0.000 000 707 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 707(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
0.000 000 707(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: 0.000 000 707.

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) 0.000 000 707 × 2 = 0 + 0.000 001 414;
  • 2) 0.000 001 414 × 2 = 0 + 0.000 002 828;
  • 3) 0.000 002 828 × 2 = 0 + 0.000 005 656;
  • 4) 0.000 005 656 × 2 = 0 + 0.000 011 312;
  • 5) 0.000 011 312 × 2 = 0 + 0.000 022 624;
  • 6) 0.000 022 624 × 2 = 0 + 0.000 045 248;
  • 7) 0.000 045 248 × 2 = 0 + 0.000 090 496;
  • 8) 0.000 090 496 × 2 = 0 + 0.000 180 992;
  • 9) 0.000 180 992 × 2 = 0 + 0.000 361 984;
  • 10) 0.000 361 984 × 2 = 0 + 0.000 723 968;
  • 11) 0.000 723 968 × 2 = 0 + 0.001 447 936;
  • 12) 0.001 447 936 × 2 = 0 + 0.002 895 872;
  • 13) 0.002 895 872 × 2 = 0 + 0.005 791 744;
  • 14) 0.005 791 744 × 2 = 0 + 0.011 583 488;
  • 15) 0.011 583 488 × 2 = 0 + 0.023 166 976;
  • 16) 0.023 166 976 × 2 = 0 + 0.046 333 952;
  • 17) 0.046 333 952 × 2 = 0 + 0.092 667 904;
  • 18) 0.092 667 904 × 2 = 0 + 0.185 335 808;
  • 19) 0.185 335 808 × 2 = 0 + 0.370 671 616;
  • 20) 0.370 671 616 × 2 = 0 + 0.741 343 232;
  • 21) 0.741 343 232 × 2 = 1 + 0.482 686 464;
  • 22) 0.482 686 464 × 2 = 0 + 0.965 372 928;
  • 23) 0.965 372 928 × 2 = 1 + 0.930 745 856;
  • 24) 0.930 745 856 × 2 = 1 + 0.861 491 712;
  • 25) 0.861 491 712 × 2 = 1 + 0.722 983 424;
  • 26) 0.722 983 424 × 2 = 1 + 0.445 966 848;
  • 27) 0.445 966 848 × 2 = 0 + 0.891 933 696;
  • 28) 0.891 933 696 × 2 = 1 + 0.783 867 392;
  • 29) 0.783 867 392 × 2 = 1 + 0.567 734 784;
  • 30) 0.567 734 784 × 2 = 1 + 0.135 469 568;
  • 31) 0.135 469 568 × 2 = 0 + 0.270 939 136;
  • 32) 0.270 939 136 × 2 = 0 + 0.541 878 272;
  • 33) 0.541 878 272 × 2 = 1 + 0.083 756 544;
  • 34) 0.083 756 544 × 2 = 0 + 0.167 513 088;
  • 35) 0.167 513 088 × 2 = 0 + 0.335 026 176;
  • 36) 0.335 026 176 × 2 = 0 + 0.670 052 352;
  • 37) 0.670 052 352 × 2 = 1 + 0.340 104 704;
  • 38) 0.340 104 704 × 2 = 0 + 0.680 209 408;
  • 39) 0.680 209 408 × 2 = 1 + 0.360 418 816;
  • 40) 0.360 418 816 × 2 = 0 + 0.720 837 632;
  • 41) 0.720 837 632 × 2 = 1 + 0.441 675 264;
  • 42) 0.441 675 264 × 2 = 0 + 0.883 350 528;
  • 43) 0.883 350 528 × 2 = 1 + 0.766 701 056;
  • 44) 0.766 701 056 × 2 = 1 + 0.533 402 112;

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:


0.000 000 707(10) =

0.0000 0000 0000 0000 0000 1011 1101 1100 1000 1010 1011(2)

5. Positive number before normalization:

0.000 000 707(10) =

0.0000 0000 0000 0000 0000 1011 1101 1100 1000 1010 1011(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:


0.000 000 707(10) =

0.0000 0000 0000 0000 0000 1011 1101 1100 1000 1010 1011(2) =

0.0000 0000 0000 0000 0000 1011 1101 1100 1000 1010 1011(2) × 20 =

1.0111 1011 1001 0001 0101 011(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 1011 1001 0001 0101 011


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 1101 1100 1000 1010 1011 =

011 1101 1100 1000 1010 1011


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 1101 1100 1000 1010 1011


Decimal number 0.000 000 707 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 011 1101 1100 1000 1010 1011

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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