-0.000 000 003 45 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 003 45(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
-0.000 000 003 45(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:

|-0.000 000 003 45| = 0.000 000 003 45


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: 0.000 000 003 45.

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) 0.000 000 003 45 × 2 = 0 + 0.000 000 006 9;
  • 2) 0.000 000 006 9 × 2 = 0 + 0.000 000 013 8;
  • 3) 0.000 000 013 8 × 2 = 0 + 0.000 000 027 6;
  • 4) 0.000 000 027 6 × 2 = 0 + 0.000 000 055 2;
  • 5) 0.000 000 055 2 × 2 = 0 + 0.000 000 110 4;
  • 6) 0.000 000 110 4 × 2 = 0 + 0.000 000 220 8;
  • 7) 0.000 000 220 8 × 2 = 0 + 0.000 000 441 6;
  • 8) 0.000 000 441 6 × 2 = 0 + 0.000 000 883 2;
  • 9) 0.000 000 883 2 × 2 = 0 + 0.000 001 766 4;
  • 10) 0.000 001 766 4 × 2 = 0 + 0.000 003 532 8;
  • 11) 0.000 003 532 8 × 2 = 0 + 0.000 007 065 6;
  • 12) 0.000 007 065 6 × 2 = 0 + 0.000 014 131 2;
  • 13) 0.000 014 131 2 × 2 = 0 + 0.000 028 262 4;
  • 14) 0.000 028 262 4 × 2 = 0 + 0.000 056 524 8;
  • 15) 0.000 056 524 8 × 2 = 0 + 0.000 113 049 6;
  • 16) 0.000 113 049 6 × 2 = 0 + 0.000 226 099 2;
  • 17) 0.000 226 099 2 × 2 = 0 + 0.000 452 198 4;
  • 18) 0.000 452 198 4 × 2 = 0 + 0.000 904 396 8;
  • 19) 0.000 904 396 8 × 2 = 0 + 0.001 808 793 6;
  • 20) 0.001 808 793 6 × 2 = 0 + 0.003 617 587 2;
  • 21) 0.003 617 587 2 × 2 = 0 + 0.007 235 174 4;
  • 22) 0.007 235 174 4 × 2 = 0 + 0.014 470 348 8;
  • 23) 0.014 470 348 8 × 2 = 0 + 0.028 940 697 6;
  • 24) 0.028 940 697 6 × 2 = 0 + 0.057 881 395 2;
  • 25) 0.057 881 395 2 × 2 = 0 + 0.115 762 790 4;
  • 26) 0.115 762 790 4 × 2 = 0 + 0.231 525 580 8;
  • 27) 0.231 525 580 8 × 2 = 0 + 0.463 051 161 6;
  • 28) 0.463 051 161 6 × 2 = 0 + 0.926 102 323 2;
  • 29) 0.926 102 323 2 × 2 = 1 + 0.852 204 646 4;
  • 30) 0.852 204 646 4 × 2 = 1 + 0.704 409 292 8;
  • 31) 0.704 409 292 8 × 2 = 1 + 0.408 818 585 6;
  • 32) 0.408 818 585 6 × 2 = 0 + 0.817 637 171 2;
  • 33) 0.817 637 171 2 × 2 = 1 + 0.635 274 342 4;
  • 34) 0.635 274 342 4 × 2 = 1 + 0.270 548 684 8;
  • 35) 0.270 548 684 8 × 2 = 0 + 0.541 097 369 6;
  • 36) 0.541 097 369 6 × 2 = 1 + 0.082 194 739 2;
  • 37) 0.082 194 739 2 × 2 = 0 + 0.164 389 478 4;
  • 38) 0.164 389 478 4 × 2 = 0 + 0.328 778 956 8;
  • 39) 0.328 778 956 8 × 2 = 0 + 0.657 557 913 6;
  • 40) 0.657 557 913 6 × 2 = 1 + 0.315 115 827 2;
  • 41) 0.315 115 827 2 × 2 = 0 + 0.630 231 654 4;
  • 42) 0.630 231 654 4 × 2 = 1 + 0.260 463 308 8;
  • 43) 0.260 463 308 8 × 2 = 0 + 0.520 926 617 6;
  • 44) 0.520 926 617 6 × 2 = 1 + 0.041 853 235 2;
  • 45) 0.041 853 235 2 × 2 = 0 + 0.083 706 470 4;
  • 46) 0.083 706 470 4 × 2 = 0 + 0.167 412 940 8;
  • 47) 0.167 412 940 8 × 2 = 0 + 0.334 825 881 6;
  • 48) 0.334 825 881 6 × 2 = 0 + 0.669 651 763 2;
  • 49) 0.669 651 763 2 × 2 = 1 + 0.339 303 526 4;
  • 50) 0.339 303 526 4 × 2 = 0 + 0.678 607 052 8;
  • 51) 0.678 607 052 8 × 2 = 1 + 0.357 214 105 6;
  • 52) 0.357 214 105 6 × 2 = 0 + 0.714 428 211 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).


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:


0.000 000 003 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 1101 0001 0101 0000 1010(2)

6. Positive number before normalization:

0.000 000 003 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 1101 0001 0101 0000 1010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 003 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 1101 0001 0101 0000 1010(2) =

0.0000 0000 0000 0000 0000 0000 0000 1110 1101 0001 0101 0000 1010(2) × 20 =

1.1101 1010 0010 1010 0001 010(2) × 2-29


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): -29

Mantissa (not normalized):
1.1101 1010 0010 1010 0001 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-29 + 2(8-1) - 1 =

(-29 + 127)(10) =

98(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;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

98(10) =

0110 0010(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. 110 1101 0001 0101 0000 1010 =

110 1101 0001 0101 0000 1010


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 0010

Mantissa (23 bits) =
110 1101 0001 0101 0000 1010


Decimal number -0.000 000 003 45 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0010 - 110 1101 0001 0101 0000 1010

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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