0.000 000 110 010 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 010 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
0.000 000 110 010 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: 0.000 000 110 010 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) 0.000 000 110 010 3 × 2 = 0 + 0.000 000 220 020 6;
  • 2) 0.000 000 220 020 6 × 2 = 0 + 0.000 000 440 041 2;
  • 3) 0.000 000 440 041 2 × 2 = 0 + 0.000 000 880 082 4;
  • 4) 0.000 000 880 082 4 × 2 = 0 + 0.000 001 760 164 8;
  • 5) 0.000 001 760 164 8 × 2 = 0 + 0.000 003 520 329 6;
  • 6) 0.000 003 520 329 6 × 2 = 0 + 0.000 007 040 659 2;
  • 7) 0.000 007 040 659 2 × 2 = 0 + 0.000 014 081 318 4;
  • 8) 0.000 014 081 318 4 × 2 = 0 + 0.000 028 162 636 8;
  • 9) 0.000 028 162 636 8 × 2 = 0 + 0.000 056 325 273 6;
  • 10) 0.000 056 325 273 6 × 2 = 0 + 0.000 112 650 547 2;
  • 11) 0.000 112 650 547 2 × 2 = 0 + 0.000 225 301 094 4;
  • 12) 0.000 225 301 094 4 × 2 = 0 + 0.000 450 602 188 8;
  • 13) 0.000 450 602 188 8 × 2 = 0 + 0.000 901 204 377 6;
  • 14) 0.000 901 204 377 6 × 2 = 0 + 0.001 802 408 755 2;
  • 15) 0.001 802 408 755 2 × 2 = 0 + 0.003 604 817 510 4;
  • 16) 0.003 604 817 510 4 × 2 = 0 + 0.007 209 635 020 8;
  • 17) 0.007 209 635 020 8 × 2 = 0 + 0.014 419 270 041 6;
  • 18) 0.014 419 270 041 6 × 2 = 0 + 0.028 838 540 083 2;
  • 19) 0.028 838 540 083 2 × 2 = 0 + 0.057 677 080 166 4;
  • 20) 0.057 677 080 166 4 × 2 = 0 + 0.115 354 160 332 8;
  • 21) 0.115 354 160 332 8 × 2 = 0 + 0.230 708 320 665 6;
  • 22) 0.230 708 320 665 6 × 2 = 0 + 0.461 416 641 331 2;
  • 23) 0.461 416 641 331 2 × 2 = 0 + 0.922 833 282 662 4;
  • 24) 0.922 833 282 662 4 × 2 = 1 + 0.845 666 565 324 8;
  • 25) 0.845 666 565 324 8 × 2 = 1 + 0.691 333 130 649 6;
  • 26) 0.691 333 130 649 6 × 2 = 1 + 0.382 666 261 299 2;
  • 27) 0.382 666 261 299 2 × 2 = 0 + 0.765 332 522 598 4;
  • 28) 0.765 332 522 598 4 × 2 = 1 + 0.530 665 045 196 8;
  • 29) 0.530 665 045 196 8 × 2 = 1 + 0.061 330 090 393 6;
  • 30) 0.061 330 090 393 6 × 2 = 0 + 0.122 660 180 787 2;
  • 31) 0.122 660 180 787 2 × 2 = 0 + 0.245 320 361 574 4;
  • 32) 0.245 320 361 574 4 × 2 = 0 + 0.490 640 723 148 8;
  • 33) 0.490 640 723 148 8 × 2 = 0 + 0.981 281 446 297 6;
  • 34) 0.981 281 446 297 6 × 2 = 1 + 0.962 562 892 595 2;
  • 35) 0.962 562 892 595 2 × 2 = 1 + 0.925 125 785 190 4;
  • 36) 0.925 125 785 190 4 × 2 = 1 + 0.850 251 570 380 8;
  • 37) 0.850 251 570 380 8 × 2 = 1 + 0.700 503 140 761 6;
  • 38) 0.700 503 140 761 6 × 2 = 1 + 0.401 006 281 523 2;
  • 39) 0.401 006 281 523 2 × 2 = 0 + 0.802 012 563 046 4;
  • 40) 0.802 012 563 046 4 × 2 = 1 + 0.604 025 126 092 8;
  • 41) 0.604 025 126 092 8 × 2 = 1 + 0.208 050 252 185 6;
  • 42) 0.208 050 252 185 6 × 2 = 0 + 0.416 100 504 371 2;
  • 43) 0.416 100 504 371 2 × 2 = 0 + 0.832 201 008 742 4;
  • 44) 0.832 201 008 742 4 × 2 = 1 + 0.664 402 017 484 8;
  • 45) 0.664 402 017 484 8 × 2 = 1 + 0.328 804 034 969 6;
  • 46) 0.328 804 034 969 6 × 2 = 0 + 0.657 608 069 939 2;
  • 47) 0.657 608 069 939 2 × 2 = 1 + 0.315 216 139 878 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:


0.000 000 110 010 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 1001 101(2)

5. Positive number before normalization:

0.000 000 110 010 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 1001 101(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:


0.000 000 110 010 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 1001 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 1001 101(2) × 20 =

1.1101 1000 0111 1101 1001 101(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 1101 1001 101


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 1100 1101 =

110 1100 0011 1110 1100 1101


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 1100 1101


Decimal number 0.000 000 110 010 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 1100 1101

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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