0.000 000 487 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 487(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
0.000 000 487(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: 0.000 000 487.

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) 0.000 000 487 × 2 = 0 + 0.000 000 974;
  • 2) 0.000 000 974 × 2 = 0 + 0.000 001 948;
  • 3) 0.000 001 948 × 2 = 0 + 0.000 003 896;
  • 4) 0.000 003 896 × 2 = 0 + 0.000 007 792;
  • 5) 0.000 007 792 × 2 = 0 + 0.000 015 584;
  • 6) 0.000 015 584 × 2 = 0 + 0.000 031 168;
  • 7) 0.000 031 168 × 2 = 0 + 0.000 062 336;
  • 8) 0.000 062 336 × 2 = 0 + 0.000 124 672;
  • 9) 0.000 124 672 × 2 = 0 + 0.000 249 344;
  • 10) 0.000 249 344 × 2 = 0 + 0.000 498 688;
  • 11) 0.000 498 688 × 2 = 0 + 0.000 997 376;
  • 12) 0.000 997 376 × 2 = 0 + 0.001 994 752;
  • 13) 0.001 994 752 × 2 = 0 + 0.003 989 504;
  • 14) 0.003 989 504 × 2 = 0 + 0.007 979 008;
  • 15) 0.007 979 008 × 2 = 0 + 0.015 958 016;
  • 16) 0.015 958 016 × 2 = 0 + 0.031 916 032;
  • 17) 0.031 916 032 × 2 = 0 + 0.063 832 064;
  • 18) 0.063 832 064 × 2 = 0 + 0.127 664 128;
  • 19) 0.127 664 128 × 2 = 0 + 0.255 328 256;
  • 20) 0.255 328 256 × 2 = 0 + 0.510 656 512;
  • 21) 0.510 656 512 × 2 = 1 + 0.021 313 024;
  • 22) 0.021 313 024 × 2 = 0 + 0.042 626 048;
  • 23) 0.042 626 048 × 2 = 0 + 0.085 252 096;
  • 24) 0.085 252 096 × 2 = 0 + 0.170 504 192;
  • 25) 0.170 504 192 × 2 = 0 + 0.341 008 384;
  • 26) 0.341 008 384 × 2 = 0 + 0.682 016 768;
  • 27) 0.682 016 768 × 2 = 1 + 0.364 033 536;
  • 28) 0.364 033 536 × 2 = 0 + 0.728 067 072;
  • 29) 0.728 067 072 × 2 = 1 + 0.456 134 144;
  • 30) 0.456 134 144 × 2 = 0 + 0.912 268 288;
  • 31) 0.912 268 288 × 2 = 1 + 0.824 536 576;
  • 32) 0.824 536 576 × 2 = 1 + 0.649 073 152;
  • 33) 0.649 073 152 × 2 = 1 + 0.298 146 304;
  • 34) 0.298 146 304 × 2 = 0 + 0.596 292 608;
  • 35) 0.596 292 608 × 2 = 1 + 0.192 585 216;
  • 36) 0.192 585 216 × 2 = 0 + 0.385 170 432;
  • 37) 0.385 170 432 × 2 = 0 + 0.770 340 864;
  • 38) 0.770 340 864 × 2 = 1 + 0.540 681 728;
  • 39) 0.540 681 728 × 2 = 1 + 0.081 363 456;
  • 40) 0.081 363 456 × 2 = 0 + 0.162 726 912;
  • 41) 0.162 726 912 × 2 = 0 + 0.325 453 824;
  • 42) 0.325 453 824 × 2 = 0 + 0.650 907 648;
  • 43) 0.650 907 648 × 2 = 1 + 0.301 815 296;
  • 44) 0.301 815 296 × 2 = 0 + 0.603 630 592;

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:


0.000 000 487(10) =

0.0000 0000 0000 0000 0000 1000 0010 1011 1010 0110 0010(2)

5. Positive number before normalization:

0.000 000 487(10) =

0.0000 0000 0000 0000 0000 1000 0010 1011 1010 0110 0010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 487(10) =

0.0000 0000 0000 0000 0000 1000 0010 1011 1010 0110 0010(2) =

0.0000 0000 0000 0000 0000 1000 0010 1011 1010 0110 0010(2) × 20 =

1.0000 0101 0111 0100 1100 010(2) × 2-21


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): -21

Mantissa (not normalized):
1.0000 0101 0111 0100 1100 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 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) =

106(10) =

0110 1010(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 0010 1011 1010 0110 0010 =

000 0010 1011 1010 0110 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 1010

Mantissa (23 bits) =
000 0010 1011 1010 0110 0010


Decimal number 0.000 000 487 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 000 0010 1011 1010 0110 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