0.000 000 423 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 423(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
0.000 000 423(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: 0.000 000 423.

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) 0.000 000 423 × 2 = 0 + 0.000 000 846;
  • 2) 0.000 000 846 × 2 = 0 + 0.000 001 692;
  • 3) 0.000 001 692 × 2 = 0 + 0.000 003 384;
  • 4) 0.000 003 384 × 2 = 0 + 0.000 006 768;
  • 5) 0.000 006 768 × 2 = 0 + 0.000 013 536;
  • 6) 0.000 013 536 × 2 = 0 + 0.000 027 072;
  • 7) 0.000 027 072 × 2 = 0 + 0.000 054 144;
  • 8) 0.000 054 144 × 2 = 0 + 0.000 108 288;
  • 9) 0.000 108 288 × 2 = 0 + 0.000 216 576;
  • 10) 0.000 216 576 × 2 = 0 + 0.000 433 152;
  • 11) 0.000 433 152 × 2 = 0 + 0.000 866 304;
  • 12) 0.000 866 304 × 2 = 0 + 0.001 732 608;
  • 13) 0.001 732 608 × 2 = 0 + 0.003 465 216;
  • 14) 0.003 465 216 × 2 = 0 + 0.006 930 432;
  • 15) 0.006 930 432 × 2 = 0 + 0.013 860 864;
  • 16) 0.013 860 864 × 2 = 0 + 0.027 721 728;
  • 17) 0.027 721 728 × 2 = 0 + 0.055 443 456;
  • 18) 0.055 443 456 × 2 = 0 + 0.110 886 912;
  • 19) 0.110 886 912 × 2 = 0 + 0.221 773 824;
  • 20) 0.221 773 824 × 2 = 0 + 0.443 547 648;
  • 21) 0.443 547 648 × 2 = 0 + 0.887 095 296;
  • 22) 0.887 095 296 × 2 = 1 + 0.774 190 592;
  • 23) 0.774 190 592 × 2 = 1 + 0.548 381 184;
  • 24) 0.548 381 184 × 2 = 1 + 0.096 762 368;
  • 25) 0.096 762 368 × 2 = 0 + 0.193 524 736;
  • 26) 0.193 524 736 × 2 = 0 + 0.387 049 472;
  • 27) 0.387 049 472 × 2 = 0 + 0.774 098 944;
  • 28) 0.774 098 944 × 2 = 1 + 0.548 197 888;
  • 29) 0.548 197 888 × 2 = 1 + 0.096 395 776;
  • 30) 0.096 395 776 × 2 = 0 + 0.192 791 552;
  • 31) 0.192 791 552 × 2 = 0 + 0.385 583 104;
  • 32) 0.385 583 104 × 2 = 0 + 0.771 166 208;
  • 33) 0.771 166 208 × 2 = 1 + 0.542 332 416;
  • 34) 0.542 332 416 × 2 = 1 + 0.084 664 832;
  • 35) 0.084 664 832 × 2 = 0 + 0.169 329 664;
  • 36) 0.169 329 664 × 2 = 0 + 0.338 659 328;
  • 37) 0.338 659 328 × 2 = 0 + 0.677 318 656;
  • 38) 0.677 318 656 × 2 = 1 + 0.354 637 312;
  • 39) 0.354 637 312 × 2 = 0 + 0.709 274 624;
  • 40) 0.709 274 624 × 2 = 1 + 0.418 549 248;
  • 41) 0.418 549 248 × 2 = 0 + 0.837 098 496;
  • 42) 0.837 098 496 × 2 = 1 + 0.674 196 992;
  • 43) 0.674 196 992 × 2 = 1 + 0.348 393 984;
  • 44) 0.348 393 984 × 2 = 0 + 0.696 787 968;
  • 45) 0.696 787 968 × 2 = 1 + 0.393 575 936;

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:


0.000 000 423(10) =

0.0000 0000 0000 0000 0000 0111 0001 1000 1100 0101 0110 1(2)

5. Positive number before normalization:

0.000 000 423(10) =

0.0000 0000 0000 0000 0000 0111 0001 1000 1100 0101 0110 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:


0.000 000 423(10) =

0.0000 0000 0000 0000 0000 0111 0001 1000 1100 0101 0110 1(2) =

0.0000 0000 0000 0000 0000 0111 0001 1000 1100 0101 0110 1(2) × 20 =

1.1100 0110 0011 0001 0101 101(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):
1.1100 0110 0011 0001 0101 101


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. 110 0011 0001 1000 1010 1101 =

110 0011 0001 1000 1010 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 1001

Mantissa (23 bits) =
110 0011 0001 1000 1010 1101


Decimal number 0.000 000 423 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 110 0011 0001 1000 1010 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