0.000 000 341 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 341(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
0.000 000 341(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: 0.000 000 341.

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) 0.000 000 341 × 2 = 0 + 0.000 000 682;
  • 2) 0.000 000 682 × 2 = 0 + 0.000 001 364;
  • 3) 0.000 001 364 × 2 = 0 + 0.000 002 728;
  • 4) 0.000 002 728 × 2 = 0 + 0.000 005 456;
  • 5) 0.000 005 456 × 2 = 0 + 0.000 010 912;
  • 6) 0.000 010 912 × 2 = 0 + 0.000 021 824;
  • 7) 0.000 021 824 × 2 = 0 + 0.000 043 648;
  • 8) 0.000 043 648 × 2 = 0 + 0.000 087 296;
  • 9) 0.000 087 296 × 2 = 0 + 0.000 174 592;
  • 10) 0.000 174 592 × 2 = 0 + 0.000 349 184;
  • 11) 0.000 349 184 × 2 = 0 + 0.000 698 368;
  • 12) 0.000 698 368 × 2 = 0 + 0.001 396 736;
  • 13) 0.001 396 736 × 2 = 0 + 0.002 793 472;
  • 14) 0.002 793 472 × 2 = 0 + 0.005 586 944;
  • 15) 0.005 586 944 × 2 = 0 + 0.011 173 888;
  • 16) 0.011 173 888 × 2 = 0 + 0.022 347 776;
  • 17) 0.022 347 776 × 2 = 0 + 0.044 695 552;
  • 18) 0.044 695 552 × 2 = 0 + 0.089 391 104;
  • 19) 0.089 391 104 × 2 = 0 + 0.178 782 208;
  • 20) 0.178 782 208 × 2 = 0 + 0.357 564 416;
  • 21) 0.357 564 416 × 2 = 0 + 0.715 128 832;
  • 22) 0.715 128 832 × 2 = 1 + 0.430 257 664;
  • 23) 0.430 257 664 × 2 = 0 + 0.860 515 328;
  • 24) 0.860 515 328 × 2 = 1 + 0.721 030 656;
  • 25) 0.721 030 656 × 2 = 1 + 0.442 061 312;
  • 26) 0.442 061 312 × 2 = 0 + 0.884 122 624;
  • 27) 0.884 122 624 × 2 = 1 + 0.768 245 248;
  • 28) 0.768 245 248 × 2 = 1 + 0.536 490 496;
  • 29) 0.536 490 496 × 2 = 1 + 0.072 980 992;
  • 30) 0.072 980 992 × 2 = 0 + 0.145 961 984;
  • 31) 0.145 961 984 × 2 = 0 + 0.291 923 968;
  • 32) 0.291 923 968 × 2 = 0 + 0.583 847 936;
  • 33) 0.583 847 936 × 2 = 1 + 0.167 695 872;
  • 34) 0.167 695 872 × 2 = 0 + 0.335 391 744;
  • 35) 0.335 391 744 × 2 = 0 + 0.670 783 488;
  • 36) 0.670 783 488 × 2 = 1 + 0.341 566 976;
  • 37) 0.341 566 976 × 2 = 0 + 0.683 133 952;
  • 38) 0.683 133 952 × 2 = 1 + 0.366 267 904;
  • 39) 0.366 267 904 × 2 = 0 + 0.732 535 808;
  • 40) 0.732 535 808 × 2 = 1 + 0.465 071 616;
  • 41) 0.465 071 616 × 2 = 0 + 0.930 143 232;
  • 42) 0.930 143 232 × 2 = 1 + 0.860 286 464;
  • 43) 0.860 286 464 × 2 = 1 + 0.720 572 928;
  • 44) 0.720 572 928 × 2 = 1 + 0.441 145 856;
  • 45) 0.441 145 856 × 2 = 0 + 0.882 291 712;

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:


0.000 000 341(10) =

0.0000 0000 0000 0000 0000 0101 1011 1000 1001 0101 0111 0(2)

5. Positive number before normalization:

0.000 000 341(10) =

0.0000 0000 0000 0000 0000 0101 1011 1000 1001 0101 0111 0(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:


0.000 000 341(10) =

0.0000 0000 0000 0000 0000 0101 1011 1000 1001 0101 0111 0(2) =

0.0000 0000 0000 0000 0000 0101 1011 1000 1001 0101 0111 0(2) × 20 =

1.0110 1110 0010 0101 0101 110(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):
1.0110 1110 0010 0101 0101 110


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. 011 0111 0001 0010 1010 1110 =

011 0111 0001 0010 1010 1110


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) =
011 0111 0001 0010 1010 1110


Decimal number 0.000 000 341 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 011 0111 0001 0010 1010 1110

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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