0.000 000 372 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 372(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
0.000 000 372(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: 0.000 000 372.

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) 0.000 000 372 × 2 = 0 + 0.000 000 744;
  • 2) 0.000 000 744 × 2 = 0 + 0.000 001 488;
  • 3) 0.000 001 488 × 2 = 0 + 0.000 002 976;
  • 4) 0.000 002 976 × 2 = 0 + 0.000 005 952;
  • 5) 0.000 005 952 × 2 = 0 + 0.000 011 904;
  • 6) 0.000 011 904 × 2 = 0 + 0.000 023 808;
  • 7) 0.000 023 808 × 2 = 0 + 0.000 047 616;
  • 8) 0.000 047 616 × 2 = 0 + 0.000 095 232;
  • 9) 0.000 095 232 × 2 = 0 + 0.000 190 464;
  • 10) 0.000 190 464 × 2 = 0 + 0.000 380 928;
  • 11) 0.000 380 928 × 2 = 0 + 0.000 761 856;
  • 12) 0.000 761 856 × 2 = 0 + 0.001 523 712;
  • 13) 0.001 523 712 × 2 = 0 + 0.003 047 424;
  • 14) 0.003 047 424 × 2 = 0 + 0.006 094 848;
  • 15) 0.006 094 848 × 2 = 0 + 0.012 189 696;
  • 16) 0.012 189 696 × 2 = 0 + 0.024 379 392;
  • 17) 0.024 379 392 × 2 = 0 + 0.048 758 784;
  • 18) 0.048 758 784 × 2 = 0 + 0.097 517 568;
  • 19) 0.097 517 568 × 2 = 0 + 0.195 035 136;
  • 20) 0.195 035 136 × 2 = 0 + 0.390 070 272;
  • 21) 0.390 070 272 × 2 = 0 + 0.780 140 544;
  • 22) 0.780 140 544 × 2 = 1 + 0.560 281 088;
  • 23) 0.560 281 088 × 2 = 1 + 0.120 562 176;
  • 24) 0.120 562 176 × 2 = 0 + 0.241 124 352;
  • 25) 0.241 124 352 × 2 = 0 + 0.482 248 704;
  • 26) 0.482 248 704 × 2 = 0 + 0.964 497 408;
  • 27) 0.964 497 408 × 2 = 1 + 0.928 994 816;
  • 28) 0.928 994 816 × 2 = 1 + 0.857 989 632;
  • 29) 0.857 989 632 × 2 = 1 + 0.715 979 264;
  • 30) 0.715 979 264 × 2 = 1 + 0.431 958 528;
  • 31) 0.431 958 528 × 2 = 0 + 0.863 917 056;
  • 32) 0.863 917 056 × 2 = 1 + 0.727 834 112;
  • 33) 0.727 834 112 × 2 = 1 + 0.455 668 224;
  • 34) 0.455 668 224 × 2 = 0 + 0.911 336 448;
  • 35) 0.911 336 448 × 2 = 1 + 0.822 672 896;
  • 36) 0.822 672 896 × 2 = 1 + 0.645 345 792;
  • 37) 0.645 345 792 × 2 = 1 + 0.290 691 584;
  • 38) 0.290 691 584 × 2 = 0 + 0.581 383 168;
  • 39) 0.581 383 168 × 2 = 1 + 0.162 766 336;
  • 40) 0.162 766 336 × 2 = 0 + 0.325 532 672;
  • 41) 0.325 532 672 × 2 = 0 + 0.651 065 344;
  • 42) 0.651 065 344 × 2 = 1 + 0.302 130 688;
  • 43) 0.302 130 688 × 2 = 0 + 0.604 261 376;
  • 44) 0.604 261 376 × 2 = 1 + 0.208 522 752;
  • 45) 0.208 522 752 × 2 = 0 + 0.417 045 504;

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:


0.000 000 372(10) =

0.0000 0000 0000 0000 0000 0110 0011 1101 1011 1010 0101 0(2)

5. Positive number before normalization:

0.000 000 372(10) =

0.0000 0000 0000 0000 0000 0110 0011 1101 1011 1010 0101 0(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:


0.000 000 372(10) =

0.0000 0000 0000 0000 0000 0110 0011 1101 1011 1010 0101 0(2) =

0.0000 0000 0000 0000 0000 0110 0011 1101 1011 1010 0101 0(2) × 20 =

1.1000 1111 0110 1110 1001 010(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):
1.1000 1111 0110 1110 1001 010


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. 100 0111 1011 0111 0100 1010 =

100 0111 1011 0111 0100 1010


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) =
100 0111 1011 0111 0100 1010


Decimal number 0.000 000 372 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 100 0111 1011 0111 0100 1010

Share

» Convert decimal 0.000 000 409 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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