-0.000 000 012 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 012 1(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
-0.000 000 012 1(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:

|-0.000 000 012 1| = 0.000 000 012 1


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: 0.000 000 012 1.

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) 0.000 000 012 1 × 2 = 0 + 0.000 000 024 2;
  • 2) 0.000 000 024 2 × 2 = 0 + 0.000 000 048 4;
  • 3) 0.000 000 048 4 × 2 = 0 + 0.000 000 096 8;
  • 4) 0.000 000 096 8 × 2 = 0 + 0.000 000 193 6;
  • 5) 0.000 000 193 6 × 2 = 0 + 0.000 000 387 2;
  • 6) 0.000 000 387 2 × 2 = 0 + 0.000 000 774 4;
  • 7) 0.000 000 774 4 × 2 = 0 + 0.000 001 548 8;
  • 8) 0.000 001 548 8 × 2 = 0 + 0.000 003 097 6;
  • 9) 0.000 003 097 6 × 2 = 0 + 0.000 006 195 2;
  • 10) 0.000 006 195 2 × 2 = 0 + 0.000 012 390 4;
  • 11) 0.000 012 390 4 × 2 = 0 + 0.000 024 780 8;
  • 12) 0.000 024 780 8 × 2 = 0 + 0.000 049 561 6;
  • 13) 0.000 049 561 6 × 2 = 0 + 0.000 099 123 2;
  • 14) 0.000 099 123 2 × 2 = 0 + 0.000 198 246 4;
  • 15) 0.000 198 246 4 × 2 = 0 + 0.000 396 492 8;
  • 16) 0.000 396 492 8 × 2 = 0 + 0.000 792 985 6;
  • 17) 0.000 792 985 6 × 2 = 0 + 0.001 585 971 2;
  • 18) 0.001 585 971 2 × 2 = 0 + 0.003 171 942 4;
  • 19) 0.003 171 942 4 × 2 = 0 + 0.006 343 884 8;
  • 20) 0.006 343 884 8 × 2 = 0 + 0.012 687 769 6;
  • 21) 0.012 687 769 6 × 2 = 0 + 0.025 375 539 2;
  • 22) 0.025 375 539 2 × 2 = 0 + 0.050 751 078 4;
  • 23) 0.050 751 078 4 × 2 = 0 + 0.101 502 156 8;
  • 24) 0.101 502 156 8 × 2 = 0 + 0.203 004 313 6;
  • 25) 0.203 004 313 6 × 2 = 0 + 0.406 008 627 2;
  • 26) 0.406 008 627 2 × 2 = 0 + 0.812 017 254 4;
  • 27) 0.812 017 254 4 × 2 = 1 + 0.624 034 508 8;
  • 28) 0.624 034 508 8 × 2 = 1 + 0.248 069 017 6;
  • 29) 0.248 069 017 6 × 2 = 0 + 0.496 138 035 2;
  • 30) 0.496 138 035 2 × 2 = 0 + 0.992 276 070 4;
  • 31) 0.992 276 070 4 × 2 = 1 + 0.984 552 140 8;
  • 32) 0.984 552 140 8 × 2 = 1 + 0.969 104 281 6;
  • 33) 0.969 104 281 6 × 2 = 1 + 0.938 208 563 2;
  • 34) 0.938 208 563 2 × 2 = 1 + 0.876 417 126 4;
  • 35) 0.876 417 126 4 × 2 = 1 + 0.752 834 252 8;
  • 36) 0.752 834 252 8 × 2 = 1 + 0.505 668 505 6;
  • 37) 0.505 668 505 6 × 2 = 1 + 0.011 337 011 2;
  • 38) 0.011 337 011 2 × 2 = 0 + 0.022 674 022 4;
  • 39) 0.022 674 022 4 × 2 = 0 + 0.045 348 044 8;
  • 40) 0.045 348 044 8 × 2 = 0 + 0.090 696 089 6;
  • 41) 0.090 696 089 6 × 2 = 0 + 0.181 392 179 2;
  • 42) 0.181 392 179 2 × 2 = 0 + 0.362 784 358 4;
  • 43) 0.362 784 358 4 × 2 = 0 + 0.725 568 716 8;
  • 44) 0.725 568 716 8 × 2 = 1 + 0.451 137 433 6;
  • 45) 0.451 137 433 6 × 2 = 0 + 0.902 274 867 2;
  • 46) 0.902 274 867 2 × 2 = 1 + 0.804 549 734 4;
  • 47) 0.804 549 734 4 × 2 = 1 + 0.609 099 468 8;
  • 48) 0.609 099 468 8 × 2 = 1 + 0.218 198 937 6;
  • 49) 0.218 198 937 6 × 2 = 0 + 0.436 397 875 2;
  • 50) 0.436 397 875 2 × 2 = 0 + 0.872 795 750 4;

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:


0.000 000 012 1(10) =

0.0000 0000 0000 0000 0000 0000 0011 0011 1111 1000 0001 0111 00(2)

6. Positive number before normalization:

0.000 000 012 1(10) =

0.0000 0000 0000 0000 0000 0000 0011 0011 1111 1000 0001 0111 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 012 1(10) =

0.0000 0000 0000 0000 0000 0000 0011 0011 1111 1000 0001 0111 00(2) =

0.0000 0000 0000 0000 0000 0000 0011 0011 1111 1000 0001 0111 00(2) × 20 =

1.1001 1111 1100 0000 1011 100(2) × 2-27


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): -27

Mantissa (not normalized):
1.1001 1111 1100 0000 1011 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(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;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

100(10) =

0110 0100(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. 100 1111 1110 0000 0101 1100 =

100 1111 1110 0000 0101 1100


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 0100

Mantissa (23 bits) =
100 1111 1110 0000 0101 1100


Decimal number -0.000 000 012 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0100 - 100 1111 1110 0000 0101 1100

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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