2 = 1 + 0.763 717 12;
  • 32) 0.763 717 12 × 2 = 1 + 0.527 434 24;
  • 33) 0.527 434 24 × 2 = 1 + 0.054 868 48;
  • 34) 0.054 868 48 × 2 = 0 + 0.109 736 96;
  • 35) 0.109 736 96 × 2 = 0 + 0.219 473 92;
  • 36) 0.219 473 92 × 2 = 0 + 0.438 947 84;
  • 37) 0.438 947 84 × 2 = 0 + 0.877 895 68;
  • 38) 0.877 895 68 × 2 = 1 + 0.755 791 36;
  • 39) 0.755 791 36 × 2 = 1 + 0.511 582 72;
  • 40) 0.511 582 72 × 2 = 1 + 0.023 165 44;
  • 41) 0.023 165 44 × 2 = 0 + 0.046 330 88;
  • 42) 0.046 330 88 × 2 = 0 + 0.092 661 76;
  • 43) 0.092 661 76 × 2 = 0 + 0.185 323 52;
  • 44) 0.185 323 52 × 2 = 0 + 0.370 647 04;

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 69(10) =

0.0000 0000 0000 0000 0000 1011 1001 0011 1000 0111 0000(2)

5. Positive number before normalization:

0.000 000 69(10) =

0.0000 0000 0000 0000 0000 1011 1001 0011 1000 0111 0000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 69(10) =

0.0000 0000 0000 0000 0000 1011 1001 0011 1000 0111 0000(2) =

0.0000 0000 0000 0000 0000 1011 1001 0011 1000 0111 0000(2) × 20 =

1.0111 0010 0111 0000 1110 000(2) × 2-21


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): -21

Mantissa (not normalized):
1.0111 0010 0111 0000 1110 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 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) =

106(10) =

0110 1010(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 0011 1000 0111 0000 =

011 1001 0011 1000 0111 0000


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 1010

Mantissa (23 bits) =
011 1001 0011 1000 0111 0000


Decimal number 0.000 000 69 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 011 1001 0011 1000 0111 0000

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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