0.000 000 110 009 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 009 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
0.000 000 110 009 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. 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: 0.000 000 110 009 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) 0.000 000 110 009 3 × 2 = 0 + 0.000 000 220 018 6;
  • 2) 0.000 000 220 018 6 × 2 = 0 + 0.000 000 440 037 2;
  • 3) 0.000 000 440 037 2 × 2 = 0 + 0.000 000 880 074 4;
  • 4) 0.000 000 880 074 4 × 2 = 0 + 0.000 001 760 148 8;
  • 5) 0.000 001 760 148 8 × 2 = 0 + 0.000 003 520 297 6;
  • 6) 0.000 003 520 297 6 × 2 = 0 + 0.000 007 040 595 2;
  • 7) 0.000 007 040 595 2 × 2 = 0 + 0.000 014 081 190 4;
  • 8) 0.000 014 081 190 4 × 2 = 0 + 0.000 028 162 380 8;
  • 9) 0.000 028 162 380 8 × 2 = 0 + 0.000 056 324 761 6;
  • 10) 0.000 056 324 761 6 × 2 = 0 + 0.000 112 649 523 2;
  • 11) 0.000 112 649 523 2 × 2 = 0 + 0.000 225 299 046 4;
  • 12) 0.000 225 299 046 4 × 2 = 0 + 0.000 450 598 092 8;
  • 13) 0.000 450 598 092 8 × 2 = 0 + 0.000 901 196 185 6;
  • 14) 0.000 901 196 185 6 × 2 = 0 + 0.001 802 392 371 2;
  • 15) 0.001 802 392 371 2 × 2 = 0 + 0.003 604 784 742 4;
  • 16) 0.003 604 784 742 4 × 2 = 0 + 0.007 209 569 484 8;
  • 17) 0.007 209 569 484 8 × 2 = 0 + 0.014 419 138 969 6;
  • 18) 0.014 419 138 969 6 × 2 = 0 + 0.028 838 277 939 2;
  • 19) 0.028 838 277 939 2 × 2 = 0 + 0.057 676 555 878 4;
  • 20) 0.057 676 555 878 4 × 2 = 0 + 0.115 353 111 756 8;
  • 21) 0.115 353 111 756 8 × 2 = 0 + 0.230 706 223 513 6;
  • 22) 0.230 706 223 513 6 × 2 = 0 + 0.461 412 447 027 2;
  • 23) 0.461 412 447 027 2 × 2 = 0 + 0.922 824 894 054 4;
  • 24) 0.922 824 894 054 4 × 2 = 1 + 0.845 649 788 108 8;
  • 25) 0.845 649 788 108 8 × 2 = 1 + 0.691 299 576 217 6;
  • 26) 0.691 299 576 217 6 × 2 = 1 + 0.382 599 152 435 2;
  • 27) 0.382 599 152 435 2 × 2 = 0 + 0.765 198 304 870 4;
  • 28) 0.765 198 304 870 4 × 2 = 1 + 0.530 396 609 740 8;
  • 29) 0.530 396 609 740 8 × 2 = 1 + 0.060 793 219 481 6;
  • 30) 0.060 793 219 481 6 × 2 = 0 + 0.121 586 438 963 2;
  • 31) 0.121 586 438 963 2 × 2 = 0 + 0.243 172 877 926 4;
  • 32) 0.243 172 877 926 4 × 2 = 0 + 0.486 345 755 852 8;
  • 33) 0.486 345 755 852 8 × 2 = 0 + 0.972 691 511 705 6;
  • 34) 0.972 691 511 705 6 × 2 = 1 + 0.945 383 023 411 2;
  • 35) 0.945 383 023 411 2 × 2 = 1 + 0.890 766 046 822 4;
  • 36) 0.890 766 046 822 4 × 2 = 1 + 0.781 532 093 644 8;
  • 37) 0.781 532 093 644 8 × 2 = 1 + 0.563 064 187 289 6;
  • 38) 0.563 064 187 289 6 × 2 = 1 + 0.126 128 374 579 2;
  • 39) 0.126 128 374 579 2 × 2 = 0 + 0.252 256 749 158 4;
  • 40) 0.252 256 749 158 4 × 2 = 0 + 0.504 513 498 316 8;
  • 41) 0.504 513 498 316 8 × 2 = 1 + 0.009 026 996 633 6;
  • 42) 0.009 026 996 633 6 × 2 = 0 + 0.018 053 993 267 2;
  • 43) 0.018 053 993 267 2 × 2 = 0 + 0.036 107 986 534 4;
  • 44) 0.036 107 986 534 4 × 2 = 0 + 0.072 215 973 068 8;
  • 45) 0.072 215 973 068 8 × 2 = 0 + 0.144 431 946 137 6;
  • 46) 0.144 431 946 137 6 × 2 = 0 + 0.288 863 892 275 2;
  • 47) 0.288 863 892 275 2 × 2 = 0 + 0.577 727 784 550 4;

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:


0.000 000 110 009 3(10) =

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

5. Positive number before normalization:

0.000 000 110 009 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 1000 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:


0.000 000 110 009 3(10) =

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

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

1.1101 1000 0111 1100 1000 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 1000 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 0100 0000 =

110 1100 0011 1110 0100 0000


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 0100 0000


Decimal number 0.000 000 110 009 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 0100 0000

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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