0.000 000 077 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 077(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
0.000 000 077(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: 0.000 000 077.

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) 0.000 000 077 × 2 = 0 + 0.000 000 154;
  • 2) 0.000 000 154 × 2 = 0 + 0.000 000 308;
  • 3) 0.000 000 308 × 2 = 0 + 0.000 000 616;
  • 4) 0.000 000 616 × 2 = 0 + 0.000 001 232;
  • 5) 0.000 001 232 × 2 = 0 + 0.000 002 464;
  • 6) 0.000 002 464 × 2 = 0 + 0.000 004 928;
  • 7) 0.000 004 928 × 2 = 0 + 0.000 009 856;
  • 8) 0.000 009 856 × 2 = 0 + 0.000 019 712;
  • 9) 0.000 019 712 × 2 = 0 + 0.000 039 424;
  • 10) 0.000 039 424 × 2 = 0 + 0.000 078 848;
  • 11) 0.000 078 848 × 2 = 0 + 0.000 157 696;
  • 12) 0.000 157 696 × 2 = 0 + 0.000 315 392;
  • 13) 0.000 315 392 × 2 = 0 + 0.000 630 784;
  • 14) 0.000 630 784 × 2 = 0 + 0.001 261 568;
  • 15) 0.001 261 568 × 2 = 0 + 0.002 523 136;
  • 16) 0.002 523 136 × 2 = 0 + 0.005 046 272;
  • 17) 0.005 046 272 × 2 = 0 + 0.010 092 544;
  • 18) 0.010 092 544 × 2 = 0 + 0.020 185 088;
  • 19) 0.020 185 088 × 2 = 0 + 0.040 370 176;
  • 20) 0.040 370 176 × 2 = 0 + 0.080 740 352;
  • 21) 0.080 740 352 × 2 = 0 + 0.161 480 704;
  • 22) 0.161 480 704 × 2 = 0 + 0.322 961 408;
  • 23) 0.322 961 408 × 2 = 0 + 0.645 922 816;
  • 24) 0.645 922 816 × 2 = 1 + 0.291 845 632;
  • 25) 0.291 845 632 × 2 = 0 + 0.583 691 264;
  • 26) 0.583 691 264 × 2 = 1 + 0.167 382 528;
  • 27) 0.167 382 528 × 2 = 0 + 0.334 765 056;
  • 28) 0.334 765 056 × 2 = 0 + 0.669 530 112;
  • 29) 0.669 530 112 × 2 = 1 + 0.339 060 224;
  • 30) 0.339 060 224 × 2 = 0 + 0.678 120 448;
  • 31) 0.678 120 448 × 2 = 1 + 0.356 240 896;
  • 32) 0.356 240 896 × 2 = 0 + 0.712 481 792;
  • 33) 0.712 481 792 × 2 = 1 + 0.424 963 584;
  • 34) 0.424 963 584 × 2 = 0 + 0.849 927 168;
  • 35) 0.849 927 168 × 2 = 1 + 0.699 854 336;
  • 36) 0.699 854 336 × 2 = 1 + 0.399 708 672;
  • 37) 0.399 708 672 × 2 = 0 + 0.799 417 344;
  • 38) 0.799 417 344 × 2 = 1 + 0.598 834 688;
  • 39) 0.598 834 688 × 2 = 1 + 0.197 669 376;
  • 40) 0.197 669 376 × 2 = 0 + 0.395 338 752;
  • 41) 0.395 338 752 × 2 = 0 + 0.790 677 504;
  • 42) 0.790 677 504 × 2 = 1 + 0.581 355 008;
  • 43) 0.581 355 008 × 2 = 1 + 0.162 710 016;
  • 44) 0.162 710 016 × 2 = 0 + 0.325 420 032;
  • 45) 0.325 420 032 × 2 = 0 + 0.650 840 064;
  • 46) 0.650 840 064 × 2 = 1 + 0.301 680 128;
  • 47) 0.301 680 128 × 2 = 0 + 0.603 360 256;

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:


0.000 000 077(10) =

0.0000 0000 0000 0000 0000 0001 0100 1010 1011 0110 0110 010(2)

5. Positive number before normalization:

0.000 000 077(10) =

0.0000 0000 0000 0000 0000 0001 0100 1010 1011 0110 0110 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 077(10) =

0.0000 0000 0000 0000 0000 0001 0100 1010 1011 0110 0110 010(2) =

0.0000 0000 0000 0000 0000 0001 0100 1010 1011 0110 0110 010(2) × 20 =

1.0100 1010 1011 0110 0110 010(2) × 2-24


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): -24

Mantissa (not normalized):
1.0100 1010 1011 0110 0110 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

103(10) =

0110 0111(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. 010 0101 0101 1011 0011 0010 =

010 0101 0101 1011 0011 0010


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 0111

Mantissa (23 bits) =
010 0101 0101 1011 0011 0010


Decimal number 0.000 000 077 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 010 0101 0101 1011 0011 0010

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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