0.000 000 451 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 451(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
0.000 000 451(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: 0.000 000 451.

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) 0.000 000 451 × 2 = 0 + 0.000 000 902;
  • 2) 0.000 000 902 × 2 = 0 + 0.000 001 804;
  • 3) 0.000 001 804 × 2 = 0 + 0.000 003 608;
  • 4) 0.000 003 608 × 2 = 0 + 0.000 007 216;
  • 5) 0.000 007 216 × 2 = 0 + 0.000 014 432;
  • 6) 0.000 014 432 × 2 = 0 + 0.000 028 864;
  • 7) 0.000 028 864 × 2 = 0 + 0.000 057 728;
  • 8) 0.000 057 728 × 2 = 0 + 0.000 115 456;
  • 9) 0.000 115 456 × 2 = 0 + 0.000 230 912;
  • 10) 0.000 230 912 × 2 = 0 + 0.000 461 824;
  • 11) 0.000 461 824 × 2 = 0 + 0.000 923 648;
  • 12) 0.000 923 648 × 2 = 0 + 0.001 847 296;
  • 13) 0.001 847 296 × 2 = 0 + 0.003 694 592;
  • 14) 0.003 694 592 × 2 = 0 + 0.007 389 184;
  • 15) 0.007 389 184 × 2 = 0 + 0.014 778 368;
  • 16) 0.014 778 368 × 2 = 0 + 0.029 556 736;
  • 17) 0.029 556 736 × 2 = 0 + 0.059 113 472;
  • 18) 0.059 113 472 × 2 = 0 + 0.118 226 944;
  • 19) 0.118 226 944 × 2 = 0 + 0.236 453 888;
  • 20) 0.236 453 888 × 2 = 0 + 0.472 907 776;
  • 21) 0.472 907 776 × 2 = 0 + 0.945 815 552;
  • 22) 0.945 815 552 × 2 = 1 + 0.891 631 104;
  • 23) 0.891 631 104 × 2 = 1 + 0.783 262 208;
  • 24) 0.783 262 208 × 2 = 1 + 0.566 524 416;
  • 25) 0.566 524 416 × 2 = 1 + 0.133 048 832;
  • 26) 0.133 048 832 × 2 = 0 + 0.266 097 664;
  • 27) 0.266 097 664 × 2 = 0 + 0.532 195 328;
  • 28) 0.532 195 328 × 2 = 1 + 0.064 390 656;
  • 29) 0.064 390 656 × 2 = 0 + 0.128 781 312;
  • 30) 0.128 781 312 × 2 = 0 + 0.257 562 624;
  • 31) 0.257 562 624 × 2 = 0 + 0.515 125 248;
  • 32) 0.515 125 248 × 2 = 1 + 0.030 250 496;
  • 33) 0.030 250 496 × 2 = 0 + 0.060 500 992;
  • 34) 0.060 500 992 × 2 = 0 + 0.121 001 984;
  • 35) 0.121 001 984 × 2 = 0 + 0.242 003 968;
  • 36) 0.242 003 968 × 2 = 0 + 0.484 007 936;
  • 37) 0.484 007 936 × 2 = 0 + 0.968 015 872;
  • 38) 0.968 015 872 × 2 = 1 + 0.936 031 744;
  • 39) 0.936 031 744 × 2 = 1 + 0.872 063 488;
  • 40) 0.872 063 488 × 2 = 1 + 0.744 126 976;
  • 41) 0.744 126 976 × 2 = 1 + 0.488 253 952;
  • 42) 0.488 253 952 × 2 = 0 + 0.976 507 904;
  • 43) 0.976 507 904 × 2 = 1 + 0.953 015 808;
  • 44) 0.953 015 808 × 2 = 1 + 0.906 031 616;
  • 45) 0.906 031 616 × 2 = 1 + 0.812 063 232;

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:


0.000 000 451(10) =

0.0000 0000 0000 0000 0000 0111 1001 0001 0000 0111 1011 1(2)

5. Positive number before normalization:

0.000 000 451(10) =

0.0000 0000 0000 0000 0000 0111 1001 0001 0000 0111 1011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 451(10) =

0.0000 0000 0000 0000 0000 0111 1001 0001 0000 0111 1011 1(2) =

0.0000 0000 0000 0000 0000 0111 1001 0001 0000 0111 1011 1(2) × 20 =

1.1110 0100 0100 0001 1110 111(2) × 2-22


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): -22

Mantissa (not normalized):
1.1110 0100 0100 0001 1110 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 52 ÷ 2 = 26 + 0;
  • 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) =

105(10) =

0110 1001(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. 111 0010 0010 0000 1111 0111 =

111 0010 0010 0000 1111 0111


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 1001

Mantissa (23 bits) =
111 0010 0010 0000 1111 0111


Decimal number 0.000 000 451 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 111 0010 0010 0000 1111 0111

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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