-0.000 000 002 62 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 002 62(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
-0.000 000 002 62(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:

|-0.000 000 002 62| = 0.000 000 002 62


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: 0.000 000 002 62.

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) 0.000 000 002 62 × 2 = 0 + 0.000 000 005 24;
  • 2) 0.000 000 005 24 × 2 = 0 + 0.000 000 010 48;
  • 3) 0.000 000 010 48 × 2 = 0 + 0.000 000 020 96;
  • 4) 0.000 000 020 96 × 2 = 0 + 0.000 000 041 92;
  • 5) 0.000 000 041 92 × 2 = 0 + 0.000 000 083 84;
  • 6) 0.000 000 083 84 × 2 = 0 + 0.000 000 167 68;
  • 7) 0.000 000 167 68 × 2 = 0 + 0.000 000 335 36;
  • 8) 0.000 000 335 36 × 2 = 0 + 0.000 000 670 72;
  • 9) 0.000 000 670 72 × 2 = 0 + 0.000 001 341 44;
  • 10) 0.000 001 341 44 × 2 = 0 + 0.000 002 682 88;
  • 11) 0.000 002 682 88 × 2 = 0 + 0.000 005 365 76;
  • 12) 0.000 005 365 76 × 2 = 0 + 0.000 010 731 52;
  • 13) 0.000 010 731 52 × 2 = 0 + 0.000 021 463 04;
  • 14) 0.000 021 463 04 × 2 = 0 + 0.000 042 926 08;
  • 15) 0.000 042 926 08 × 2 = 0 + 0.000 085 852 16;
  • 16) 0.000 085 852 16 × 2 = 0 + 0.000 171 704 32;
  • 17) 0.000 171 704 32 × 2 = 0 + 0.000 343 408 64;
  • 18) 0.000 343 408 64 × 2 = 0 + 0.000 686 817 28;
  • 19) 0.000 686 817 28 × 2 = 0 + 0.001 373 634 56;
  • 20) 0.001 373 634 56 × 2 = 0 + 0.002 747 269 12;
  • 21) 0.002 747 269 12 × 2 = 0 + 0.005 494 538 24;
  • 22) 0.005 494 538 24 × 2 = 0 + 0.010 989 076 48;
  • 23) 0.010 989 076 48 × 2 = 0 + 0.021 978 152 96;
  • 24) 0.021 978 152 96 × 2 = 0 + 0.043 956 305 92;
  • 25) 0.043 956 305 92 × 2 = 0 + 0.087 912 611 84;
  • 26) 0.087 912 611 84 × 2 = 0 + 0.175 825 223 68;
  • 27) 0.175 825 223 68 × 2 = 0 + 0.351 650 447 36;
  • 28) 0.351 650 447 36 × 2 = 0 + 0.703 300 894 72;
  • 29) 0.703 300 894 72 × 2 = 1 + 0.406 601 789 44;
  • 30) 0.406 601 789 44 × 2 = 0 + 0.813 203 578 88;
  • 31) 0.813 203 578 88 × 2 = 1 + 0.626 407 157 76;
  • 32) 0.626 407 157 76 × 2 = 1 + 0.252 814 315 52;
  • 33) 0.252 814 315 52 × 2 = 0 + 0.505 628 631 04;
  • 34) 0.505 628 631 04 × 2 = 1 + 0.011 257 262 08;
  • 35) 0.011 257 262 08 × 2 = 0 + 0.022 514 524 16;
  • 36) 0.022 514 524 16 × 2 = 0 + 0.045 029 048 32;
  • 37) 0.045 029 048 32 × 2 = 0 + 0.090 058 096 64;
  • 38) 0.090 058 096 64 × 2 = 0 + 0.180 116 193 28;
  • 39) 0.180 116 193 28 × 2 = 0 + 0.360 232 386 56;
  • 40) 0.360 232 386 56 × 2 = 0 + 0.720 464 773 12;
  • 41) 0.720 464 773 12 × 2 = 1 + 0.440 929 546 24;
  • 42) 0.440 929 546 24 × 2 = 0 + 0.881 859 092 48;
  • 43) 0.881 859 092 48 × 2 = 1 + 0.763 718 184 96;
  • 44) 0.763 718 184 96 × 2 = 1 + 0.527 436 369 92;
  • 45) 0.527 436 369 92 × 2 = 1 + 0.054 872 739 84;
  • 46) 0.054 872 739 84 × 2 = 0 + 0.109 745 479 68;
  • 47) 0.109 745 479 68 × 2 = 0 + 0.219 490 959 36;
  • 48) 0.219 490 959 36 × 2 = 0 + 0.438 981 918 72;
  • 49) 0.438 981 918 72 × 2 = 0 + 0.877 963 837 44;
  • 50) 0.877 963 837 44 × 2 = 1 + 0.755 927 674 88;
  • 51) 0.755 927 674 88 × 2 = 1 + 0.511 855 349 76;
  • 52) 0.511 855 349 76 × 2 = 1 + 0.023 710 699 52;

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:


0.000 000 002 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 1011 0100 0000 1011 1000 0111(2)

6. Positive number before normalization:

0.000 000 002 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 1011 0100 0000 1011 1000 0111(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:


0.000 000 002 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 1011 0100 0000 1011 1000 0111(2) =

0.0000 0000 0000 0000 0000 0000 0000 1011 0100 0000 1011 1000 0111(2) × 20 =

1.0110 1000 0001 0111 0000 111(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):
1.0110 1000 0001 0111 0000 111


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. 011 0100 0000 1011 1000 0111 =

011 0100 0000 1011 1000 0111


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) =
011 0100 0000 1011 1000 0111


Decimal number -0.000 000 002 62 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0010 - 011 0100 0000 1011 1000 0111

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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