-0.000 000 004 09 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 004 09(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
-0.000 000 004 09(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:

|-0.000 000 004 09| = 0.000 000 004 09


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: 0.000 000 004 09.

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) 0.000 000 004 09 × 2 = 0 + 0.000 000 008 18;
  • 2) 0.000 000 008 18 × 2 = 0 + 0.000 000 016 36;
  • 3) 0.000 000 016 36 × 2 = 0 + 0.000 000 032 72;
  • 4) 0.000 000 032 72 × 2 = 0 + 0.000 000 065 44;
  • 5) 0.000 000 065 44 × 2 = 0 + 0.000 000 130 88;
  • 6) 0.000 000 130 88 × 2 = 0 + 0.000 000 261 76;
  • 7) 0.000 000 261 76 × 2 = 0 + 0.000 000 523 52;
  • 8) 0.000 000 523 52 × 2 = 0 + 0.000 001 047 04;
  • 9) 0.000 001 047 04 × 2 = 0 + 0.000 002 094 08;
  • 10) 0.000 002 094 08 × 2 = 0 + 0.000 004 188 16;
  • 11) 0.000 004 188 16 × 2 = 0 + 0.000 008 376 32;
  • 12) 0.000 008 376 32 × 2 = 0 + 0.000 016 752 64;
  • 13) 0.000 016 752 64 × 2 = 0 + 0.000 033 505 28;
  • 14) 0.000 033 505 28 × 2 = 0 + 0.000 067 010 56;
  • 15) 0.000 067 010 56 × 2 = 0 + 0.000 134 021 12;
  • 16) 0.000 134 021 12 × 2 = 0 + 0.000 268 042 24;
  • 17) 0.000 268 042 24 × 2 = 0 + 0.000 536 084 48;
  • 18) 0.000 536 084 48 × 2 = 0 + 0.001 072 168 96;
  • 19) 0.001 072 168 96 × 2 = 0 + 0.002 144 337 92;
  • 20) 0.002 144 337 92 × 2 = 0 + 0.004 288 675 84;
  • 21) 0.004 288 675 84 × 2 = 0 + 0.008 577 351 68;
  • 22) 0.008 577 351 68 × 2 = 0 + 0.017 154 703 36;
  • 23) 0.017 154 703 36 × 2 = 0 + 0.034 309 406 72;
  • 24) 0.034 309 406 72 × 2 = 0 + 0.068 618 813 44;
  • 25) 0.068 618 813 44 × 2 = 0 + 0.137 237 626 88;
  • 26) 0.137 237 626 88 × 2 = 0 + 0.274 475 253 76;
  • 27) 0.274 475 253 76 × 2 = 0 + 0.548 950 507 52;
  • 28) 0.548 950 507 52 × 2 = 1 + 0.097 901 015 04;
  • 29) 0.097 901 015 04 × 2 = 0 + 0.195 802 030 08;
  • 30) 0.195 802 030 08 × 2 = 0 + 0.391 604 060 16;
  • 31) 0.391 604 060 16 × 2 = 0 + 0.783 208 120 32;
  • 32) 0.783 208 120 32 × 2 = 1 + 0.566 416 240 64;
  • 33) 0.566 416 240 64 × 2 = 1 + 0.132 832 481 28;
  • 34) 0.132 832 481 28 × 2 = 0 + 0.265 664 962 56;
  • 35) 0.265 664 962 56 × 2 = 0 + 0.531 329 925 12;
  • 36) 0.531 329 925 12 × 2 = 1 + 0.062 659 850 24;
  • 37) 0.062 659 850 24 × 2 = 0 + 0.125 319 700 48;
  • 38) 0.125 319 700 48 × 2 = 0 + 0.250 639 400 96;
  • 39) 0.250 639 400 96 × 2 = 0 + 0.501 278 801 92;
  • 40) 0.501 278 801 92 × 2 = 1 + 0.002 557 603 84;
  • 41) 0.002 557 603 84 × 2 = 0 + 0.005 115 207 68;
  • 42) 0.005 115 207 68 × 2 = 0 + 0.010 230 415 36;
  • 43) 0.010 230 415 36 × 2 = 0 + 0.020 460 830 72;
  • 44) 0.020 460 830 72 × 2 = 0 + 0.040 921 661 44;
  • 45) 0.040 921 661 44 × 2 = 0 + 0.081 843 322 88;
  • 46) 0.081 843 322 88 × 2 = 0 + 0.163 686 645 76;
  • 47) 0.163 686 645 76 × 2 = 0 + 0.327 373 291 52;
  • 48) 0.327 373 291 52 × 2 = 0 + 0.654 746 583 04;
  • 49) 0.654 746 583 04 × 2 = 1 + 0.309 493 166 08;
  • 50) 0.309 493 166 08 × 2 = 0 + 0.618 986 332 16;
  • 51) 0.618 986 332 16 × 2 = 1 + 0.237 972 664 32;

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:


0.000 000 004 09(10) =

0.0000 0000 0000 0000 0000 0000 0001 0001 1001 0001 0000 0000 101(2)

6. Positive number before normalization:

0.000 000 004 09(10) =

0.0000 0000 0000 0000 0000 0000 0001 0001 1001 0001 0000 0000 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 004 09(10) =

0.0000 0000 0000 0000 0000 0000 0001 0001 1001 0001 0000 0000 101(2) =

0.0000 0000 0000 0000 0000 0000 0001 0001 1001 0001 0000 0000 101(2) × 20 =

1.0001 1001 0001 0000 0000 101(2) × 2-28


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): -28

Mantissa (not normalized):
1.0001 1001 0001 0000 0000 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

99(10) =

0110 0011(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 1000 1000 0000 0101 =

000 1100 1000 1000 0000 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 0011

Mantissa (23 bits) =
000 1100 1000 1000 0000 0101


Decimal number -0.000 000 004 09 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0011 - 000 1100 1000 1000 0000 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