-0.000 000 019 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 019 8(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
-0.000 000 019 8(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:

|-0.000 000 019 8| = 0.000 000 019 8


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: 0.000 000 019 8.

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) 0.000 000 019 8 × 2 = 0 + 0.000 000 039 6;
  • 2) 0.000 000 039 6 × 2 = 0 + 0.000 000 079 2;
  • 3) 0.000 000 079 2 × 2 = 0 + 0.000 000 158 4;
  • 4) 0.000 000 158 4 × 2 = 0 + 0.000 000 316 8;
  • 5) 0.000 000 316 8 × 2 = 0 + 0.000 000 633 6;
  • 6) 0.000 000 633 6 × 2 = 0 + 0.000 001 267 2;
  • 7) 0.000 001 267 2 × 2 = 0 + 0.000 002 534 4;
  • 8) 0.000 002 534 4 × 2 = 0 + 0.000 005 068 8;
  • 9) 0.000 005 068 8 × 2 = 0 + 0.000 010 137 6;
  • 10) 0.000 010 137 6 × 2 = 0 + 0.000 020 275 2;
  • 11) 0.000 020 275 2 × 2 = 0 + 0.000 040 550 4;
  • 12) 0.000 040 550 4 × 2 = 0 + 0.000 081 100 8;
  • 13) 0.000 081 100 8 × 2 = 0 + 0.000 162 201 6;
  • 14) 0.000 162 201 6 × 2 = 0 + 0.000 324 403 2;
  • 15) 0.000 324 403 2 × 2 = 0 + 0.000 648 806 4;
  • 16) 0.000 648 806 4 × 2 = 0 + 0.001 297 612 8;
  • 17) 0.001 297 612 8 × 2 = 0 + 0.002 595 225 6;
  • 18) 0.002 595 225 6 × 2 = 0 + 0.005 190 451 2;
  • 19) 0.005 190 451 2 × 2 = 0 + 0.010 380 902 4;
  • 20) 0.010 380 902 4 × 2 = 0 + 0.020 761 804 8;
  • 21) 0.020 761 804 8 × 2 = 0 + 0.041 523 609 6;
  • 22) 0.041 523 609 6 × 2 = 0 + 0.083 047 219 2;
  • 23) 0.083 047 219 2 × 2 = 0 + 0.166 094 438 4;
  • 24) 0.166 094 438 4 × 2 = 0 + 0.332 188 876 8;
  • 25) 0.332 188 876 8 × 2 = 0 + 0.664 377 753 6;
  • 26) 0.664 377 753 6 × 2 = 1 + 0.328 755 507 2;
  • 27) 0.328 755 507 2 × 2 = 0 + 0.657 511 014 4;
  • 28) 0.657 511 014 4 × 2 = 1 + 0.315 022 028 8;
  • 29) 0.315 022 028 8 × 2 = 0 + 0.630 044 057 6;
  • 30) 0.630 044 057 6 × 2 = 1 + 0.260 088 115 2;
  • 31) 0.260 088 115 2 × 2 = 0 + 0.520 176 230 4;
  • 32) 0.520 176 230 4 × 2 = 1 + 0.040 352 460 8;
  • 33) 0.040 352 460 8 × 2 = 0 + 0.080 704 921 6;
  • 34) 0.080 704 921 6 × 2 = 0 + 0.161 409 843 2;
  • 35) 0.161 409 843 2 × 2 = 0 + 0.322 819 686 4;
  • 36) 0.322 819 686 4 × 2 = 0 + 0.645 639 372 8;
  • 37) 0.645 639 372 8 × 2 = 1 + 0.291 278 745 6;
  • 38) 0.291 278 745 6 × 2 = 0 + 0.582 557 491 2;
  • 39) 0.582 557 491 2 × 2 = 1 + 0.165 114 982 4;
  • 40) 0.165 114 982 4 × 2 = 0 + 0.330 229 964 8;
  • 41) 0.330 229 964 8 × 2 = 0 + 0.660 459 929 6;
  • 42) 0.660 459 929 6 × 2 = 1 + 0.320 919 859 2;
  • 43) 0.320 919 859 2 × 2 = 0 + 0.641 839 718 4;
  • 44) 0.641 839 718 4 × 2 = 1 + 0.283 679 436 8;
  • 45) 0.283 679 436 8 × 2 = 0 + 0.567 358 873 6;
  • 46) 0.567 358 873 6 × 2 = 1 + 0.134 717 747 2;
  • 47) 0.134 717 747 2 × 2 = 0 + 0.269 435 494 4;
  • 48) 0.269 435 494 4 × 2 = 0 + 0.538 870 988 8;
  • 49) 0.538 870 988 8 × 2 = 1 + 0.077 741 977 6;

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:


0.000 000 019 8(10) =

0.0000 0000 0000 0000 0000 0000 0101 0101 0000 1010 0101 0100 1(2)

6. Positive number before normalization:

0.000 000 019 8(10) =

0.0000 0000 0000 0000 0000 0000 0101 0101 0000 1010 0101 0100 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:


0.000 000 019 8(10) =

0.0000 0000 0000 0000 0000 0000 0101 0101 0000 1010 0101 0100 1(2) =

0.0000 0000 0000 0000 0000 0000 0101 0101 0000 1010 0101 0100 1(2) × 20 =

1.0101 0100 0010 1001 0101 001(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):
1.0101 0100 0010 1001 0101 001


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. 010 1010 0001 0100 1010 1001 =

010 1010 0001 0100 1010 1001


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) =
010 1010 0001 0100 1010 1001


Decimal number -0.000 000 019 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0101 - 010 1010 0001 0100 1010 1001

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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