0.000 000 137 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 137(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
0.000 000 137(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: 0.000 000 137.

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) 0.000 000 137 × 2 = 0 + 0.000 000 274;
  • 2) 0.000 000 274 × 2 = 0 + 0.000 000 548;
  • 3) 0.000 000 548 × 2 = 0 + 0.000 001 096;
  • 4) 0.000 001 096 × 2 = 0 + 0.000 002 192;
  • 5) 0.000 002 192 × 2 = 0 + 0.000 004 384;
  • 6) 0.000 004 384 × 2 = 0 + 0.000 008 768;
  • 7) 0.000 008 768 × 2 = 0 + 0.000 017 536;
  • 8) 0.000 017 536 × 2 = 0 + 0.000 035 072;
  • 9) 0.000 035 072 × 2 = 0 + 0.000 070 144;
  • 10) 0.000 070 144 × 2 = 0 + 0.000 140 288;
  • 11) 0.000 140 288 × 2 = 0 + 0.000 280 576;
  • 12) 0.000 280 576 × 2 = 0 + 0.000 561 152;
  • 13) 0.000 561 152 × 2 = 0 + 0.001 122 304;
  • 14) 0.001 122 304 × 2 = 0 + 0.002 244 608;
  • 15) 0.002 244 608 × 2 = 0 + 0.004 489 216;
  • 16) 0.004 489 216 × 2 = 0 + 0.008 978 432;
  • 17) 0.008 978 432 × 2 = 0 + 0.017 956 864;
  • 18) 0.017 956 864 × 2 = 0 + 0.035 913 728;
  • 19) 0.035 913 728 × 2 = 0 + 0.071 827 456;
  • 20) 0.071 827 456 × 2 = 0 + 0.143 654 912;
  • 21) 0.143 654 912 × 2 = 0 + 0.287 309 824;
  • 22) 0.287 309 824 × 2 = 0 + 0.574 619 648;
  • 23) 0.574 619 648 × 2 = 1 + 0.149 239 296;
  • 24) 0.149 239 296 × 2 = 0 + 0.298 478 592;
  • 25) 0.298 478 592 × 2 = 0 + 0.596 957 184;
  • 26) 0.596 957 184 × 2 = 1 + 0.193 914 368;
  • 27) 0.193 914 368 × 2 = 0 + 0.387 828 736;
  • 28) 0.387 828 736 × 2 = 0 + 0.775 657 472;
  • 29) 0.775 657 472 × 2 = 1 + 0.551 314 944;
  • 30) 0.551 314 944 × 2 = 1 + 0.102 629 888;
  • 31) 0.102 629 888 × 2 = 0 + 0.205 259 776;
  • 32) 0.205 259 776 × 2 = 0 + 0.410 519 552;
  • 33) 0.410 519 552 × 2 = 0 + 0.821 039 104;
  • 34) 0.821 039 104 × 2 = 1 + 0.642 078 208;
  • 35) 0.642 078 208 × 2 = 1 + 0.284 156 416;
  • 36) 0.284 156 416 × 2 = 0 + 0.568 312 832;
  • 37) 0.568 312 832 × 2 = 1 + 0.136 625 664;
  • 38) 0.136 625 664 × 2 = 0 + 0.273 251 328;
  • 39) 0.273 251 328 × 2 = 0 + 0.546 502 656;
  • 40) 0.546 502 656 × 2 = 1 + 0.093 005 312;
  • 41) 0.093 005 312 × 2 = 0 + 0.186 010 624;
  • 42) 0.186 010 624 × 2 = 0 + 0.372 021 248;
  • 43) 0.372 021 248 × 2 = 0 + 0.744 042 496;
  • 44) 0.744 042 496 × 2 = 1 + 0.488 084 992;
  • 45) 0.488 084 992 × 2 = 0 + 0.976 169 984;
  • 46) 0.976 169 984 × 2 = 1 + 0.952 339 968;

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:


0.000 000 137(10) =

0.0000 0000 0000 0000 0000 0010 0100 1100 0110 1001 0001 01(2)

5. Positive number before normalization:

0.000 000 137(10) =

0.0000 0000 0000 0000 0000 0010 0100 1100 0110 1001 0001 01(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:


0.000 000 137(10) =

0.0000 0000 0000 0000 0000 0010 0100 1100 0110 1001 0001 01(2) =

0.0000 0000 0000 0000 0000 0010 0100 1100 0110 1001 0001 01(2) × 20 =

1.0010 0110 0011 0100 1000 101(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 0110 0011 0100 1000 101


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 0011 0001 1010 0100 0101 =

001 0011 0001 1010 0100 0101


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 0011 0001 1010 0100 0101


Decimal number 0.000 000 137 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 001 0011 0001 1010 0100 0101

Share

» Convert decimal 0.000 000 15 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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