0.000 000 439 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 439(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
0.000 000 439(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: 0.000 000 439.

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) 0.000 000 439 × 2 = 0 + 0.000 000 878;
  • 2) 0.000 000 878 × 2 = 0 + 0.000 001 756;
  • 3) 0.000 001 756 × 2 = 0 + 0.000 003 512;
  • 4) 0.000 003 512 × 2 = 0 + 0.000 007 024;
  • 5) 0.000 007 024 × 2 = 0 + 0.000 014 048;
  • 6) 0.000 014 048 × 2 = 0 + 0.000 028 096;
  • 7) 0.000 028 096 × 2 = 0 + 0.000 056 192;
  • 8) 0.000 056 192 × 2 = 0 + 0.000 112 384;
  • 9) 0.000 112 384 × 2 = 0 + 0.000 224 768;
  • 10) 0.000 224 768 × 2 = 0 + 0.000 449 536;
  • 11) 0.000 449 536 × 2 = 0 + 0.000 899 072;
  • 12) 0.000 899 072 × 2 = 0 + 0.001 798 144;
  • 13) 0.001 798 144 × 2 = 0 + 0.003 596 288;
  • 14) 0.003 596 288 × 2 = 0 + 0.007 192 576;
  • 15) 0.007 192 576 × 2 = 0 + 0.014 385 152;
  • 16) 0.014 385 152 × 2 = 0 + 0.028 770 304;
  • 17) 0.028 770 304 × 2 = 0 + 0.057 540 608;
  • 18) 0.057 540 608 × 2 = 0 + 0.115 081 216;
  • 19) 0.115 081 216 × 2 = 0 + 0.230 162 432;
  • 20) 0.230 162 432 × 2 = 0 + 0.460 324 864;
  • 21) 0.460 324 864 × 2 = 0 + 0.920 649 728;
  • 22) 0.920 649 728 × 2 = 1 + 0.841 299 456;
  • 23) 0.841 299 456 × 2 = 1 + 0.682 598 912;
  • 24) 0.682 598 912 × 2 = 1 + 0.365 197 824;
  • 25) 0.365 197 824 × 2 = 0 + 0.730 395 648;
  • 26) 0.730 395 648 × 2 = 1 + 0.460 791 296;
  • 27) 0.460 791 296 × 2 = 0 + 0.921 582 592;
  • 28) 0.921 582 592 × 2 = 1 + 0.843 165 184;
  • 29) 0.843 165 184 × 2 = 1 + 0.686 330 368;
  • 30) 0.686 330 368 × 2 = 1 + 0.372 660 736;
  • 31) 0.372 660 736 × 2 = 0 + 0.745 321 472;
  • 32) 0.745 321 472 × 2 = 1 + 0.490 642 944;
  • 33) 0.490 642 944 × 2 = 0 + 0.981 285 888;
  • 34) 0.981 285 888 × 2 = 1 + 0.962 571 776;
  • 35) 0.962 571 776 × 2 = 1 + 0.925 143 552;
  • 36) 0.925 143 552 × 2 = 1 + 0.850 287 104;
  • 37) 0.850 287 104 × 2 = 1 + 0.700 574 208;
  • 38) 0.700 574 208 × 2 = 1 + 0.401 148 416;
  • 39) 0.401 148 416 × 2 = 0 + 0.802 296 832;
  • 40) 0.802 296 832 × 2 = 1 + 0.604 593 664;
  • 41) 0.604 593 664 × 2 = 1 + 0.209 187 328;
  • 42) 0.209 187 328 × 2 = 0 + 0.418 374 656;
  • 43) 0.418 374 656 × 2 = 0 + 0.836 749 312;
  • 44) 0.836 749 312 × 2 = 1 + 0.673 498 624;
  • 45) 0.673 498 624 × 2 = 1 + 0.346 997 248;

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:


0.000 000 439(10) =

0.0000 0000 0000 0000 0000 0111 0101 1101 0111 1101 1001 1(2)

5. Positive number before normalization:

0.000 000 439(10) =

0.0000 0000 0000 0000 0000 0111 0101 1101 0111 1101 1001 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:


0.000 000 439(10) =

0.0000 0000 0000 0000 0000 0111 0101 1101 0111 1101 1001 1(2) =

0.0000 0000 0000 0000 0000 0111 0101 1101 0111 1101 1001 1(2) × 20 =

1.1101 0111 0101 1111 0110 011(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):
1.1101 0111 0101 1111 0110 011


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 1011 1010 1111 1011 0011 =

110 1011 1010 1111 1011 0011


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 1011 1010 1111 1011 0011


Decimal number 0.000 000 439 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 110 1011 1010 1111 1011 0011

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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