-0.000 000 001 14 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 001 14(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 14(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 14| = 0.000 000 001 14


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 14.

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 14 × 2 = 0 + 0.000 000 002 28;
  • 2) 0.000 000 002 28 × 2 = 0 + 0.000 000 004 56;
  • 3) 0.000 000 004 56 × 2 = 0 + 0.000 000 009 12;
  • 4) 0.000 000 009 12 × 2 = 0 + 0.000 000 018 24;
  • 5) 0.000 000 018 24 × 2 = 0 + 0.000 000 036 48;
  • 6) 0.000 000 036 48 × 2 = 0 + 0.000 000 072 96;
  • 7) 0.000 000 072 96 × 2 = 0 + 0.000 000 145 92;
  • 8) 0.000 000 145 92 × 2 = 0 + 0.000 000 291 84;
  • 9) 0.000 000 291 84 × 2 = 0 + 0.000 000 583 68;
  • 10) 0.000 000 583 68 × 2 = 0 + 0.000 001 167 36;
  • 11) 0.000 001 167 36 × 2 = 0 + 0.000 002 334 72;
  • 12) 0.000 002 334 72 × 2 = 0 + 0.000 004 669 44;
  • 13) 0.000 004 669 44 × 2 = 0 + 0.000 009 338 88;
  • 14) 0.000 009 338 88 × 2 = 0 + 0.000 018 677 76;
  • 15) 0.000 018 677 76 × 2 = 0 + 0.000 037 355 52;
  • 16) 0.000 037 355 52 × 2 = 0 + 0.000 074 711 04;
  • 17) 0.000 074 711 04 × 2 = 0 + 0.000 149 422 08;
  • 18) 0.000 149 422 08 × 2 = 0 + 0.000 298 844 16;
  • 19) 0.000 298 844 16 × 2 = 0 + 0.000 597 688 32;
  • 20) 0.000 597 688 32 × 2 = 0 + 0.001 195 376 64;
  • 21) 0.001 195 376 64 × 2 = 0 + 0.002 390 753 28;
  • 22) 0.002 390 753 28 × 2 = 0 + 0.004 781 506 56;
  • 23) 0.004 781 506 56 × 2 = 0 + 0.009 563 013 12;
  • 24) 0.009 563 013 12 × 2 = 0 + 0.019 126 026 24;
  • 25) 0.019 126 026 24 × 2 = 0 + 0.038 252 052 48;
  • 26) 0.038 252 052 48 × 2 = 0 + 0.076 504 104 96;
  • 27) 0.076 504 104 96 × 2 = 0 + 0.153 008 209 92;
  • 28) 0.153 008 209 92 × 2 = 0 + 0.306 016 419 84;
  • 29) 0.306 016 419 84 × 2 = 0 + 0.612 032 839 68;
  • 30) 0.612 032 839 68 × 2 = 1 + 0.224 065 679 36;
  • 31) 0.224 065 679 36 × 2 = 0 + 0.448 131 358 72;
  • 32) 0.448 131 358 72 × 2 = 0 + 0.896 262 717 44;
  • 33) 0.896 262 717 44 × 2 = 1 + 0.792 525 434 88;
  • 34) 0.792 525 434 88 × 2 = 1 + 0.585 050 869 76;
  • 35) 0.585 050 869 76 × 2 = 1 + 0.170 101 739 52;
  • 36) 0.170 101 739 52 × 2 = 0 + 0.340 203 479 04;
  • 37) 0.340 203 479 04 × 2 = 0 + 0.680 406 958 08;
  • 38) 0.680 406 958 08 × 2 = 1 + 0.360 813 916 16;
  • 39) 0.360 813 916 16 × 2 = 0 + 0.721 627 832 32;
  • 40) 0.721 627 832 32 × 2 = 1 + 0.443 255 664 64;
  • 41) 0.443 255 664 64 × 2 = 0 + 0.886 511 329 28;
  • 42) 0.886 511 329 28 × 2 = 1 + 0.773 022 658 56;
  • 43) 0.773 022 658 56 × 2 = 1 + 0.546 045 317 12;
  • 44) 0.546 045 317 12 × 2 = 1 + 0.092 090 634 24;
  • 45) 0.092 090 634 24 × 2 = 0 + 0.184 181 268 48;
  • 46) 0.184 181 268 48 × 2 = 0 + 0.368 362 536 96;
  • 47) 0.368 362 536 96 × 2 = 0 + 0.736 725 073 92;
  • 48) 0.736 725 073 92 × 2 = 1 + 0.473 450 147 84;
  • 49) 0.473 450 147 84 × 2 = 0 + 0.946 900 295 68;
  • 50) 0.946 900 295 68 × 2 = 1 + 0.893 800 591 36;
  • 51) 0.893 800 591 36 × 2 = 1 + 0.787 601 182 72;
  • 52) 0.787 601 182 72 × 2 = 1 + 0.575 202 365 44;
  • 53) 0.575 202 365 44 × 2 = 1 + 0.150 404 730 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:


0.000 000 001 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 1110 0101 0111 0001 0111 1(2)

6. Positive number before normalization:

0.000 000 001 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 1110 0101 0111 0001 0111 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 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 1110 0101 0111 0001 0111 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 1110 0101 0111 0001 0111 1(2) × 20 =

1.0011 1001 0101 1100 0101 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 1001 0101 1100 0101 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 1100 1010 1110 0010 1111 =

001 1100 1010 1110 0010 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 0001

Mantissa (23 bits) =
001 1100 1010 1110 0010 1111


Decimal number -0.000 000 001 14 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0001 - 001 1100 1010 1110 0010 1111

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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