-0.000 000 000 54 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 54(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
-0.000 000 000 54(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:

|-0.000 000 000 54| = 0.000 000 000 54


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: 0.000 000 000 54.

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) 0.000 000 000 54 × 2 = 0 + 0.000 000 001 08;
  • 2) 0.000 000 001 08 × 2 = 0 + 0.000 000 002 16;
  • 3) 0.000 000 002 16 × 2 = 0 + 0.000 000 004 32;
  • 4) 0.000 000 004 32 × 2 = 0 + 0.000 000 008 64;
  • 5) 0.000 000 008 64 × 2 = 0 + 0.000 000 017 28;
  • 6) 0.000 000 017 28 × 2 = 0 + 0.000 000 034 56;
  • 7) 0.000 000 034 56 × 2 = 0 + 0.000 000 069 12;
  • 8) 0.000 000 069 12 × 2 = 0 + 0.000 000 138 24;
  • 9) 0.000 000 138 24 × 2 = 0 + 0.000 000 276 48;
  • 10) 0.000 000 276 48 × 2 = 0 + 0.000 000 552 96;
  • 11) 0.000 000 552 96 × 2 = 0 + 0.000 001 105 92;
  • 12) 0.000 001 105 92 × 2 = 0 + 0.000 002 211 84;
  • 13) 0.000 002 211 84 × 2 = 0 + 0.000 004 423 68;
  • 14) 0.000 004 423 68 × 2 = 0 + 0.000 008 847 36;
  • 15) 0.000 008 847 36 × 2 = 0 + 0.000 017 694 72;
  • 16) 0.000 017 694 72 × 2 = 0 + 0.000 035 389 44;
  • 17) 0.000 035 389 44 × 2 = 0 + 0.000 070 778 88;
  • 18) 0.000 070 778 88 × 2 = 0 + 0.000 141 557 76;
  • 19) 0.000 141 557 76 × 2 = 0 + 0.000 283 115 52;
  • 20) 0.000 283 115 52 × 2 = 0 + 0.000 566 231 04;
  • 21) 0.000 566 231 04 × 2 = 0 + 0.001 132 462 08;
  • 22) 0.001 132 462 08 × 2 = 0 + 0.002 264 924 16;
  • 23) 0.002 264 924 16 × 2 = 0 + 0.004 529 848 32;
  • 24) 0.004 529 848 32 × 2 = 0 + 0.009 059 696 64;
  • 25) 0.009 059 696 64 × 2 = 0 + 0.018 119 393 28;
  • 26) 0.018 119 393 28 × 2 = 0 + 0.036 238 786 56;
  • 27) 0.036 238 786 56 × 2 = 0 + 0.072 477 573 12;
  • 28) 0.072 477 573 12 × 2 = 0 + 0.144 955 146 24;
  • 29) 0.144 955 146 24 × 2 = 0 + 0.289 910 292 48;
  • 30) 0.289 910 292 48 × 2 = 0 + 0.579 820 584 96;
  • 31) 0.579 820 584 96 × 2 = 1 + 0.159 641 169 92;
  • 32) 0.159 641 169 92 × 2 = 0 + 0.319 282 339 84;
  • 33) 0.319 282 339 84 × 2 = 0 + 0.638 564 679 68;
  • 34) 0.638 564 679 68 × 2 = 1 + 0.277 129 359 36;
  • 35) 0.277 129 359 36 × 2 = 0 + 0.554 258 718 72;
  • 36) 0.554 258 718 72 × 2 = 1 + 0.108 517 437 44;
  • 37) 0.108 517 437 44 × 2 = 0 + 0.217 034 874 88;
  • 38) 0.217 034 874 88 × 2 = 0 + 0.434 069 749 76;
  • 39) 0.434 069 749 76 × 2 = 0 + 0.868 139 499 52;
  • 40) 0.868 139 499 52 × 2 = 1 + 0.736 278 999 04;
  • 41) 0.736 278 999 04 × 2 = 1 + 0.472 557 998 08;
  • 42) 0.472 557 998 08 × 2 = 0 + 0.945 115 996 16;
  • 43) 0.945 115 996 16 × 2 = 1 + 0.890 231 992 32;
  • 44) 0.890 231 992 32 × 2 = 1 + 0.780 463 984 64;
  • 45) 0.780 463 984 64 × 2 = 1 + 0.560 927 969 28;
  • 46) 0.560 927 969 28 × 2 = 1 + 0.121 855 938 56;
  • 47) 0.121 855 938 56 × 2 = 0 + 0.243 711 877 12;
  • 48) 0.243 711 877 12 × 2 = 0 + 0.487 423 754 24;
  • 49) 0.487 423 754 24 × 2 = 0 + 0.974 847 508 48;
  • 50) 0.974 847 508 48 × 2 = 1 + 0.949 695 016 96;
  • 51) 0.949 695 016 96 × 2 = 1 + 0.899 390 033 92;
  • 52) 0.899 390 033 92 × 2 = 1 + 0.798 780 067 84;
  • 53) 0.798 780 067 84 × 2 = 1 + 0.597 560 135 68;
  • 54) 0.597 560 135 68 × 2 = 1 + 0.195 120 271 36;

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:


0.000 000 000 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0101 0001 1011 1100 0111 11(2)

6. Positive number before normalization:

0.000 000 000 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0101 0001 1011 1100 0111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0101 0001 1011 1100 0111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0101 0001 1011 1100 0111 11(2) × 20 =

1.0010 1000 1101 1110 0011 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.0010 1000 1101 1110 0011 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 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) =

96(10) =

0110 0000(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 0100 0110 1111 0001 1111 =

001 0100 0110 1111 0001 1111


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 0000

Mantissa (23 bits) =
001 0100 0110 1111 0001 1111


Decimal number -0.000 000 000 54 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 001 0100 0110 1111 0001 1111

Share

» Convert decimal 0.000 000 000 1 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: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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