-0.000 000 099 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 099(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
-0.000 000 099(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:

|-0.000 000 099| = 0.000 000 099


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: 0.000 000 099.

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) 0.000 000 099 × 2 = 0 + 0.000 000 198;
  • 2) 0.000 000 198 × 2 = 0 + 0.000 000 396;
  • 3) 0.000 000 396 × 2 = 0 + 0.000 000 792;
  • 4) 0.000 000 792 × 2 = 0 + 0.000 001 584;
  • 5) 0.000 001 584 × 2 = 0 + 0.000 003 168;
  • 6) 0.000 003 168 × 2 = 0 + 0.000 006 336;
  • 7) 0.000 006 336 × 2 = 0 + 0.000 012 672;
  • 8) 0.000 012 672 × 2 = 0 + 0.000 025 344;
  • 9) 0.000 025 344 × 2 = 0 + 0.000 050 688;
  • 10) 0.000 050 688 × 2 = 0 + 0.000 101 376;
  • 11) 0.000 101 376 × 2 = 0 + 0.000 202 752;
  • 12) 0.000 202 752 × 2 = 0 + 0.000 405 504;
  • 13) 0.000 405 504 × 2 = 0 + 0.000 811 008;
  • 14) 0.000 811 008 × 2 = 0 + 0.001 622 016;
  • 15) 0.001 622 016 × 2 = 0 + 0.003 244 032;
  • 16) 0.003 244 032 × 2 = 0 + 0.006 488 064;
  • 17) 0.006 488 064 × 2 = 0 + 0.012 976 128;
  • 18) 0.012 976 128 × 2 = 0 + 0.025 952 256;
  • 19) 0.025 952 256 × 2 = 0 + 0.051 904 512;
  • 20) 0.051 904 512 × 2 = 0 + 0.103 809 024;
  • 21) 0.103 809 024 × 2 = 0 + 0.207 618 048;
  • 22) 0.207 618 048 × 2 = 0 + 0.415 236 096;
  • 23) 0.415 236 096 × 2 = 0 + 0.830 472 192;
  • 24) 0.830 472 192 × 2 = 1 + 0.660 944 384;
  • 25) 0.660 944 384 × 2 = 1 + 0.321 888 768;
  • 26) 0.321 888 768 × 2 = 0 + 0.643 777 536;
  • 27) 0.643 777 536 × 2 = 1 + 0.287 555 072;
  • 28) 0.287 555 072 × 2 = 0 + 0.575 110 144;
  • 29) 0.575 110 144 × 2 = 1 + 0.150 220 288;
  • 30) 0.150 220 288 × 2 = 0 + 0.300 440 576;
  • 31) 0.300 440 576 × 2 = 0 + 0.600 881 152;
  • 32) 0.600 881 152 × 2 = 1 + 0.201 762 304;
  • 33) 0.201 762 304 × 2 = 0 + 0.403 524 608;
  • 34) 0.403 524 608 × 2 = 0 + 0.807 049 216;
  • 35) 0.807 049 216 × 2 = 1 + 0.614 098 432;
  • 36) 0.614 098 432 × 2 = 1 + 0.228 196 864;
  • 37) 0.228 196 864 × 2 = 0 + 0.456 393 728;
  • 38) 0.456 393 728 × 2 = 0 + 0.912 787 456;
  • 39) 0.912 787 456 × 2 = 1 + 0.825 574 912;
  • 40) 0.825 574 912 × 2 = 1 + 0.651 149 824;
  • 41) 0.651 149 824 × 2 = 1 + 0.302 299 648;
  • 42) 0.302 299 648 × 2 = 0 + 0.604 599 296;
  • 43) 0.604 599 296 × 2 = 1 + 0.209 198 592;
  • 44) 0.209 198 592 × 2 = 0 + 0.418 397 184;
  • 45) 0.418 397 184 × 2 = 0 + 0.836 794 368;
  • 46) 0.836 794 368 × 2 = 1 + 0.673 588 736;
  • 47) 0.673 588 736 × 2 = 1 + 0.347 177 472;

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:


0.000 000 099(10) =

0.0000 0000 0000 0000 0000 0001 1010 1001 0011 0011 1010 011(2)

6. Positive number before normalization:

0.000 000 099(10) =

0.0000 0000 0000 0000 0000 0001 1010 1001 0011 0011 1010 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 099(10) =

0.0000 0000 0000 0000 0000 0001 1010 1001 0011 0011 1010 011(2) =

0.0000 0000 0000 0000 0000 0001 1010 1001 0011 0011 1010 011(2) × 20 =

1.1010 1001 0011 0011 1010 011(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1010 1001 0011 0011 1010 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 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) =

103(10) =

0110 0111(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 0100 1001 1001 1101 0011 =

101 0100 1001 1001 1101 0011


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 0111

Mantissa (23 bits) =
101 0100 1001 1001 1101 0011


Decimal number -0.000 000 099 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 101 0100 1001 1001 1101 0011

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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