-0.000 000 001 95 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 001 95(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
-0.000 000 001 95(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:

|-0.000 000 001 95| = 0.000 000 001 95


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: 0.000 000 001 95.

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) 0.000 000 001 95 × 2 = 0 + 0.000 000 003 9;
  • 2) 0.000 000 003 9 × 2 = 0 + 0.000 000 007 8;
  • 3) 0.000 000 007 8 × 2 = 0 + 0.000 000 015 6;
  • 4) 0.000 000 015 6 × 2 = 0 + 0.000 000 031 2;
  • 5) 0.000 000 031 2 × 2 = 0 + 0.000 000 062 4;
  • 6) 0.000 000 062 4 × 2 = 0 + 0.000 000 124 8;
  • 7) 0.000 000 124 8 × 2 = 0 + 0.000 000 249 6;
  • 8) 0.000 000 249 6 × 2 = 0 + 0.000 000 499 2;
  • 9) 0.000 000 499 2 × 2 = 0 + 0.000 000 998 4;
  • 10) 0.000 000 998 4 × 2 = 0 + 0.000 001 996 8;
  • 11) 0.000 001 996 8 × 2 = 0 + 0.000 003 993 6;
  • 12) 0.000 003 993 6 × 2 = 0 + 0.000 007 987 2;
  • 13) 0.000 007 987 2 × 2 = 0 + 0.000 015 974 4;
  • 14) 0.000 015 974 4 × 2 = 0 + 0.000 031 948 8;
  • 15) 0.000 031 948 8 × 2 = 0 + 0.000 063 897 6;
  • 16) 0.000 063 897 6 × 2 = 0 + 0.000 127 795 2;
  • 17) 0.000 127 795 2 × 2 = 0 + 0.000 255 590 4;
  • 18) 0.000 255 590 4 × 2 = 0 + 0.000 511 180 8;
  • 19) 0.000 511 180 8 × 2 = 0 + 0.001 022 361 6;
  • 20) 0.001 022 361 6 × 2 = 0 + 0.002 044 723 2;
  • 21) 0.002 044 723 2 × 2 = 0 + 0.004 089 446 4;
  • 22) 0.004 089 446 4 × 2 = 0 + 0.008 178 892 8;
  • 23) 0.008 178 892 8 × 2 = 0 + 0.016 357 785 6;
  • 24) 0.016 357 785 6 × 2 = 0 + 0.032 715 571 2;
  • 25) 0.032 715 571 2 × 2 = 0 + 0.065 431 142 4;
  • 26) 0.065 431 142 4 × 2 = 0 + 0.130 862 284 8;
  • 27) 0.130 862 284 8 × 2 = 0 + 0.261 724 569 6;
  • 28) 0.261 724 569 6 × 2 = 0 + 0.523 449 139 2;
  • 29) 0.523 449 139 2 × 2 = 1 + 0.046 898 278 4;
  • 30) 0.046 898 278 4 × 2 = 0 + 0.093 796 556 8;
  • 31) 0.093 796 556 8 × 2 = 0 + 0.187 593 113 6;
  • 32) 0.187 593 113 6 × 2 = 0 + 0.375 186 227 2;
  • 33) 0.375 186 227 2 × 2 = 0 + 0.750 372 454 4;
  • 34) 0.750 372 454 4 × 2 = 1 + 0.500 744 908 8;
  • 35) 0.500 744 908 8 × 2 = 1 + 0.001 489 817 6;
  • 36) 0.001 489 817 6 × 2 = 0 + 0.002 979 635 2;
  • 37) 0.002 979 635 2 × 2 = 0 + 0.005 959 270 4;
  • 38) 0.005 959 270 4 × 2 = 0 + 0.011 918 540 8;
  • 39) 0.011 918 540 8 × 2 = 0 + 0.023 837 081 6;
  • 40) 0.023 837 081 6 × 2 = 0 + 0.047 674 163 2;
  • 41) 0.047 674 163 2 × 2 = 0 + 0.095 348 326 4;
  • 42) 0.095 348 326 4 × 2 = 0 + 0.190 696 652 8;
  • 43) 0.190 696 652 8 × 2 = 0 + 0.381 393 305 6;
  • 44) 0.381 393 305 6 × 2 = 0 + 0.762 786 611 2;
  • 45) 0.762 786 611 2 × 2 = 1 + 0.525 573 222 4;
  • 46) 0.525 573 222 4 × 2 = 1 + 0.051 146 444 8;
  • 47) 0.051 146 444 8 × 2 = 0 + 0.102 292 889 6;
  • 48) 0.102 292 889 6 × 2 = 0 + 0.204 585 779 2;
  • 49) 0.204 585 779 2 × 2 = 0 + 0.409 171 558 4;
  • 50) 0.409 171 558 4 × 2 = 0 + 0.818 343 116 8;
  • 51) 0.818 343 116 8 × 2 = 1 + 0.636 686 233 6;
  • 52) 0.636 686 233 6 × 2 = 1 + 0.273 372 467 2;

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:


0.000 000 001 95(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0110 0000 0000 1100 0011(2)

6. Positive number before normalization:

0.000 000 001 95(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0110 0000 0000 1100 0011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 001 95(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0110 0000 0000 1100 0011(2) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0110 0000 0000 1100 0011(2) × 20 =

1.0000 1100 0000 0001 1000 011(2) × 2-29


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): -29

Mantissa (not normalized):
1.0000 1100 0000 0001 1000 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-29 + 2(8-1) - 1 =

(-29 + 127)(10) =

98(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;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

98(10) =

0110 0010(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. 000 0110 0000 0000 1100 0011 =

000 0110 0000 0000 1100 0011


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) =
0110 0010

Mantissa (23 bits) =
000 0110 0000 0000 1100 0011


Decimal number -0.000 000 001 95 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0010 - 000 0110 0000 0000 1100 0011

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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