-0.000 000 016 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 016 3(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
-0.000 000 016 3(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:

|-0.000 000 016 3| = 0.000 000 016 3


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: 0.000 000 016 3.

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) 0.000 000 016 3 × 2 = 0 + 0.000 000 032 6;
  • 2) 0.000 000 032 6 × 2 = 0 + 0.000 000 065 2;
  • 3) 0.000 000 065 2 × 2 = 0 + 0.000 000 130 4;
  • 4) 0.000 000 130 4 × 2 = 0 + 0.000 000 260 8;
  • 5) 0.000 000 260 8 × 2 = 0 + 0.000 000 521 6;
  • 6) 0.000 000 521 6 × 2 = 0 + 0.000 001 043 2;
  • 7) 0.000 001 043 2 × 2 = 0 + 0.000 002 086 4;
  • 8) 0.000 002 086 4 × 2 = 0 + 0.000 004 172 8;
  • 9) 0.000 004 172 8 × 2 = 0 + 0.000 008 345 6;
  • 10) 0.000 008 345 6 × 2 = 0 + 0.000 016 691 2;
  • 11) 0.000 016 691 2 × 2 = 0 + 0.000 033 382 4;
  • 12) 0.000 033 382 4 × 2 = 0 + 0.000 066 764 8;
  • 13) 0.000 066 764 8 × 2 = 0 + 0.000 133 529 6;
  • 14) 0.000 133 529 6 × 2 = 0 + 0.000 267 059 2;
  • 15) 0.000 267 059 2 × 2 = 0 + 0.000 534 118 4;
  • 16) 0.000 534 118 4 × 2 = 0 + 0.001 068 236 8;
  • 17) 0.001 068 236 8 × 2 = 0 + 0.002 136 473 6;
  • 18) 0.002 136 473 6 × 2 = 0 + 0.004 272 947 2;
  • 19) 0.004 272 947 2 × 2 = 0 + 0.008 545 894 4;
  • 20) 0.008 545 894 4 × 2 = 0 + 0.017 091 788 8;
  • 21) 0.017 091 788 8 × 2 = 0 + 0.034 183 577 6;
  • 22) 0.034 183 577 6 × 2 = 0 + 0.068 367 155 2;
  • 23) 0.068 367 155 2 × 2 = 0 + 0.136 734 310 4;
  • 24) 0.136 734 310 4 × 2 = 0 + 0.273 468 620 8;
  • 25) 0.273 468 620 8 × 2 = 0 + 0.546 937 241 6;
  • 26) 0.546 937 241 6 × 2 = 1 + 0.093 874 483 2;
  • 27) 0.093 874 483 2 × 2 = 0 + 0.187 748 966 4;
  • 28) 0.187 748 966 4 × 2 = 0 + 0.375 497 932 8;
  • 29) 0.375 497 932 8 × 2 = 0 + 0.750 995 865 6;
  • 30) 0.750 995 865 6 × 2 = 1 + 0.501 991 731 2;
  • 31) 0.501 991 731 2 × 2 = 1 + 0.003 983 462 4;
  • 32) 0.003 983 462 4 × 2 = 0 + 0.007 966 924 8;
  • 33) 0.007 966 924 8 × 2 = 0 + 0.015 933 849 6;
  • 34) 0.015 933 849 6 × 2 = 0 + 0.031 867 699 2;
  • 35) 0.031 867 699 2 × 2 = 0 + 0.063 735 398 4;
  • 36) 0.063 735 398 4 × 2 = 0 + 0.127 470 796 8;
  • 37) 0.127 470 796 8 × 2 = 0 + 0.254 941 593 6;
  • 38) 0.254 941 593 6 × 2 = 0 + 0.509 883 187 2;
  • 39) 0.509 883 187 2 × 2 = 1 + 0.019 766 374 4;
  • 40) 0.019 766 374 4 × 2 = 0 + 0.039 532 748 8;
  • 41) 0.039 532 748 8 × 2 = 0 + 0.079 065 497 6;
  • 42) 0.079 065 497 6 × 2 = 0 + 0.158 130 995 2;
  • 43) 0.158 130 995 2 × 2 = 0 + 0.316 261 990 4;
  • 44) 0.316 261 990 4 × 2 = 0 + 0.632 523 980 8;
  • 45) 0.632 523 980 8 × 2 = 1 + 0.265 047 961 6;
  • 46) 0.265 047 961 6 × 2 = 0 + 0.530 095 923 2;
  • 47) 0.530 095 923 2 × 2 = 1 + 0.060 191 846 4;
  • 48) 0.060 191 846 4 × 2 = 0 + 0.120 383 692 8;
  • 49) 0.120 383 692 8 × 2 = 0 + 0.240 767 385 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:


0.000 000 016 3(10) =

0.0000 0000 0000 0000 0000 0000 0100 0110 0000 0010 0000 1010 0(2)

6. Positive number before normalization:

0.000 000 016 3(10) =

0.0000 0000 0000 0000 0000 0000 0100 0110 0000 0010 0000 1010 0(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:


0.000 000 016 3(10) =

0.0000 0000 0000 0000 0000 0000 0100 0110 0000 0010 0000 1010 0(2) =

0.0000 0000 0000 0000 0000 0000 0100 0110 0000 0010 0000 1010 0(2) × 20 =

1.0001 1000 0000 1000 0010 100(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):
1.0001 1000 0000 1000 0010 100


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. 000 1100 0000 0100 0001 0100 =

000 1100 0000 0100 0001 0100


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) =
000 1100 0000 0100 0001 0100


Decimal number -0.000 000 016 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0101 - 000 1100 0000 0100 0001 0100

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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