0.000 000 262 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 262(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
0.000 000 262(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: 0.000 000 262.

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) 0.000 000 262 × 2 = 0 + 0.000 000 524;
  • 2) 0.000 000 524 × 2 = 0 + 0.000 001 048;
  • 3) 0.000 001 048 × 2 = 0 + 0.000 002 096;
  • 4) 0.000 002 096 × 2 = 0 + 0.000 004 192;
  • 5) 0.000 004 192 × 2 = 0 + 0.000 008 384;
  • 6) 0.000 008 384 × 2 = 0 + 0.000 016 768;
  • 7) 0.000 016 768 × 2 = 0 + 0.000 033 536;
  • 8) 0.000 033 536 × 2 = 0 + 0.000 067 072;
  • 9) 0.000 067 072 × 2 = 0 + 0.000 134 144;
  • 10) 0.000 134 144 × 2 = 0 + 0.000 268 288;
  • 11) 0.000 268 288 × 2 = 0 + 0.000 536 576;
  • 12) 0.000 536 576 × 2 = 0 + 0.001 073 152;
  • 13) 0.001 073 152 × 2 = 0 + 0.002 146 304;
  • 14) 0.002 146 304 × 2 = 0 + 0.004 292 608;
  • 15) 0.004 292 608 × 2 = 0 + 0.008 585 216;
  • 16) 0.008 585 216 × 2 = 0 + 0.017 170 432;
  • 17) 0.017 170 432 × 2 = 0 + 0.034 340 864;
  • 18) 0.034 340 864 × 2 = 0 + 0.068 681 728;
  • 19) 0.068 681 728 × 2 = 0 + 0.137 363 456;
  • 20) 0.137 363 456 × 2 = 0 + 0.274 726 912;
  • 21) 0.274 726 912 × 2 = 0 + 0.549 453 824;
  • 22) 0.549 453 824 × 2 = 1 + 0.098 907 648;
  • 23) 0.098 907 648 × 2 = 0 + 0.197 815 296;
  • 24) 0.197 815 296 × 2 = 0 + 0.395 630 592;
  • 25) 0.395 630 592 × 2 = 0 + 0.791 261 184;
  • 26) 0.791 261 184 × 2 = 1 + 0.582 522 368;
  • 27) 0.582 522 368 × 2 = 1 + 0.165 044 736;
  • 28) 0.165 044 736 × 2 = 0 + 0.330 089 472;
  • 29) 0.330 089 472 × 2 = 0 + 0.660 178 944;
  • 30) 0.660 178 944 × 2 = 1 + 0.320 357 888;
  • 31) 0.320 357 888 × 2 = 0 + 0.640 715 776;
  • 32) 0.640 715 776 × 2 = 1 + 0.281 431 552;
  • 33) 0.281 431 552 × 2 = 0 + 0.562 863 104;
  • 34) 0.562 863 104 × 2 = 1 + 0.125 726 208;
  • 35) 0.125 726 208 × 2 = 0 + 0.251 452 416;
  • 36) 0.251 452 416 × 2 = 0 + 0.502 904 832;
  • 37) 0.502 904 832 × 2 = 1 + 0.005 809 664;
  • 38) 0.005 809 664 × 2 = 0 + 0.011 619 328;
  • 39) 0.011 619 328 × 2 = 0 + 0.023 238 656;
  • 40) 0.023 238 656 × 2 = 0 + 0.046 477 312;
  • 41) 0.046 477 312 × 2 = 0 + 0.092 954 624;
  • 42) 0.092 954 624 × 2 = 0 + 0.185 909 248;
  • 43) 0.185 909 248 × 2 = 0 + 0.371 818 496;
  • 44) 0.371 818 496 × 2 = 0 + 0.743 636 992;
  • 45) 0.743 636 992 × 2 = 1 + 0.487 273 984;

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:


0.000 000 262(10) =

0.0000 0000 0000 0000 0000 0100 0110 0101 0100 1000 0000 1(2)

5. Positive number before normalization:

0.000 000 262(10) =

0.0000 0000 0000 0000 0000 0100 0110 0101 0100 1000 0000 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:


0.000 000 262(10) =

0.0000 0000 0000 0000 0000 0100 0110 0101 0100 1000 0000 1(2) =

0.0000 0000 0000 0000 0000 0100 0110 0101 0100 1000 0000 1(2) × 20 =

1.0001 1001 0101 0010 0000 001(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):
1.0001 1001 0101 0010 0000 001


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. 000 1100 1010 1001 0000 0001 =

000 1100 1010 1001 0000 0001


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) =
000 1100 1010 1001 0000 0001


Decimal number 0.000 000 262 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 000 1100 1010 1001 0000 0001

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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