0.000 000 277 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 277(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
0.000 000 277(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: 0.000 000 277.

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) 0.000 000 277 × 2 = 0 + 0.000 000 554;
  • 2) 0.000 000 554 × 2 = 0 + 0.000 001 108;
  • 3) 0.000 001 108 × 2 = 0 + 0.000 002 216;
  • 4) 0.000 002 216 × 2 = 0 + 0.000 004 432;
  • 5) 0.000 004 432 × 2 = 0 + 0.000 008 864;
  • 6) 0.000 008 864 × 2 = 0 + 0.000 017 728;
  • 7) 0.000 017 728 × 2 = 0 + 0.000 035 456;
  • 8) 0.000 035 456 × 2 = 0 + 0.000 070 912;
  • 9) 0.000 070 912 × 2 = 0 + 0.000 141 824;
  • 10) 0.000 141 824 × 2 = 0 + 0.000 283 648;
  • 11) 0.000 283 648 × 2 = 0 + 0.000 567 296;
  • 12) 0.000 567 296 × 2 = 0 + 0.001 134 592;
  • 13) 0.001 134 592 × 2 = 0 + 0.002 269 184;
  • 14) 0.002 269 184 × 2 = 0 + 0.004 538 368;
  • 15) 0.004 538 368 × 2 = 0 + 0.009 076 736;
  • 16) 0.009 076 736 × 2 = 0 + 0.018 153 472;
  • 17) 0.018 153 472 × 2 = 0 + 0.036 306 944;
  • 18) 0.036 306 944 × 2 = 0 + 0.072 613 888;
  • 19) 0.072 613 888 × 2 = 0 + 0.145 227 776;
  • 20) 0.145 227 776 × 2 = 0 + 0.290 455 552;
  • 21) 0.290 455 552 × 2 = 0 + 0.580 911 104;
  • 22) 0.580 911 104 × 2 = 1 + 0.161 822 208;
  • 23) 0.161 822 208 × 2 = 0 + 0.323 644 416;
  • 24) 0.323 644 416 × 2 = 0 + 0.647 288 832;
  • 25) 0.647 288 832 × 2 = 1 + 0.294 577 664;
  • 26) 0.294 577 664 × 2 = 0 + 0.589 155 328;
  • 27) 0.589 155 328 × 2 = 1 + 0.178 310 656;
  • 28) 0.178 310 656 × 2 = 0 + 0.356 621 312;
  • 29) 0.356 621 312 × 2 = 0 + 0.713 242 624;
  • 30) 0.713 242 624 × 2 = 1 + 0.426 485 248;
  • 31) 0.426 485 248 × 2 = 0 + 0.852 970 496;
  • 32) 0.852 970 496 × 2 = 1 + 0.705 940 992;
  • 33) 0.705 940 992 × 2 = 1 + 0.411 881 984;
  • 34) 0.411 881 984 × 2 = 0 + 0.823 763 968;
  • 35) 0.823 763 968 × 2 = 1 + 0.647 527 936;
  • 36) 0.647 527 936 × 2 = 1 + 0.295 055 872;
  • 37) 0.295 055 872 × 2 = 0 + 0.590 111 744;
  • 38) 0.590 111 744 × 2 = 1 + 0.180 223 488;
  • 39) 0.180 223 488 × 2 = 0 + 0.360 446 976;
  • 40) 0.360 446 976 × 2 = 0 + 0.720 893 952;
  • 41) 0.720 893 952 × 2 = 1 + 0.441 787 904;
  • 42) 0.441 787 904 × 2 = 0 + 0.883 575 808;
  • 43) 0.883 575 808 × 2 = 1 + 0.767 151 616;
  • 44) 0.767 151 616 × 2 = 1 + 0.534 303 232;
  • 45) 0.534 303 232 × 2 = 1 + 0.068 606 464;

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:


0.000 000 277(10) =

0.0000 0000 0000 0000 0000 0100 1010 0101 1011 0100 1011 1(2)

5. Positive number before normalization:

0.000 000 277(10) =

0.0000 0000 0000 0000 0000 0100 1010 0101 1011 0100 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:


0.000 000 277(10) =

0.0000 0000 0000 0000 0000 0100 1010 0101 1011 0100 1011 1(2) =

0.0000 0000 0000 0000 0000 0100 1010 0101 1011 0100 1011 1(2) × 20 =

1.0010 1001 0110 1101 0010 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):
1.0010 1001 0110 1101 0010 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. 001 0100 1011 0110 1001 0111 =

001 0100 1011 0110 1001 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) =
001 0100 1011 0110 1001 0111


Decimal number 0.000 000 277 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 001 0100 1011 0110 1001 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