-0.000 000 019 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 019 7(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
-0.000 000 019 7(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:

|-0.000 000 019 7| = 0.000 000 019 7


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: 0.000 000 019 7.

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) 0.000 000 019 7 × 2 = 0 + 0.000 000 039 4;
  • 2) 0.000 000 039 4 × 2 = 0 + 0.000 000 078 8;
  • 3) 0.000 000 078 8 × 2 = 0 + 0.000 000 157 6;
  • 4) 0.000 000 157 6 × 2 = 0 + 0.000 000 315 2;
  • 5) 0.000 000 315 2 × 2 = 0 + 0.000 000 630 4;
  • 6) 0.000 000 630 4 × 2 = 0 + 0.000 001 260 8;
  • 7) 0.000 001 260 8 × 2 = 0 + 0.000 002 521 6;
  • 8) 0.000 002 521 6 × 2 = 0 + 0.000 005 043 2;
  • 9) 0.000 005 043 2 × 2 = 0 + 0.000 010 086 4;
  • 10) 0.000 010 086 4 × 2 = 0 + 0.000 020 172 8;
  • 11) 0.000 020 172 8 × 2 = 0 + 0.000 040 345 6;
  • 12) 0.000 040 345 6 × 2 = 0 + 0.000 080 691 2;
  • 13) 0.000 080 691 2 × 2 = 0 + 0.000 161 382 4;
  • 14) 0.000 161 382 4 × 2 = 0 + 0.000 322 764 8;
  • 15) 0.000 322 764 8 × 2 = 0 + 0.000 645 529 6;
  • 16) 0.000 645 529 6 × 2 = 0 + 0.001 291 059 2;
  • 17) 0.001 291 059 2 × 2 = 0 + 0.002 582 118 4;
  • 18) 0.002 582 118 4 × 2 = 0 + 0.005 164 236 8;
  • 19) 0.005 164 236 8 × 2 = 0 + 0.010 328 473 6;
  • 20) 0.010 328 473 6 × 2 = 0 + 0.020 656 947 2;
  • 21) 0.020 656 947 2 × 2 = 0 + 0.041 313 894 4;
  • 22) 0.041 313 894 4 × 2 = 0 + 0.082 627 788 8;
  • 23) 0.082 627 788 8 × 2 = 0 + 0.165 255 577 6;
  • 24) 0.165 255 577 6 × 2 = 0 + 0.330 511 155 2;
  • 25) 0.330 511 155 2 × 2 = 0 + 0.661 022 310 4;
  • 26) 0.661 022 310 4 × 2 = 1 + 0.322 044 620 8;
  • 27) 0.322 044 620 8 × 2 = 0 + 0.644 089 241 6;
  • 28) 0.644 089 241 6 × 2 = 1 + 0.288 178 483 2;
  • 29) 0.288 178 483 2 × 2 = 0 + 0.576 356 966 4;
  • 30) 0.576 356 966 4 × 2 = 1 + 0.152 713 932 8;
  • 31) 0.152 713 932 8 × 2 = 0 + 0.305 427 865 6;
  • 32) 0.305 427 865 6 × 2 = 0 + 0.610 855 731 2;
  • 33) 0.610 855 731 2 × 2 = 1 + 0.221 711 462 4;
  • 34) 0.221 711 462 4 × 2 = 0 + 0.443 422 924 8;
  • 35) 0.443 422 924 8 × 2 = 0 + 0.886 845 849 6;
  • 36) 0.886 845 849 6 × 2 = 1 + 0.773 691 699 2;
  • 37) 0.773 691 699 2 × 2 = 1 + 0.547 383 398 4;
  • 38) 0.547 383 398 4 × 2 = 1 + 0.094 766 796 8;
  • 39) 0.094 766 796 8 × 2 = 0 + 0.189 533 593 6;
  • 40) 0.189 533 593 6 × 2 = 0 + 0.379 067 187 2;
  • 41) 0.379 067 187 2 × 2 = 0 + 0.758 134 374 4;
  • 42) 0.758 134 374 4 × 2 = 1 + 0.516 268 748 8;
  • 43) 0.516 268 748 8 × 2 = 1 + 0.032 537 497 6;
  • 44) 0.032 537 497 6 × 2 = 0 + 0.065 074 995 2;
  • 45) 0.065 074 995 2 × 2 = 0 + 0.130 149 990 4;
  • 46) 0.130 149 990 4 × 2 = 0 + 0.260 299 980 8;
  • 47) 0.260 299 980 8 × 2 = 0 + 0.520 599 961 6;
  • 48) 0.520 599 961 6 × 2 = 1 + 0.041 199 923 2;
  • 49) 0.041 199 923 2 × 2 = 0 + 0.082 399 846 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:


0.000 000 019 7(10) =

0.0000 0000 0000 0000 0000 0000 0101 0100 1001 1100 0110 0001 0(2)

6. Positive number before normalization:

0.000 000 019 7(10) =

0.0000 0000 0000 0000 0000 0000 0101 0100 1001 1100 0110 0001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 019 7(10) =

0.0000 0000 0000 0000 0000 0000 0101 0100 1001 1100 0110 0001 0(2) =

0.0000 0000 0000 0000 0000 0000 0101 0100 1001 1100 0110 0001 0(2) × 20 =

1.0101 0010 0111 0001 1000 010(2) × 2-26


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): -26

Mantissa (not normalized):
1.0101 0010 0111 0001 1000 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(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;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 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) =

101(10) =

0110 0101(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. 010 1001 0011 1000 1100 0010 =

010 1001 0011 1000 1100 0010


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 0101

Mantissa (23 bits) =
010 1001 0011 1000 1100 0010


Decimal number -0.000 000 019 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0101 - 010 1001 0011 1000 1100 0010

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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