-0.000 000 025 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 025 7(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
-0.000 000 025 7(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:

|-0.000 000 025 7| = 0.000 000 025 7


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: 0.000 000 025 7.

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) 0.000 000 025 7 × 2 = 0 + 0.000 000 051 4;
  • 2) 0.000 000 051 4 × 2 = 0 + 0.000 000 102 8;
  • 3) 0.000 000 102 8 × 2 = 0 + 0.000 000 205 6;
  • 4) 0.000 000 205 6 × 2 = 0 + 0.000 000 411 2;
  • 5) 0.000 000 411 2 × 2 = 0 + 0.000 000 822 4;
  • 6) 0.000 000 822 4 × 2 = 0 + 0.000 001 644 8;
  • 7) 0.000 001 644 8 × 2 = 0 + 0.000 003 289 6;
  • 8) 0.000 003 289 6 × 2 = 0 + 0.000 006 579 2;
  • 9) 0.000 006 579 2 × 2 = 0 + 0.000 013 158 4;
  • 10) 0.000 013 158 4 × 2 = 0 + 0.000 026 316 8;
  • 11) 0.000 026 316 8 × 2 = 0 + 0.000 052 633 6;
  • 12) 0.000 052 633 6 × 2 = 0 + 0.000 105 267 2;
  • 13) 0.000 105 267 2 × 2 = 0 + 0.000 210 534 4;
  • 14) 0.000 210 534 4 × 2 = 0 + 0.000 421 068 8;
  • 15) 0.000 421 068 8 × 2 = 0 + 0.000 842 137 6;
  • 16) 0.000 842 137 6 × 2 = 0 + 0.001 684 275 2;
  • 17) 0.001 684 275 2 × 2 = 0 + 0.003 368 550 4;
  • 18) 0.003 368 550 4 × 2 = 0 + 0.006 737 100 8;
  • 19) 0.006 737 100 8 × 2 = 0 + 0.013 474 201 6;
  • 20) 0.013 474 201 6 × 2 = 0 + 0.026 948 403 2;
  • 21) 0.026 948 403 2 × 2 = 0 + 0.053 896 806 4;
  • 22) 0.053 896 806 4 × 2 = 0 + 0.107 793 612 8;
  • 23) 0.107 793 612 8 × 2 = 0 + 0.215 587 225 6;
  • 24) 0.215 587 225 6 × 2 = 0 + 0.431 174 451 2;
  • 25) 0.431 174 451 2 × 2 = 0 + 0.862 348 902 4;
  • 26) 0.862 348 902 4 × 2 = 1 + 0.724 697 804 8;
  • 27) 0.724 697 804 8 × 2 = 1 + 0.449 395 609 6;
  • 28) 0.449 395 609 6 × 2 = 0 + 0.898 791 219 2;
  • 29) 0.898 791 219 2 × 2 = 1 + 0.797 582 438 4;
  • 30) 0.797 582 438 4 × 2 = 1 + 0.595 164 876 8;
  • 31) 0.595 164 876 8 × 2 = 1 + 0.190 329 753 6;
  • 32) 0.190 329 753 6 × 2 = 0 + 0.380 659 507 2;
  • 33) 0.380 659 507 2 × 2 = 0 + 0.761 319 014 4;
  • 34) 0.761 319 014 4 × 2 = 1 + 0.522 638 028 8;
  • 35) 0.522 638 028 8 × 2 = 1 + 0.045 276 057 6;
  • 36) 0.045 276 057 6 × 2 = 0 + 0.090 552 115 2;
  • 37) 0.090 552 115 2 × 2 = 0 + 0.181 104 230 4;
  • 38) 0.181 104 230 4 × 2 = 0 + 0.362 208 460 8;
  • 39) 0.362 208 460 8 × 2 = 0 + 0.724 416 921 6;
  • 40) 0.724 416 921 6 × 2 = 1 + 0.448 833 843 2;
  • 41) 0.448 833 843 2 × 2 = 0 + 0.897 667 686 4;
  • 42) 0.897 667 686 4 × 2 = 1 + 0.795 335 372 8;
  • 43) 0.795 335 372 8 × 2 = 1 + 0.590 670 745 6;
  • 44) 0.590 670 745 6 × 2 = 1 + 0.181 341 491 2;
  • 45) 0.181 341 491 2 × 2 = 0 + 0.362 682 982 4;
  • 46) 0.362 682 982 4 × 2 = 0 + 0.725 365 964 8;
  • 47) 0.725 365 964 8 × 2 = 1 + 0.450 731 929 6;
  • 48) 0.450 731 929 6 × 2 = 0 + 0.901 463 859 2;
  • 49) 0.901 463 859 2 × 2 = 1 + 0.802 927 718 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:


0.000 000 025 7(10) =

0.0000 0000 0000 0000 0000 0000 0110 1110 0110 0001 0111 0010 1(2)

6. Positive number before normalization:

0.000 000 025 7(10) =

0.0000 0000 0000 0000 0000 0000 0110 1110 0110 0001 0111 0010 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 025 7(10) =

0.0000 0000 0000 0000 0000 0000 0110 1110 0110 0001 0111 0010 1(2) =

0.0000 0000 0000 0000 0000 0000 0110 1110 0110 0001 0111 0010 1(2) × 20 =

1.1011 1001 1000 0101 1100 101(2) × 2-26


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): -26

Mantissa (not normalized):
1.1011 1001 1000 0101 1100 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(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;
  • 101 ÷ 2 = 50 + 1;
  • 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) =

101(10) =

0110 0101(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 1100 1100 0010 1110 0101 =

101 1100 1100 0010 1110 0101


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 0101

Mantissa (23 bits) =
101 1100 1100 0010 1110 0101


Decimal number -0.000 000 025 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0101 - 101 1100 1100 0010 1110 0101

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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