0.000 000 110 002 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 002 6(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
0.000 000 110 002 6(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: 0.000 000 110 002 6.

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) 0.000 000 110 002 6 × 2 = 0 + 0.000 000 220 005 2;
  • 2) 0.000 000 220 005 2 × 2 = 0 + 0.000 000 440 010 4;
  • 3) 0.000 000 440 010 4 × 2 = 0 + 0.000 000 880 020 8;
  • 4) 0.000 000 880 020 8 × 2 = 0 + 0.000 001 760 041 6;
  • 5) 0.000 001 760 041 6 × 2 = 0 + 0.000 003 520 083 2;
  • 6) 0.000 003 520 083 2 × 2 = 0 + 0.000 007 040 166 4;
  • 7) 0.000 007 040 166 4 × 2 = 0 + 0.000 014 080 332 8;
  • 8) 0.000 014 080 332 8 × 2 = 0 + 0.000 028 160 665 6;
  • 9) 0.000 028 160 665 6 × 2 = 0 + 0.000 056 321 331 2;
  • 10) 0.000 056 321 331 2 × 2 = 0 + 0.000 112 642 662 4;
  • 11) 0.000 112 642 662 4 × 2 = 0 + 0.000 225 285 324 8;
  • 12) 0.000 225 285 324 8 × 2 = 0 + 0.000 450 570 649 6;
  • 13) 0.000 450 570 649 6 × 2 = 0 + 0.000 901 141 299 2;
  • 14) 0.000 901 141 299 2 × 2 = 0 + 0.001 802 282 598 4;
  • 15) 0.001 802 282 598 4 × 2 = 0 + 0.003 604 565 196 8;
  • 16) 0.003 604 565 196 8 × 2 = 0 + 0.007 209 130 393 6;
  • 17) 0.007 209 130 393 6 × 2 = 0 + 0.014 418 260 787 2;
  • 18) 0.014 418 260 787 2 × 2 = 0 + 0.028 836 521 574 4;
  • 19) 0.028 836 521 574 4 × 2 = 0 + 0.057 673 043 148 8;
  • 20) 0.057 673 043 148 8 × 2 = 0 + 0.115 346 086 297 6;
  • 21) 0.115 346 086 297 6 × 2 = 0 + 0.230 692 172 595 2;
  • 22) 0.230 692 172 595 2 × 2 = 0 + 0.461 384 345 190 4;
  • 23) 0.461 384 345 190 4 × 2 = 0 + 0.922 768 690 380 8;
  • 24) 0.922 768 690 380 8 × 2 = 1 + 0.845 537 380 761 6;
  • 25) 0.845 537 380 761 6 × 2 = 1 + 0.691 074 761 523 2;
  • 26) 0.691 074 761 523 2 × 2 = 1 + 0.382 149 523 046 4;
  • 27) 0.382 149 523 046 4 × 2 = 0 + 0.764 299 046 092 8;
  • 28) 0.764 299 046 092 8 × 2 = 1 + 0.528 598 092 185 6;
  • 29) 0.528 598 092 185 6 × 2 = 1 + 0.057 196 184 371 2;
  • 30) 0.057 196 184 371 2 × 2 = 0 + 0.114 392 368 742 4;
  • 31) 0.114 392 368 742 4 × 2 = 0 + 0.228 784 737 484 8;
  • 32) 0.228 784 737 484 8 × 2 = 0 + 0.457 569 474 969 6;
  • 33) 0.457 569 474 969 6 × 2 = 0 + 0.915 138 949 939 2;
  • 34) 0.915 138 949 939 2 × 2 = 1 + 0.830 277 899 878 4;
  • 35) 0.830 277 899 878 4 × 2 = 1 + 0.660 555 799 756 8;
  • 36) 0.660 555 799 756 8 × 2 = 1 + 0.321 111 599 513 6;
  • 37) 0.321 111 599 513 6 × 2 = 0 + 0.642 223 199 027 2;
  • 38) 0.642 223 199 027 2 × 2 = 1 + 0.284 446 398 054 4;
  • 39) 0.284 446 398 054 4 × 2 = 0 + 0.568 892 796 108 8;
  • 40) 0.568 892 796 108 8 × 2 = 1 + 0.137 785 592 217 6;
  • 41) 0.137 785 592 217 6 × 2 = 0 + 0.275 571 184 435 2;
  • 42) 0.275 571 184 435 2 × 2 = 0 + 0.551 142 368 870 4;
  • 43) 0.551 142 368 870 4 × 2 = 1 + 0.102 284 737 740 8;
  • 44) 0.102 284 737 740 8 × 2 = 0 + 0.204 569 475 481 6;
  • 45) 0.204 569 475 481 6 × 2 = 0 + 0.409 138 950 963 2;
  • 46) 0.409 138 950 963 2 × 2 = 0 + 0.818 277 901 926 4;
  • 47) 0.818 277 901 926 4 × 2 = 1 + 0.636 555 803 852 8;

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:


0.000 000 110 002 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0101 0010 001(2)

5. Positive number before normalization:

0.000 000 110 002 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0101 0010 001(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:


0.000 000 110 002 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0101 0010 001(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0101 0010 001(2) × 20 =

1.1101 1000 0111 0101 0010 001(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 0101 0010 001


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 1010 1001 0001 =

110 1100 0011 1010 1001 0001


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 1010 1001 0001


Decimal number 0.000 000 110 002 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1010 1001 0001

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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