0.000 005 49 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 005 49(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
0.000 005 49(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: 0.000 005 49.

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) 0.000 005 49 × 2 = 0 + 0.000 010 98;
  • 2) 0.000 010 98 × 2 = 0 + 0.000 021 96;
  • 3) 0.000 021 96 × 2 = 0 + 0.000 043 92;
  • 4) 0.000 043 92 × 2 = 0 + 0.000 087 84;
  • 5) 0.000 087 84 × 2 = 0 + 0.000 175 68;
  • 6) 0.000 175 68 × 2 = 0 + 0.000 351 36;
  • 7) 0.000 351 36 × 2 = 0 + 0.000 702 72;
  • 8) 0.000 702 72 × 2 = 0 + 0.001 405 44;
  • 9) 0.001 405 44 × 2 = 0 + 0.002 810 88;
  • 10) 0.002 810 88 × 2 = 0 + 0.005 621 76;
  • 11) 0.005 621 76 × 2 = 0 + 0.011 243 52;
  • 12) 0.011 243 52 × 2 = 0 + 0.022 487 04;
  • 13) 0.022 487 04 × 2 = 0 + 0.044 974 08;
  • 14) 0.044 974 08 × 2 = 0 + 0.089 948 16;
  • 15) 0.089 948 16 × 2 = 0 + 0.179 896 32;
  • 16) 0.179 896 32 × 2 = 0 + 0.359 792 64;
  • 17) 0.359 792 64 × 2 = 0 + 0.719 585 28;
  • 18) 0.719 585 28 × 2 = 1 + 0.439 170 56;
  • 19) 0.439 170 56 × 2 = 0 + 0.878 341 12;
  • 20) 0.878 341 12 × 2 = 1 + 0.756 682 24;
  • 21) 0.756 682 24 × 2 = 1 + 0.513 364 48;
  • 22) 0.513 364 48 × 2 = 1 + 0.026 728 96;
  • 23) 0.026 728 96 × 2 = 0 + 0.053 457 92;
  • 24) 0.053 457 92 × 2 = 0 + 0.106 915 84;
  • 25) 0.106 915 84 × 2 = 0 + 0.213 831 68;
  • 26) 0.213 831 68 × 2 = 0 + 0.427 663 36;
  • 27) 0.427 663 36 × 2 = 0 + 0.855 326 72;
  • 28) 0.855 326 72 × 2 = 1 + 0.710 653 44;
  • 29) 0.710 653 44 × 2 = 1 + 0.421 306 88;
  • 30) 0.421 306 88 × 2 = 0 + 0.842 613 76;
  • 31) 0.842 613 76 × 2 = 1 + 0.685 227 52;
  • 32) 0.685 227 52 × 2 = 1 + 0.370 455 04;
  • 33) 0.370 455 04 × 2 = 0 + 0.740 910 08;
  • 34) 0.740 910 08 × 2 = 1 + 0.481 820 16;
  • 35) 0.481 820 16 × 2 = 0 + 0.963 640 32;
  • 36) 0.963 640 32 × 2 = 1 + 0.927 280 64;
  • 37) 0.927 280 64 × 2 = 1 + 0.854 561 28;
  • 38) 0.854 561 28 × 2 = 1 + 0.709 122 56;
  • 39) 0.709 122 56 × 2 = 1 + 0.418 245 12;
  • 40) 0.418 245 12 × 2 = 0 + 0.836 490 24;
  • 41) 0.836 490 24 × 2 = 1 + 0.672 980 48;

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:


0.000 005 49(10) =

0.0000 0000 0000 0000 0101 1100 0001 1011 0101 1110 1(2)

5. Positive number before normalization:

0.000 005 49(10) =

0.0000 0000 0000 0000 0101 1100 0001 1011 0101 1110 1(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:


0.000 005 49(10) =

0.0000 0000 0000 0000 0101 1100 0001 1011 0101 1110 1(2) =

0.0000 0000 0000 0000 0101 1100 0001 1011 0101 1110 1(2) × 20 =

1.0111 0000 0110 1101 0111 101(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):
1.0111 0000 0110 1101 0111 101


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. 011 1000 0011 0110 1011 1101 =

011 1000 0011 0110 1011 1101


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) =
011 1000 0011 0110 1011 1101


Decimal number 0.000 005 49 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1101 - 011 1000 0011 0110 1011 1101

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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