-0.000 000 004 18 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 004 18(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
-0.000 000 004 18(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:

|-0.000 000 004 18| = 0.000 000 004 18


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: 0.000 000 004 18.

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) 0.000 000 004 18 × 2 = 0 + 0.000 000 008 36;
  • 2) 0.000 000 008 36 × 2 = 0 + 0.000 000 016 72;
  • 3) 0.000 000 016 72 × 2 = 0 + 0.000 000 033 44;
  • 4) 0.000 000 033 44 × 2 = 0 + 0.000 000 066 88;
  • 5) 0.000 000 066 88 × 2 = 0 + 0.000 000 133 76;
  • 6) 0.000 000 133 76 × 2 = 0 + 0.000 000 267 52;
  • 7) 0.000 000 267 52 × 2 = 0 + 0.000 000 535 04;
  • 8) 0.000 000 535 04 × 2 = 0 + 0.000 001 070 08;
  • 9) 0.000 001 070 08 × 2 = 0 + 0.000 002 140 16;
  • 10) 0.000 002 140 16 × 2 = 0 + 0.000 004 280 32;
  • 11) 0.000 004 280 32 × 2 = 0 + 0.000 008 560 64;
  • 12) 0.000 008 560 64 × 2 = 0 + 0.000 017 121 28;
  • 13) 0.000 017 121 28 × 2 = 0 + 0.000 034 242 56;
  • 14) 0.000 034 242 56 × 2 = 0 + 0.000 068 485 12;
  • 15) 0.000 068 485 12 × 2 = 0 + 0.000 136 970 24;
  • 16) 0.000 136 970 24 × 2 = 0 + 0.000 273 940 48;
  • 17) 0.000 273 940 48 × 2 = 0 + 0.000 547 880 96;
  • 18) 0.000 547 880 96 × 2 = 0 + 0.001 095 761 92;
  • 19) 0.001 095 761 92 × 2 = 0 + 0.002 191 523 84;
  • 20) 0.002 191 523 84 × 2 = 0 + 0.004 383 047 68;
  • 21) 0.004 383 047 68 × 2 = 0 + 0.008 766 095 36;
  • 22) 0.008 766 095 36 × 2 = 0 + 0.017 532 190 72;
  • 23) 0.017 532 190 72 × 2 = 0 + 0.035 064 381 44;
  • 24) 0.035 064 381 44 × 2 = 0 + 0.070 128 762 88;
  • 25) 0.070 128 762 88 × 2 = 0 + 0.140 257 525 76;
  • 26) 0.140 257 525 76 × 2 = 0 + 0.280 515 051 52;
  • 27) 0.280 515 051 52 × 2 = 0 + 0.561 030 103 04;
  • 28) 0.561 030 103 04 × 2 = 1 + 0.122 060 206 08;
  • 29) 0.122 060 206 08 × 2 = 0 + 0.244 120 412 16;
  • 30) 0.244 120 412 16 × 2 = 0 + 0.488 240 824 32;
  • 31) 0.488 240 824 32 × 2 = 0 + 0.976 481 648 64;
  • 32) 0.976 481 648 64 × 2 = 1 + 0.952 963 297 28;
  • 33) 0.952 963 297 28 × 2 = 1 + 0.905 926 594 56;
  • 34) 0.905 926 594 56 × 2 = 1 + 0.811 853 189 12;
  • 35) 0.811 853 189 12 × 2 = 1 + 0.623 706 378 24;
  • 36) 0.623 706 378 24 × 2 = 1 + 0.247 412 756 48;
  • 37) 0.247 412 756 48 × 2 = 0 + 0.494 825 512 96;
  • 38) 0.494 825 512 96 × 2 = 0 + 0.989 651 025 92;
  • 39) 0.989 651 025 92 × 2 = 1 + 0.979 302 051 84;
  • 40) 0.979 302 051 84 × 2 = 1 + 0.958 604 103 68;
  • 41) 0.958 604 103 68 × 2 = 1 + 0.917 208 207 36;
  • 42) 0.917 208 207 36 × 2 = 1 + 0.834 416 414 72;
  • 43) 0.834 416 414 72 × 2 = 1 + 0.668 832 829 44;
  • 44) 0.668 832 829 44 × 2 = 1 + 0.337 665 658 88;
  • 45) 0.337 665 658 88 × 2 = 0 + 0.675 331 317 76;
  • 46) 0.675 331 317 76 × 2 = 1 + 0.350 662 635 52;
  • 47) 0.350 662 635 52 × 2 = 0 + 0.701 325 271 04;
  • 48) 0.701 325 271 04 × 2 = 1 + 0.402 650 542 08;
  • 49) 0.402 650 542 08 × 2 = 0 + 0.805 301 084 16;
  • 50) 0.805 301 084 16 × 2 = 1 + 0.610 602 168 32;
  • 51) 0.610 602 168 32 × 2 = 1 + 0.221 204 336 64;

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:


0.000 000 004 18(10) =

0.0000 0000 0000 0000 0000 0000 0001 0001 1111 0011 1111 0101 011(2)

6. Positive number before normalization:

0.000 000 004 18(10) =

0.0000 0000 0000 0000 0000 0000 0001 0001 1111 0011 1111 0101 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 004 18(10) =

0.0000 0000 0000 0000 0000 0000 0001 0001 1111 0011 1111 0101 011(2) =

0.0000 0000 0000 0000 0000 0000 0001 0001 1111 0011 1111 0101 011(2) × 20 =

1.0001 1111 0011 1111 0101 011(2) × 2-28


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): -28

Mantissa (not normalized):
1.0001 1111 0011 1111 0101 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 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) =

99(10) =

0110 0011(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. 000 1111 1001 1111 1010 1011 =

000 1111 1001 1111 1010 1011


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 0011

Mantissa (23 bits) =
000 1111 1001 1111 1010 1011


Decimal number -0.000 000 004 18 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0011 - 000 1111 1001 1111 1010 1011

Share

» Convert decimal -0.000 000 003 2 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 Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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