-0.000 000 000 38 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 38(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
-0.000 000 000 38(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. Start with the positive version of the number:

|-0.000 000 000 38| = 0.000 000 000 38


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 38.

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) 0.000 000 000 38 × 2 = 0 + 0.000 000 000 76;
  • 2) 0.000 000 000 76 × 2 = 0 + 0.000 000 001 52;
  • 3) 0.000 000 001 52 × 2 = 0 + 0.000 000 003 04;
  • 4) 0.000 000 003 04 × 2 = 0 + 0.000 000 006 08;
  • 5) 0.000 000 006 08 × 2 = 0 + 0.000 000 012 16;
  • 6) 0.000 000 012 16 × 2 = 0 + 0.000 000 024 32;
  • 7) 0.000 000 024 32 × 2 = 0 + 0.000 000 048 64;
  • 8) 0.000 000 048 64 × 2 = 0 + 0.000 000 097 28;
  • 9) 0.000 000 097 28 × 2 = 0 + 0.000 000 194 56;
  • 10) 0.000 000 194 56 × 2 = 0 + 0.000 000 389 12;
  • 11) 0.000 000 389 12 × 2 = 0 + 0.000 000 778 24;
  • 12) 0.000 000 778 24 × 2 = 0 + 0.000 001 556 48;
  • 13) 0.000 001 556 48 × 2 = 0 + 0.000 003 112 96;
  • 14) 0.000 003 112 96 × 2 = 0 + 0.000 006 225 92;
  • 15) 0.000 006 225 92 × 2 = 0 + 0.000 012 451 84;
  • 16) 0.000 012 451 84 × 2 = 0 + 0.000 024 903 68;
  • 17) 0.000 024 903 68 × 2 = 0 + 0.000 049 807 36;
  • 18) 0.000 049 807 36 × 2 = 0 + 0.000 099 614 72;
  • 19) 0.000 099 614 72 × 2 = 0 + 0.000 199 229 44;
  • 20) 0.000 199 229 44 × 2 = 0 + 0.000 398 458 88;
  • 21) 0.000 398 458 88 × 2 = 0 + 0.000 796 917 76;
  • 22) 0.000 796 917 76 × 2 = 0 + 0.001 593 835 52;
  • 23) 0.001 593 835 52 × 2 = 0 + 0.003 187 671 04;
  • 24) 0.003 187 671 04 × 2 = 0 + 0.006 375 342 08;
  • 25) 0.006 375 342 08 × 2 = 0 + 0.012 750 684 16;
  • 26) 0.012 750 684 16 × 2 = 0 + 0.025 501 368 32;
  • 27) 0.025 501 368 32 × 2 = 0 + 0.051 002 736 64;
  • 28) 0.051 002 736 64 × 2 = 0 + 0.102 005 473 28;
  • 29) 0.102 005 473 28 × 2 = 0 + 0.204 010 946 56;
  • 30) 0.204 010 946 56 × 2 = 0 + 0.408 021 893 12;
  • 31) 0.408 021 893 12 × 2 = 0 + 0.816 043 786 24;
  • 32) 0.816 043 786 24 × 2 = 1 + 0.632 087 572 48;
  • 33) 0.632 087 572 48 × 2 = 1 + 0.264 175 144 96;
  • 34) 0.264 175 144 96 × 2 = 0 + 0.528 350 289 92;
  • 35) 0.528 350 289 92 × 2 = 1 + 0.056 700 579 84;
  • 36) 0.056 700 579 84 × 2 = 0 + 0.113 401 159 68;
  • 37) 0.113 401 159 68 × 2 = 0 + 0.226 802 319 36;
  • 38) 0.226 802 319 36 × 2 = 0 + 0.453 604 638 72;
  • 39) 0.453 604 638 72 × 2 = 0 + 0.907 209 277 44;
  • 40) 0.907 209 277 44 × 2 = 1 + 0.814 418 554 88;
  • 41) 0.814 418 554 88 × 2 = 1 + 0.628 837 109 76;
  • 42) 0.628 837 109 76 × 2 = 1 + 0.257 674 219 52;
  • 43) 0.257 674 219 52 × 2 = 0 + 0.515 348 439 04;
  • 44) 0.515 348 439 04 × 2 = 1 + 0.030 696 878 08;
  • 45) 0.030 696 878 08 × 2 = 0 + 0.061 393 756 16;
  • 46) 0.061 393 756 16 × 2 = 0 + 0.122 787 512 32;
  • 47) 0.122 787 512 32 × 2 = 0 + 0.245 575 024 64;
  • 48) 0.245 575 024 64 × 2 = 0 + 0.491 150 049 28;
  • 49) 0.491 150 049 28 × 2 = 0 + 0.982 300 098 56;
  • 50) 0.982 300 098 56 × 2 = 1 + 0.964 600 197 12;
  • 51) 0.964 600 197 12 × 2 = 1 + 0.929 200 394 24;
  • 52) 0.929 200 394 24 × 2 = 1 + 0.858 400 788 48;
  • 53) 0.858 400 788 48 × 2 = 1 + 0.716 801 576 96;
  • 54) 0.716 801 576 96 × 2 = 1 + 0.433 603 153 92;
  • 55) 0.433 603 153 92 × 2 = 0 + 0.867 206 307 84;

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).


5. 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:


0.000 000 000 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1010 0001 1101 0000 0111 110(2)

6. Positive number before normalization:

0.000 000 000 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1010 0001 1101 0000 0111 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1010 0001 1101 0000 0111 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1010 0001 1101 0000 0111 110(2) × 20 =

1.1010 0001 1101 0000 0111 110(2) × 2-32


8. 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 1 (a negative number)

Exponent (unadjusted): -32

Mantissa (not normalized):
1.1010 0001 1101 0000 0111 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(10)


10. 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;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

95(10) =

0101 1111(2)


12. 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. 101 0000 1110 1000 0011 1110 =

101 0000 1110 1000 0011 1110


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0101 1111

Mantissa (23 bits) =
101 0000 1110 1000 0011 1110


Decimal number -0.000 000 000 38 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1111 - 101 0000 1110 1000 0011 1110

Share

» Convert decimal -0.000 000 001 17 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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