0.000 001 77 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 77(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
0.000 001 77(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: 0.000 001 77.

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) 0.000 001 77 × 2 = 0 + 0.000 003 54;
  • 2) 0.000 003 54 × 2 = 0 + 0.000 007 08;
  • 3) 0.000 007 08 × 2 = 0 + 0.000 014 16;
  • 4) 0.000 014 16 × 2 = 0 + 0.000 028 32;
  • 5) 0.000 028 32 × 2 = 0 + 0.000 056 64;
  • 6) 0.000 056 64 × 2 = 0 + 0.000 113 28;
  • 7) 0.000 113 28 × 2 = 0 + 0.000 226 56;
  • 8) 0.000 226 56 × 2 = 0 + 0.000 453 12;
  • 9) 0.000 453 12 × 2 = 0 + 0.000 906 24;
  • 10) 0.000 906 24 × 2 = 0 + 0.001 812 48;
  • 11) 0.001 812 48 × 2 = 0 + 0.003 624 96;
  • 12) 0.003 624 96 × 2 = 0 + 0.007 249 92;
  • 13) 0.007 249 92 × 2 = 0 + 0.014 499 84;
  • 14) 0.014 499 84 × 2 = 0 + 0.028 999 68;
  • 15) 0.028 999 68 × 2 = 0 + 0.057 999 36;
  • 16) 0.057 999 36 × 2 = 0 + 0.115 998 72;
  • 17) 0.115 998 72 × 2 = 0 + 0.231 997 44;
  • 18) 0.231 997 44 × 2 = 0 + 0.463 994 88;
  • 19) 0.463 994 88 × 2 = 0 + 0.927 989 76;
  • 20) 0.927 989 76 × 2 = 1 + 0.855 979 52;
  • 21) 0.855 979 52 × 2 = 1 + 0.711 959 04;
  • 22) 0.711 959 04 × 2 = 1 + 0.423 918 08;
  • 23) 0.423 918 08 × 2 = 0 + 0.847 836 16;
  • 24) 0.847 836 16 × 2 = 1 + 0.695 672 32;
  • 25) 0.695 672 32 × 2 = 1 + 0.391 344 64;
  • 26) 0.391 344 64 × 2 = 0 + 0.782 689 28;
  • 27) 0.782 689 28 × 2 = 1 + 0.565 378 56;
  • 28) 0.565 378 56 × 2 = 1 + 0.130 757 12;
  • 29) 0.130 757 12 × 2 = 0 + 0.261 514 24;
  • 30) 0.261 514 24 × 2 = 0 + 0.523 028 48;
  • 31) 0.523 028 48 × 2 = 1 + 0.046 056 96;
  • 32) 0.046 056 96 × 2 = 0 + 0.092 113 92;
  • 33) 0.092 113 92 × 2 = 0 + 0.184 227 84;
  • 34) 0.184 227 84 × 2 = 0 + 0.368 455 68;
  • 35) 0.368 455 68 × 2 = 0 + 0.736 911 36;
  • 36) 0.736 911 36 × 2 = 1 + 0.473 822 72;
  • 37) 0.473 822 72 × 2 = 0 + 0.947 645 44;
  • 38) 0.947 645 44 × 2 = 1 + 0.895 290 88;
  • 39) 0.895 290 88 × 2 = 1 + 0.790 581 76;
  • 40) 0.790 581 76 × 2 = 1 + 0.581 163 52;
  • 41) 0.581 163 52 × 2 = 1 + 0.162 327 04;
  • 42) 0.162 327 04 × 2 = 0 + 0.324 654 08;
  • 43) 0.324 654 08 × 2 = 0 + 0.649 308 16;

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:


0.000 001 77(10) =

0.0000 0000 0000 0000 0001 1101 1011 0010 0001 0111 100(2)

5. Positive number before normalization:

0.000 001 77(10) =

0.0000 0000 0000 0000 0001 1101 1011 0010 0001 0111 100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 77(10) =

0.0000 0000 0000 0000 0001 1101 1011 0010 0001 0111 100(2) =

0.0000 0000 0000 0000 0001 1101 1011 0010 0001 0111 100(2) × 20 =

1.1101 1011 0010 0001 0111 100(2) × 2-20


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): -20

Mantissa (not normalized):
1.1101 1011 0010 0001 0111 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 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) =

107(10) =

0110 1011(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 1101 1001 0000 1011 1100 =

110 1101 1001 0000 1011 1100


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 1011

Mantissa (23 bits) =
110 1101 1001 0000 1011 1100


Decimal number 0.000 001 77 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1011 - 110 1101 1001 0000 1011 1100

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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