0.000 000 139 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 139(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
0.000 000 139(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: 0.000 000 139.

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) 0.000 000 139 × 2 = 0 + 0.000 000 278;
  • 2) 0.000 000 278 × 2 = 0 + 0.000 000 556;
  • 3) 0.000 000 556 × 2 = 0 + 0.000 001 112;
  • 4) 0.000 001 112 × 2 = 0 + 0.000 002 224;
  • 5) 0.000 002 224 × 2 = 0 + 0.000 004 448;
  • 6) 0.000 004 448 × 2 = 0 + 0.000 008 896;
  • 7) 0.000 008 896 × 2 = 0 + 0.000 017 792;
  • 8) 0.000 017 792 × 2 = 0 + 0.000 035 584;
  • 9) 0.000 035 584 × 2 = 0 + 0.000 071 168;
  • 10) 0.000 071 168 × 2 = 0 + 0.000 142 336;
  • 11) 0.000 142 336 × 2 = 0 + 0.000 284 672;
  • 12) 0.000 284 672 × 2 = 0 + 0.000 569 344;
  • 13) 0.000 569 344 × 2 = 0 + 0.001 138 688;
  • 14) 0.001 138 688 × 2 = 0 + 0.002 277 376;
  • 15) 0.002 277 376 × 2 = 0 + 0.004 554 752;
  • 16) 0.004 554 752 × 2 = 0 + 0.009 109 504;
  • 17) 0.009 109 504 × 2 = 0 + 0.018 219 008;
  • 18) 0.018 219 008 × 2 = 0 + 0.036 438 016;
  • 19) 0.036 438 016 × 2 = 0 + 0.072 876 032;
  • 20) 0.072 876 032 × 2 = 0 + 0.145 752 064;
  • 21) 0.145 752 064 × 2 = 0 + 0.291 504 128;
  • 22) 0.291 504 128 × 2 = 0 + 0.583 008 256;
  • 23) 0.583 008 256 × 2 = 1 + 0.166 016 512;
  • 24) 0.166 016 512 × 2 = 0 + 0.332 033 024;
  • 25) 0.332 033 024 × 2 = 0 + 0.664 066 048;
  • 26) 0.664 066 048 × 2 = 1 + 0.328 132 096;
  • 27) 0.328 132 096 × 2 = 0 + 0.656 264 192;
  • 28) 0.656 264 192 × 2 = 1 + 0.312 528 384;
  • 29) 0.312 528 384 × 2 = 0 + 0.625 056 768;
  • 30) 0.625 056 768 × 2 = 1 + 0.250 113 536;
  • 31) 0.250 113 536 × 2 = 0 + 0.500 227 072;
  • 32) 0.500 227 072 × 2 = 1 + 0.000 454 144;
  • 33) 0.000 454 144 × 2 = 0 + 0.000 908 288;
  • 34) 0.000 908 288 × 2 = 0 + 0.001 816 576;
  • 35) 0.001 816 576 × 2 = 0 + 0.003 633 152;
  • 36) 0.003 633 152 × 2 = 0 + 0.007 266 304;
  • 37) 0.007 266 304 × 2 = 0 + 0.014 532 608;
  • 38) 0.014 532 608 × 2 = 0 + 0.029 065 216;
  • 39) 0.029 065 216 × 2 = 0 + 0.058 130 432;
  • 40) 0.058 130 432 × 2 = 0 + 0.116 260 864;
  • 41) 0.116 260 864 × 2 = 0 + 0.232 521 728;
  • 42) 0.232 521 728 × 2 = 0 + 0.465 043 456;
  • 43) 0.465 043 456 × 2 = 0 + 0.930 086 912;
  • 44) 0.930 086 912 × 2 = 1 + 0.860 173 824;
  • 45) 0.860 173 824 × 2 = 1 + 0.720 347 648;
  • 46) 0.720 347 648 × 2 = 1 + 0.440 695 296;

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:


0.000 000 139(10) =

0.0000 0000 0000 0000 0000 0010 0101 0101 0000 0000 0001 11(2)

5. Positive number before normalization:

0.000 000 139(10) =

0.0000 0000 0000 0000 0000 0010 0101 0101 0000 0000 0001 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 139(10) =

0.0000 0000 0000 0000 0000 0010 0101 0101 0000 0000 0001 11(2) =

0.0000 0000 0000 0000 0000 0010 0101 0101 0000 0000 0001 11(2) × 20 =

1.0010 1010 1000 0000 0000 111(2) × 2-23


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): -23

Mantissa (not normalized):
1.0010 1010 1000 0000 0000 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(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;
  • 104 ÷ 2 = 52 + 0;
  • 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) =

104(10) =

0110 1000(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 0101 0100 0000 0000 0111 =

001 0101 0100 0000 0000 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 1000

Mantissa (23 bits) =
001 0101 0100 0000 0000 0111


Decimal number 0.000 000 139 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 001 0101 0100 0000 0000 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