0.000 000 110 008 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 008 9(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
0.000 000 110 008 9(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 110 008 9.

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) 0.000 000 110 008 9 × 2 = 0 + 0.000 000 220 017 8;
  • 2) 0.000 000 220 017 8 × 2 = 0 + 0.000 000 440 035 6;
  • 3) 0.000 000 440 035 6 × 2 = 0 + 0.000 000 880 071 2;
  • 4) 0.000 000 880 071 2 × 2 = 0 + 0.000 001 760 142 4;
  • 5) 0.000 001 760 142 4 × 2 = 0 + 0.000 003 520 284 8;
  • 6) 0.000 003 520 284 8 × 2 = 0 + 0.000 007 040 569 6;
  • 7) 0.000 007 040 569 6 × 2 = 0 + 0.000 014 081 139 2;
  • 8) 0.000 014 081 139 2 × 2 = 0 + 0.000 028 162 278 4;
  • 9) 0.000 028 162 278 4 × 2 = 0 + 0.000 056 324 556 8;
  • 10) 0.000 056 324 556 8 × 2 = 0 + 0.000 112 649 113 6;
  • 11) 0.000 112 649 113 6 × 2 = 0 + 0.000 225 298 227 2;
  • 12) 0.000 225 298 227 2 × 2 = 0 + 0.000 450 596 454 4;
  • 13) 0.000 450 596 454 4 × 2 = 0 + 0.000 901 192 908 8;
  • 14) 0.000 901 192 908 8 × 2 = 0 + 0.001 802 385 817 6;
  • 15) 0.001 802 385 817 6 × 2 = 0 + 0.003 604 771 635 2;
  • 16) 0.003 604 771 635 2 × 2 = 0 + 0.007 209 543 270 4;
  • 17) 0.007 209 543 270 4 × 2 = 0 + 0.014 419 086 540 8;
  • 18) 0.014 419 086 540 8 × 2 = 0 + 0.028 838 173 081 6;
  • 19) 0.028 838 173 081 6 × 2 = 0 + 0.057 676 346 163 2;
  • 20) 0.057 676 346 163 2 × 2 = 0 + 0.115 352 692 326 4;
  • 21) 0.115 352 692 326 4 × 2 = 0 + 0.230 705 384 652 8;
  • 22) 0.230 705 384 652 8 × 2 = 0 + 0.461 410 769 305 6;
  • 23) 0.461 410 769 305 6 × 2 = 0 + 0.922 821 538 611 2;
  • 24) 0.922 821 538 611 2 × 2 = 1 + 0.845 643 077 222 4;
  • 25) 0.845 643 077 222 4 × 2 = 1 + 0.691 286 154 444 8;
  • 26) 0.691 286 154 444 8 × 2 = 1 + 0.382 572 308 889 6;
  • 27) 0.382 572 308 889 6 × 2 = 0 + 0.765 144 617 779 2;
  • 28) 0.765 144 617 779 2 × 2 = 1 + 0.530 289 235 558 4;
  • 29) 0.530 289 235 558 4 × 2 = 1 + 0.060 578 471 116 8;
  • 30) 0.060 578 471 116 8 × 2 = 0 + 0.121 156 942 233 6;
  • 31) 0.121 156 942 233 6 × 2 = 0 + 0.242 313 884 467 2;
  • 32) 0.242 313 884 467 2 × 2 = 0 + 0.484 627 768 934 4;
  • 33) 0.484 627 768 934 4 × 2 = 0 + 0.969 255 537 868 8;
  • 34) 0.969 255 537 868 8 × 2 = 1 + 0.938 511 075 737 6;
  • 35) 0.938 511 075 737 6 × 2 = 1 + 0.877 022 151 475 2;
  • 36) 0.877 022 151 475 2 × 2 = 1 + 0.754 044 302 950 4;
  • 37) 0.754 044 302 950 4 × 2 = 1 + 0.508 088 605 900 8;
  • 38) 0.508 088 605 900 8 × 2 = 1 + 0.016 177 211 801 6;
  • 39) 0.016 177 211 801 6 × 2 = 0 + 0.032 354 423 603 2;
  • 40) 0.032 354 423 603 2 × 2 = 0 + 0.064 708 847 206 4;
  • 41) 0.064 708 847 206 4 × 2 = 0 + 0.129 417 694 412 8;
  • 42) 0.129 417 694 412 8 × 2 = 0 + 0.258 835 388 825 6;
  • 43) 0.258 835 388 825 6 × 2 = 0 + 0.517 670 777 651 2;
  • 44) 0.517 670 777 651 2 × 2 = 1 + 0.035 341 555 302 4;
  • 45) 0.035 341 555 302 4 × 2 = 0 + 0.070 683 110 604 8;
  • 46) 0.070 683 110 604 8 × 2 = 0 + 0.141 366 221 209 6;
  • 47) 0.141 366 221 209 6 × 2 = 0 + 0.282 732 442 419 2;

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).


4. 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:


0.000 000 110 008 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 000(2)

5. Positive number before normalization:

0.000 000 110 008 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 000(2)

6. 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:


0.000 000 110 008 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 000(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 000(2) × 20 =

1.1101 1000 0111 1100 0001 000(2) × 2-24


7. 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 0 (a positive number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1000 0111 1100 0001 000


8. 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)


9. 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;

10. 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)


11. 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. 110 1100 0011 1110 0000 1000 =

110 1100 0011 1110 0000 1000


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0111

Mantissa (23 bits) =
110 1100 0011 1110 0000 1000


Decimal number 0.000 000 110 008 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 0000 1000

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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