-0.000 000 007 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 007 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
-0.000 000 007 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. Start with the positive version of the number:

|-0.000 000 007 3| = 0.000 000 007 3


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: 0.000 000 007 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) 0.000 000 007 3 × 2 = 0 + 0.000 000 014 6;
  • 2) 0.000 000 014 6 × 2 = 0 + 0.000 000 029 2;
  • 3) 0.000 000 029 2 × 2 = 0 + 0.000 000 058 4;
  • 4) 0.000 000 058 4 × 2 = 0 + 0.000 000 116 8;
  • 5) 0.000 000 116 8 × 2 = 0 + 0.000 000 233 6;
  • 6) 0.000 000 233 6 × 2 = 0 + 0.000 000 467 2;
  • 7) 0.000 000 467 2 × 2 = 0 + 0.000 000 934 4;
  • 8) 0.000 000 934 4 × 2 = 0 + 0.000 001 868 8;
  • 9) 0.000 001 868 8 × 2 = 0 + 0.000 003 737 6;
  • 10) 0.000 003 737 6 × 2 = 0 + 0.000 007 475 2;
  • 11) 0.000 007 475 2 × 2 = 0 + 0.000 014 950 4;
  • 12) 0.000 014 950 4 × 2 = 0 + 0.000 029 900 8;
  • 13) 0.000 029 900 8 × 2 = 0 + 0.000 059 801 6;
  • 14) 0.000 059 801 6 × 2 = 0 + 0.000 119 603 2;
  • 15) 0.000 119 603 2 × 2 = 0 + 0.000 239 206 4;
  • 16) 0.000 239 206 4 × 2 = 0 + 0.000 478 412 8;
  • 17) 0.000 478 412 8 × 2 = 0 + 0.000 956 825 6;
  • 18) 0.000 956 825 6 × 2 = 0 + 0.001 913 651 2;
  • 19) 0.001 913 651 2 × 2 = 0 + 0.003 827 302 4;
  • 20) 0.003 827 302 4 × 2 = 0 + 0.007 654 604 8;
  • 21) 0.007 654 604 8 × 2 = 0 + 0.015 309 209 6;
  • 22) 0.015 309 209 6 × 2 = 0 + 0.030 618 419 2;
  • 23) 0.030 618 419 2 × 2 = 0 + 0.061 236 838 4;
  • 24) 0.061 236 838 4 × 2 = 0 + 0.122 473 676 8;
  • 25) 0.122 473 676 8 × 2 = 0 + 0.244 947 353 6;
  • 26) 0.244 947 353 6 × 2 = 0 + 0.489 894 707 2;
  • 27) 0.489 894 707 2 × 2 = 0 + 0.979 789 414 4;
  • 28) 0.979 789 414 4 × 2 = 1 + 0.959 578 828 8;
  • 29) 0.959 578 828 8 × 2 = 1 + 0.919 157 657 6;
  • 30) 0.919 157 657 6 × 2 = 1 + 0.838 315 315 2;
  • 31) 0.838 315 315 2 × 2 = 1 + 0.676 630 630 4;
  • 32) 0.676 630 630 4 × 2 = 1 + 0.353 261 260 8;
  • 33) 0.353 261 260 8 × 2 = 0 + 0.706 522 521 6;
  • 34) 0.706 522 521 6 × 2 = 1 + 0.413 045 043 2;
  • 35) 0.413 045 043 2 × 2 = 0 + 0.826 090 086 4;
  • 36) 0.826 090 086 4 × 2 = 1 + 0.652 180 172 8;
  • 37) 0.652 180 172 8 × 2 = 1 + 0.304 360 345 6;
  • 38) 0.304 360 345 6 × 2 = 0 + 0.608 720 691 2;
  • 39) 0.608 720 691 2 × 2 = 1 + 0.217 441 382 4;
  • 40) 0.217 441 382 4 × 2 = 0 + 0.434 882 764 8;
  • 41) 0.434 882 764 8 × 2 = 0 + 0.869 765 529 6;
  • 42) 0.869 765 529 6 × 2 = 1 + 0.739 531 059 2;
  • 43) 0.739 531 059 2 × 2 = 1 + 0.479 062 118 4;
  • 44) 0.479 062 118 4 × 2 = 0 + 0.958 124 236 8;
  • 45) 0.958 124 236 8 × 2 = 1 + 0.916 248 473 6;
  • 46) 0.916 248 473 6 × 2 = 1 + 0.832 496 947 2;
  • 47) 0.832 496 947 2 × 2 = 1 + 0.664 993 894 4;
  • 48) 0.664 993 894 4 × 2 = 1 + 0.329 987 788 8;
  • 49) 0.329 987 788 8 × 2 = 0 + 0.659 975 577 6;
  • 50) 0.659 975 577 6 × 2 = 1 + 0.319 951 155 2;
  • 51) 0.319 951 155 2 × 2 = 0 + 0.639 902 310 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).


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:


0.000 000 007 3(10) =

0.0000 0000 0000 0000 0000 0000 0001 1111 0101 1010 0110 1111 010(2)

6. Positive number before normalization:

0.000 000 007 3(10) =

0.0000 0000 0000 0000 0000 0000 0001 1111 0101 1010 0110 1111 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 007 3(10) =

0.0000 0000 0000 0000 0000 0000 0001 1111 0101 1010 0110 1111 010(2) =

0.0000 0000 0000 0000 0000 0000 0001 1111 0101 1010 0110 1111 010(2) × 20 =

1.1111 0101 1010 0110 1111 010(2) × 2-28


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): -28

Mantissa (not normalized):
1.1111 0101 1010 0110 1111 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 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) =

99(10) =

0110 0011(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. 111 1010 1101 0011 0111 1010 =

111 1010 1101 0011 0111 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 0011

Mantissa (23 bits) =
111 1010 1101 0011 0111 1010


Decimal number -0.000 000 007 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0011 - 111 1010 1101 0011 0111 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