-0.000 000 78 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 78(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 78(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 78| = 0.000 000 78


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 78.

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 78 × 2 = 0 + 0.000 001 56;
  • 2) 0.000 001 56 × 2 = 0 + 0.000 003 12;
  • 3) 0.000 003 12 × 2 = 0 + 0.000 006 24;
  • 4) 0.000 006 24 × 2 = 0 + 0.000 012 48;
  • 5) 0.000 012 48 × 2 = 0 + 0.000 024 96;
  • 6) 0.000 024 96 × 2 = 0 + 0.000 049 92;
  • 7) 0.000 049 92 × 2 = 0 + 0.000 099 84;
  • 8) 0.000 099 84 × 2 = 0 + 0.000 199 68;
  • 9) 0.000 199 68 × 2 = 0 + 0.000 399 36;
  • 10) 0.000 399 36 × 2 = 0 + 0.000 798 72;
  • 11) 0.000 798 72 × 2 = 0 + 0.001 597 44;
  • 12) 0.001 597 44 × 2 = 0 + 0.003 194 88;
  • 13) 0.003 194 88 × 2 = 0 + 0.006 389 76;
  • 14) 0.006 389 76 × 2 = 0 + 0.012 779 52;
  • 15) 0.012 779 52 × 2 = 0 + 0.025 559 04;
  • 16) 0.025 559 04 × 2 = 0 + 0.051 118 08;
  • 17) 0.051 118 08 × 2 = 0 + 0.102 236 16;
  • 18) 0.102 236 16 × 2 = 0 + 0.204 472 32;
  • 19) 0.204 472 32 × 2 = 0 + 0.408 944 64;
  • 20) 0.408 944 64 × 2 = 0 + 0.817 889 28;
  • 21) 0.817 889 28 × 2 = 1 + 0.635 778 56;
  • 22) 0.635 778 56 × 2 = 1 + 0.271 557 12;
  • 23) 0.271 557 12 × 2 = 0 + 0.543 114 24;
  • 24) 0.543 114 24 × 2 = 1 + 0.086 228 48;
  • 25) 0.086 228 48 × 2 = 0 + 0.172 456 96;
  • 26) 0.172 456 96 × 2 = 0 + 0.344 913 92;
  • 27) 0.344 913 92 × 2 = 0 + 0.689 827 84;
  • 28) 0.689 827 84 × 2 = 1 + 0.379 655 68;
  • 29) 0.379 655 68 × 2 = 0 + 0.759 311 36;
  • 30) 0.759 311 36 × 2 = 1 + 0.518 622 72;
  • 31) 0.518 622 72 × 2 = 1 + 0.037 245 44;
  • 32) 0.037 245 44 × 2 = 0 + 0.074 490 88;
  • 33) 0.074 490 88 × 2 = 0 + 0.148 981 76;
  • 34) 0.148 981 76 × 2 = 0 + 0.297 963 52;
  • 35) 0.297 963 52 × 2 = 0 + 0.595 927 04;
  • 36) 0.595 927 04 × 2 = 1 + 0.191 854 08;
  • 37) 0.191 854 08 × 2 = 0 + 0.383 708 16;
  • 38) 0.383 708 16 × 2 = 0 + 0.767 416 32;
  • 39) 0.767 416 32 × 2 = 1 + 0.534 832 64;
  • 40) 0.534 832 64 × 2 = 1 + 0.069 665 28;
  • 41) 0.069 665 28 × 2 = 0 + 0.139 330 56;
  • 42) 0.139 330 56 × 2 = 0 + 0.278 661 12;
  • 43) 0.278 661 12 × 2 = 0 + 0.557 322 24;
  • 44) 0.557 322 24 × 2 = 1 + 0.114 644 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 78(10) =

0.0000 0000 0000 0000 0000 1101 0001 0110 0001 0011 0001(2)

6. Positive number before normalization:

0.000 000 78(10) =

0.0000 0000 0000 0000 0000 1101 0001 0110 0001 0011 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 78(10) =

0.0000 0000 0000 0000 0000 1101 0001 0110 0001 0011 0001(2) =

0.0000 0000 0000 0000 0000 1101 0001 0110 0001 0011 0001(2) × 20 =

1.1010 0010 1100 0010 0110 001(2) × 2-21


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): -21

Mantissa (not normalized):
1.1010 0010 1100 0010 0110 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

106(10) =

0110 1010(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. 101 0001 0110 0001 0011 0001 =

101 0001 0110 0001 0011 0001


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 1010

Mantissa (23 bits) =
101 0001 0110 0001 0011 0001


Decimal number -0.000 000 78 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1010 - 101 0001 0110 0001 0011 0001

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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