-0.000 000 032 422 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 032 422 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
-0.000 000 032 422 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. Start with the positive version of the number:

|-0.000 000 032 422 6| = 0.000 000 032 422 6


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: 0.000 000 032 422 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) 0.000 000 032 422 6 × 2 = 0 + 0.000 000 064 845 2;
  • 2) 0.000 000 064 845 2 × 2 = 0 + 0.000 000 129 690 4;
  • 3) 0.000 000 129 690 4 × 2 = 0 + 0.000 000 259 380 8;
  • 4) 0.000 000 259 380 8 × 2 = 0 + 0.000 000 518 761 6;
  • 5) 0.000 000 518 761 6 × 2 = 0 + 0.000 001 037 523 2;
  • 6) 0.000 001 037 523 2 × 2 = 0 + 0.000 002 075 046 4;
  • 7) 0.000 002 075 046 4 × 2 = 0 + 0.000 004 150 092 8;
  • 8) 0.000 004 150 092 8 × 2 = 0 + 0.000 008 300 185 6;
  • 9) 0.000 008 300 185 6 × 2 = 0 + 0.000 016 600 371 2;
  • 10) 0.000 016 600 371 2 × 2 = 0 + 0.000 033 200 742 4;
  • 11) 0.000 033 200 742 4 × 2 = 0 + 0.000 066 401 484 8;
  • 12) 0.000 066 401 484 8 × 2 = 0 + 0.000 132 802 969 6;
  • 13) 0.000 132 802 969 6 × 2 = 0 + 0.000 265 605 939 2;
  • 14) 0.000 265 605 939 2 × 2 = 0 + 0.000 531 211 878 4;
  • 15) 0.000 531 211 878 4 × 2 = 0 + 0.001 062 423 756 8;
  • 16) 0.001 062 423 756 8 × 2 = 0 + 0.002 124 847 513 6;
  • 17) 0.002 124 847 513 6 × 2 = 0 + 0.004 249 695 027 2;
  • 18) 0.004 249 695 027 2 × 2 = 0 + 0.008 499 390 054 4;
  • 19) 0.008 499 390 054 4 × 2 = 0 + 0.016 998 780 108 8;
  • 20) 0.016 998 780 108 8 × 2 = 0 + 0.033 997 560 217 6;
  • 21) 0.033 997 560 217 6 × 2 = 0 + 0.067 995 120 435 2;
  • 22) 0.067 995 120 435 2 × 2 = 0 + 0.135 990 240 870 4;
  • 23) 0.135 990 240 870 4 × 2 = 0 + 0.271 980 481 740 8;
  • 24) 0.271 980 481 740 8 × 2 = 0 + 0.543 960 963 481 6;
  • 25) 0.543 960 963 481 6 × 2 = 1 + 0.087 921 926 963 2;
  • 26) 0.087 921 926 963 2 × 2 = 0 + 0.175 843 853 926 4;
  • 27) 0.175 843 853 926 4 × 2 = 0 + 0.351 687 707 852 8;
  • 28) 0.351 687 707 852 8 × 2 = 0 + 0.703 375 415 705 6;
  • 29) 0.703 375 415 705 6 × 2 = 1 + 0.406 750 831 411 2;
  • 30) 0.406 750 831 411 2 × 2 = 0 + 0.813 501 662 822 4;
  • 31) 0.813 501 662 822 4 × 2 = 1 + 0.627 003 325 644 8;
  • 32) 0.627 003 325 644 8 × 2 = 1 + 0.254 006 651 289 6;
  • 33) 0.254 006 651 289 6 × 2 = 0 + 0.508 013 302 579 2;
  • 34) 0.508 013 302 579 2 × 2 = 1 + 0.016 026 605 158 4;
  • 35) 0.016 026 605 158 4 × 2 = 0 + 0.032 053 210 316 8;
  • 36) 0.032 053 210 316 8 × 2 = 0 + 0.064 106 420 633 6;
  • 37) 0.064 106 420 633 6 × 2 = 0 + 0.128 212 841 267 2;
  • 38) 0.128 212 841 267 2 × 2 = 0 + 0.256 425 682 534 4;
  • 39) 0.256 425 682 534 4 × 2 = 0 + 0.512 851 365 068 8;
  • 40) 0.512 851 365 068 8 × 2 = 1 + 0.025 702 730 137 6;
  • 41) 0.025 702 730 137 6 × 2 = 0 + 0.051 405 460 275 2;
  • 42) 0.051 405 460 275 2 × 2 = 0 + 0.102 810 920 550 4;
  • 43) 0.102 810 920 550 4 × 2 = 0 + 0.205 621 841 100 8;
  • 44) 0.205 621 841 100 8 × 2 = 0 + 0.411 243 682 201 6;
  • 45) 0.411 243 682 201 6 × 2 = 0 + 0.822 487 364 403 2;
  • 46) 0.822 487 364 403 2 × 2 = 1 + 0.644 974 728 806 4;
  • 47) 0.644 974 728 806 4 × 2 = 1 + 0.289 949 457 612 8;
  • 48) 0.289 949 457 612 8 × 2 = 0 + 0.579 898 915 225 6;

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:


0.000 000 032 422 6(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0001 0000 0110(2)

6. Positive number before normalization:

0.000 000 032 422 6(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0001 0000 0110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 032 422 6(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0001 0000 0110(2) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0001 0000 0110(2) × 20 =

1.0001 0110 1000 0010 0000 110(2) × 2-25


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): -25

Mantissa (not normalized):
1.0001 0110 1000 0010 0000 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(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;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

102(10) =

0110 0110(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 1011 0100 0001 0000 0110 =

000 1011 0100 0001 0000 0110


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 0110

Mantissa (23 bits) =
000 1011 0100 0001 0000 0110


Decimal number -0.000 000 032 422 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 000 1011 0100 0001 0000 0110

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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