-0.000 000 001 13 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 001 13(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 13(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 13| = 0.000 000 001 13


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 13.

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 13 × 2 = 0 + 0.000 000 002 26;
  • 2) 0.000 000 002 26 × 2 = 0 + 0.000 000 004 52;
  • 3) 0.000 000 004 52 × 2 = 0 + 0.000 000 009 04;
  • 4) 0.000 000 009 04 × 2 = 0 + 0.000 000 018 08;
  • 5) 0.000 000 018 08 × 2 = 0 + 0.000 000 036 16;
  • 6) 0.000 000 036 16 × 2 = 0 + 0.000 000 072 32;
  • 7) 0.000 000 072 32 × 2 = 0 + 0.000 000 144 64;
  • 8) 0.000 000 144 64 × 2 = 0 + 0.000 000 289 28;
  • 9) 0.000 000 289 28 × 2 = 0 + 0.000 000 578 56;
  • 10) 0.000 000 578 56 × 2 = 0 + 0.000 001 157 12;
  • 11) 0.000 001 157 12 × 2 = 0 + 0.000 002 314 24;
  • 12) 0.000 002 314 24 × 2 = 0 + 0.000 004 628 48;
  • 13) 0.000 004 628 48 × 2 = 0 + 0.000 009 256 96;
  • 14) 0.000 009 256 96 × 2 = 0 + 0.000 018 513 92;
  • 15) 0.000 018 513 92 × 2 = 0 + 0.000 037 027 84;
  • 16) 0.000 037 027 84 × 2 = 0 + 0.000 074 055 68;
  • 17) 0.000 074 055 68 × 2 = 0 + 0.000 148 111 36;
  • 18) 0.000 148 111 36 × 2 = 0 + 0.000 296 222 72;
  • 19) 0.000 296 222 72 × 2 = 0 + 0.000 592 445 44;
  • 20) 0.000 592 445 44 × 2 = 0 + 0.001 184 890 88;
  • 21) 0.001 184 890 88 × 2 = 0 + 0.002 369 781 76;
  • 22) 0.002 369 781 76 × 2 = 0 + 0.004 739 563 52;
  • 23) 0.004 739 563 52 × 2 = 0 + 0.009 479 127 04;
  • 24) 0.009 479 127 04 × 2 = 0 + 0.018 958 254 08;
  • 25) 0.018 958 254 08 × 2 = 0 + 0.037 916 508 16;
  • 26) 0.037 916 508 16 × 2 = 0 + 0.075 833 016 32;
  • 27) 0.075 833 016 32 × 2 = 0 + 0.151 666 032 64;
  • 28) 0.151 666 032 64 × 2 = 0 + 0.303 332 065 28;
  • 29) 0.303 332 065 28 × 2 = 0 + 0.606 664 130 56;
  • 30) 0.606 664 130 56 × 2 = 1 + 0.213 328 261 12;
  • 31) 0.213 328 261 12 × 2 = 0 + 0.426 656 522 24;
  • 32) 0.426 656 522 24 × 2 = 0 + 0.853 313 044 48;
  • 33) 0.853 313 044 48 × 2 = 1 + 0.706 626 088 96;
  • 34) 0.706 626 088 96 × 2 = 1 + 0.413 252 177 92;
  • 35) 0.413 252 177 92 × 2 = 0 + 0.826 504 355 84;
  • 36) 0.826 504 355 84 × 2 = 1 + 0.653 008 711 68;
  • 37) 0.653 008 711 68 × 2 = 1 + 0.306 017 423 36;
  • 38) 0.306 017 423 36 × 2 = 0 + 0.612 034 846 72;
  • 39) 0.612 034 846 72 × 2 = 1 + 0.224 069 693 44;
  • 40) 0.224 069 693 44 × 2 = 0 + 0.448 139 386 88;
  • 41) 0.448 139 386 88 × 2 = 0 + 0.896 278 773 76;
  • 42) 0.896 278 773 76 × 2 = 1 + 0.792 557 547 52;
  • 43) 0.792 557 547 52 × 2 = 1 + 0.585 115 095 04;
  • 44) 0.585 115 095 04 × 2 = 1 + 0.170 230 190 08;
  • 45) 0.170 230 190 08 × 2 = 0 + 0.340 460 380 16;
  • 46) 0.340 460 380 16 × 2 = 0 + 0.680 920 760 32;
  • 47) 0.680 920 760 32 × 2 = 1 + 0.361 841 520 64;
  • 48) 0.361 841 520 64 × 2 = 0 + 0.723 683 041 28;
  • 49) 0.723 683 041 28 × 2 = 1 + 0.447 366 082 56;
  • 50) 0.447 366 082 56 × 2 = 0 + 0.894 732 165 12;
  • 51) 0.894 732 165 12 × 2 = 1 + 0.789 464 330 24;
  • 52) 0.789 464 330 24 × 2 = 1 + 0.578 928 660 48;
  • 53) 0.578 928 660 48 × 2 = 1 + 0.157 857 320 96;

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 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 1101 1010 0111 0010 1011 1(2)

6. Positive number before normalization:

0.000 000 001 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 1101 1010 0111 0010 1011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 001 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 1101 1010 0111 0010 1011 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 1101 1010 0111 0010 1011 1(2) × 20 =

1.0011 0110 1001 1100 1010 111(2) × 2-30


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): -30

Mantissa (not normalized):
1.0011 0110 1001 1100 1010 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-30 + 2(8-1) - 1 =

(-30 + 127)(10) =

97(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;
  • 97 ÷ 2 = 48 + 1;
  • 48 ÷ 2 = 24 + 0;
  • 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) =

97(10) =

0110 0001(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. 001 1011 0100 1110 0101 0111 =

001 1011 0100 1110 0101 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 0001

Mantissa (23 bits) =
001 1011 0100 1110 0101 0111


Decimal number -0.000 000 001 13 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0001 - 001 1011 0100 1110 0101 0111

Share

» Convert decimal -0.000 000 001 69 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