-0.000 000 68 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 68(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 68(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 68| = 0.000 000 68


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 68.

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 68 × 2 = 0 + 0.000 001 36;
  • 2) 0.000 001 36 × 2 = 0 + 0.000 002 72;
  • 3) 0.000 002 72 × 2 = 0 + 0.000 005 44;
  • 4) 0.000 005 44 × 2 = 0 + 0.000 010 88;
  • 5) 0.000 010 88 × 2 = 0 + 0.000 021 76;
  • 6) 0.000 021 76 × 2 = 0 + 0.000 043 52;
  • 7) 0.000 043 52 × 2 = 0 + 0.000 087 04;
  • 8) 0.000 087 04 × 2 = 0 + 0.000 174 08;
  • 9) 0.000 174 08 × 2 = 0 + 0.000 348 16;
  • 10) 0.000 348 16 × 2 = 0 + 0.000 696 32;
  • 11) 0.000 696 32 × 2 = 0 + 0.001 392 64;
  • 12) 0.001 392 64 × 2 = 0 + 0.002 785 28;
  • 13) 0.002 785 28 × 2 = 0 + 0.005 570 56;
  • 14) 0.005 570 56 × 2 = 0 + 0.011 141 12;
  • 15) 0.011 141 12 × 2 = 0 + 0.022 282 24;
  • 16) 0.022 282 24 × 2 = 0 + 0.044 564 48;
  • 17) 0.044 564 48 × 2 = 0 + 0.089 128 96;
  • 18) 0.089 128 96 × 2 = 0 + 0.178 257 92;
  • 19) 0.178 257 92 × 2 = 0 + 0.356 515 84;
  • 20) 0.356 515 84 × 2 = 0 + 0.713 031 68;
  • 21) 0.713 031 68 × 2 = 1 + 0.426 063 36;
  • 22) 0.426 063 36 × 2 = 0 + 0.852 126 72;
  • 23) 0.852 126 72 × 2 = 1 + 0.704 253 44;
  • 24) 0.704 253 44 × 2 = 1 + 0.408 506 88;
  • 25) 0.408 506 88 × 2 = 0 + 0.817 013 76;
  • 26) 0.817 013 76 × 2 = 1 + 0.634 027 52;
  • 27) 0.634 027 52 × 2 = 1 + 0.268 055 04;
  • 28) 0.268 055 04 × 2 = 0 + 0.536 110 08;
  • 29) 0.536 110 08 × 2 = 1 + 0.072 220 16;
  • 30) 0.072 220 16 × 2 = 0 + 0.144 440 32;
  • 31) 0.144 440 32 × 2 = 0 + 0.288 880 64;
  • 32) 0.288 880 64 × 2 = 0 + 0.577 761 28;
  • 33) 0.577 761 28 × 2 = 1 + 0.155 522 56;
  • 34) 0.155 522 56 × 2 = 0 + 0.311 045 12;
  • 35) 0.311 045 12 × 2 = 0 + 0.622 090 24;
  • 36) 0.622 090 24 × 2 = 1 + 0.244 180 48;
  • 37) 0.244 180 48 × 2 = 0 + 0.488 360 96;
  • 38) 0.488 360 96 × 2 = 0 + 0.976 721 92;
  • 39) 0.976 721 92 × 2 = 1 + 0.953 443 84;
  • 40) 0.953 443 84 × 2 = 1 + 0.906 887 68;
  • 41) 0.906 887 68 × 2 = 1 + 0.813 775 36;
  • 42) 0.813 775 36 × 2 = 1 + 0.627 550 72;
  • 43) 0.627 550 72 × 2 = 1 + 0.255 101 44;
  • 44) 0.255 101 44 × 2 = 0 + 0.510 202 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:


0.000 000 68(10) =

0.0000 0000 0000 0000 0000 1011 0110 1000 1001 0011 1110(2)

6. Positive number before normalization:

0.000 000 68(10) =

0.0000 0000 0000 0000 0000 1011 0110 1000 1001 0011 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 68(10) =

0.0000 0000 0000 0000 0000 1011 0110 1000 1001 0011 1110(2) =

0.0000 0000 0000 0000 0000 1011 0110 1000 1001 0011 1110(2) × 20 =

1.0110 1101 0001 0010 0111 110(2) × 2-21


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): -21

Mantissa (not normalized):
1.0110 1101 0001 0010 0111 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 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) =

106(10) =

0110 1010(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. 011 0110 1000 1001 0011 1110 =

011 0110 1000 1001 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) =
0110 1010

Mantissa (23 bits) =
011 0110 1000 1001 0011 1110


Decimal number -0.000 000 68 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1010 - 011 0110 1000 1001 0011 1110

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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