-0.000 000 001 19 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 001 19(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 19(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 19| = 0.000 000 001 19


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 19.

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 19 × 2 = 0 + 0.000 000 002 38;
  • 2) 0.000 000 002 38 × 2 = 0 + 0.000 000 004 76;
  • 3) 0.000 000 004 76 × 2 = 0 + 0.000 000 009 52;
  • 4) 0.000 000 009 52 × 2 = 0 + 0.000 000 019 04;
  • 5) 0.000 000 019 04 × 2 = 0 + 0.000 000 038 08;
  • 6) 0.000 000 038 08 × 2 = 0 + 0.000 000 076 16;
  • 7) 0.000 000 076 16 × 2 = 0 + 0.000 000 152 32;
  • 8) 0.000 000 152 32 × 2 = 0 + 0.000 000 304 64;
  • 9) 0.000 000 304 64 × 2 = 0 + 0.000 000 609 28;
  • 10) 0.000 000 609 28 × 2 = 0 + 0.000 001 218 56;
  • 11) 0.000 001 218 56 × 2 = 0 + 0.000 002 437 12;
  • 12) 0.000 002 437 12 × 2 = 0 + 0.000 004 874 24;
  • 13) 0.000 004 874 24 × 2 = 0 + 0.000 009 748 48;
  • 14) 0.000 009 748 48 × 2 = 0 + 0.000 019 496 96;
  • 15) 0.000 019 496 96 × 2 = 0 + 0.000 038 993 92;
  • 16) 0.000 038 993 92 × 2 = 0 + 0.000 077 987 84;
  • 17) 0.000 077 987 84 × 2 = 0 + 0.000 155 975 68;
  • 18) 0.000 155 975 68 × 2 = 0 + 0.000 311 951 36;
  • 19) 0.000 311 951 36 × 2 = 0 + 0.000 623 902 72;
  • 20) 0.000 623 902 72 × 2 = 0 + 0.001 247 805 44;
  • 21) 0.001 247 805 44 × 2 = 0 + 0.002 495 610 88;
  • 22) 0.002 495 610 88 × 2 = 0 + 0.004 991 221 76;
  • 23) 0.004 991 221 76 × 2 = 0 + 0.009 982 443 52;
  • 24) 0.009 982 443 52 × 2 = 0 + 0.019 964 887 04;
  • 25) 0.019 964 887 04 × 2 = 0 + 0.039 929 774 08;
  • 26) 0.039 929 774 08 × 2 = 0 + 0.079 859 548 16;
  • 27) 0.079 859 548 16 × 2 = 0 + 0.159 719 096 32;
  • 28) 0.159 719 096 32 × 2 = 0 + 0.319 438 192 64;
  • 29) 0.319 438 192 64 × 2 = 0 + 0.638 876 385 28;
  • 30) 0.638 876 385 28 × 2 = 1 + 0.277 752 770 56;
  • 31) 0.277 752 770 56 × 2 = 0 + 0.555 505 541 12;
  • 32) 0.555 505 541 12 × 2 = 1 + 0.111 011 082 24;
  • 33) 0.111 011 082 24 × 2 = 0 + 0.222 022 164 48;
  • 34) 0.222 022 164 48 × 2 = 0 + 0.444 044 328 96;
  • 35) 0.444 044 328 96 × 2 = 0 + 0.888 088 657 92;
  • 36) 0.888 088 657 92 × 2 = 1 + 0.776 177 315 84;
  • 37) 0.776 177 315 84 × 2 = 1 + 0.552 354 631 68;
  • 38) 0.552 354 631 68 × 2 = 1 + 0.104 709 263 36;
  • 39) 0.104 709 263 36 × 2 = 0 + 0.209 418 526 72;
  • 40) 0.209 418 526 72 × 2 = 0 + 0.418 837 053 44;
  • 41) 0.418 837 053 44 × 2 = 0 + 0.837 674 106 88;
  • 42) 0.837 674 106 88 × 2 = 1 + 0.675 348 213 76;
  • 43) 0.675 348 213 76 × 2 = 1 + 0.350 696 427 52;
  • 44) 0.350 696 427 52 × 2 = 0 + 0.701 392 855 04;
  • 45) 0.701 392 855 04 × 2 = 1 + 0.402 785 710 08;
  • 46) 0.402 785 710 08 × 2 = 0 + 0.805 571 420 16;
  • 47) 0.805 571 420 16 × 2 = 1 + 0.611 142 840 32;
  • 48) 0.611 142 840 32 × 2 = 1 + 0.222 285 680 64;
  • 49) 0.222 285 680 64 × 2 = 0 + 0.444 571 361 28;
  • 50) 0.444 571 361 28 × 2 = 0 + 0.889 142 722 56;
  • 51) 0.889 142 722 56 × 2 = 1 + 0.778 285 445 12;
  • 52) 0.778 285 445 12 × 2 = 1 + 0.556 570 890 24;
  • 53) 0.556 570 890 24 × 2 = 1 + 0.113 141 780 48;

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 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0101 0001 1100 0110 1011 0011 1(2)

6. Positive number before normalization:

0.000 000 001 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0101 0001 1100 0110 1011 0011 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 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0101 0001 1100 0110 1011 0011 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0101 0001 1100 0110 1011 0011 1(2) × 20 =

1.0100 0111 0001 1010 1100 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):
1.0100 0111 0001 1010 1100 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. 010 0011 1000 1101 0110 0111 =

010 0011 1000 1101 0110 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) =
010 0011 1000 1101 0110 0111


Decimal number -0.000 000 001 19 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0001 - 010 0011 1000 1101 0110 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