-0.000 001 61 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 61(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
-0.000 001 61(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:

|-0.000 001 61| = 0.000 001 61


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: 0.000 001 61.

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) 0.000 001 61 × 2 = 0 + 0.000 003 22;
  • 2) 0.000 003 22 × 2 = 0 + 0.000 006 44;
  • 3) 0.000 006 44 × 2 = 0 + 0.000 012 88;
  • 4) 0.000 012 88 × 2 = 0 + 0.000 025 76;
  • 5) 0.000 025 76 × 2 = 0 + 0.000 051 52;
  • 6) 0.000 051 52 × 2 = 0 + 0.000 103 04;
  • 7) 0.000 103 04 × 2 = 0 + 0.000 206 08;
  • 8) 0.000 206 08 × 2 = 0 + 0.000 412 16;
  • 9) 0.000 412 16 × 2 = 0 + 0.000 824 32;
  • 10) 0.000 824 32 × 2 = 0 + 0.001 648 64;
  • 11) 0.001 648 64 × 2 = 0 + 0.003 297 28;
  • 12) 0.003 297 28 × 2 = 0 + 0.006 594 56;
  • 13) 0.006 594 56 × 2 = 0 + 0.013 189 12;
  • 14) 0.013 189 12 × 2 = 0 + 0.026 378 24;
  • 15) 0.026 378 24 × 2 = 0 + 0.052 756 48;
  • 16) 0.052 756 48 × 2 = 0 + 0.105 512 96;
  • 17) 0.105 512 96 × 2 = 0 + 0.211 025 92;
  • 18) 0.211 025 92 × 2 = 0 + 0.422 051 84;
  • 19) 0.422 051 84 × 2 = 0 + 0.844 103 68;
  • 20) 0.844 103 68 × 2 = 1 + 0.688 207 36;
  • 21) 0.688 207 36 × 2 = 1 + 0.376 414 72;
  • 22) 0.376 414 72 × 2 = 0 + 0.752 829 44;
  • 23) 0.752 829 44 × 2 = 1 + 0.505 658 88;
  • 24) 0.505 658 88 × 2 = 1 + 0.011 317 76;
  • 25) 0.011 317 76 × 2 = 0 + 0.022 635 52;
  • 26) 0.022 635 52 × 2 = 0 + 0.045 271 04;
  • 27) 0.045 271 04 × 2 = 0 + 0.090 542 08;
  • 28) 0.090 542 08 × 2 = 0 + 0.181 084 16;
  • 29) 0.181 084 16 × 2 = 0 + 0.362 168 32;
  • 30) 0.362 168 32 × 2 = 0 + 0.724 336 64;
  • 31) 0.724 336 64 × 2 = 1 + 0.448 673 28;
  • 32) 0.448 673 28 × 2 = 0 + 0.897 346 56;
  • 33) 0.897 346 56 × 2 = 1 + 0.794 693 12;
  • 34) 0.794 693 12 × 2 = 1 + 0.589 386 24;
  • 35) 0.589 386 24 × 2 = 1 + 0.178 772 48;
  • 36) 0.178 772 48 × 2 = 0 + 0.357 544 96;
  • 37) 0.357 544 96 × 2 = 0 + 0.715 089 92;
  • 38) 0.715 089 92 × 2 = 1 + 0.430 179 84;
  • 39) 0.430 179 84 × 2 = 0 + 0.860 359 68;
  • 40) 0.860 359 68 × 2 = 1 + 0.720 719 36;
  • 41) 0.720 719 36 × 2 = 1 + 0.441 438 72;
  • 42) 0.441 438 72 × 2 = 0 + 0.882 877 44;
  • 43) 0.882 877 44 × 2 = 1 + 0.765 754 88;

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:


0.000 001 61(10) =

0.0000 0000 0000 0000 0001 1011 0000 0010 1110 0101 101(2)

6. Positive number before normalization:

0.000 001 61(10) =

0.0000 0000 0000 0000 0001 1011 0000 0010 1110 0101 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 61(10) =

0.0000 0000 0000 0000 0001 1011 0000 0010 1110 0101 101(2) =

0.0000 0000 0000 0000 0001 1011 0000 0010 1110 0101 101(2) × 20 =

1.1011 0000 0010 1110 0101 101(2) × 2-20


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): -20

Mantissa (not normalized):
1.1011 0000 0010 1110 0101 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

107(10) =

0110 1011(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 1000 0001 0111 0010 1101 =

101 1000 0001 0111 0010 1101


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 1011

Mantissa (23 bits) =
101 1000 0001 0111 0010 1101


Decimal number -0.000 001 61 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1011 - 101 1000 0001 0111 0010 1101

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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