-0.000 000 000 741 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 4(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
-0.000 000 000 741 4(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:

|-0.000 000 000 741 4| = 0.000 000 000 741 4


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: 0.000 000 000 741 4.

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) 0.000 000 000 741 4 × 2 = 0 + 0.000 000 001 482 8;
  • 2) 0.000 000 001 482 8 × 2 = 0 + 0.000 000 002 965 6;
  • 3) 0.000 000 002 965 6 × 2 = 0 + 0.000 000 005 931 2;
  • 4) 0.000 000 005 931 2 × 2 = 0 + 0.000 000 011 862 4;
  • 5) 0.000 000 011 862 4 × 2 = 0 + 0.000 000 023 724 8;
  • 6) 0.000 000 023 724 8 × 2 = 0 + 0.000 000 047 449 6;
  • 7) 0.000 000 047 449 6 × 2 = 0 + 0.000 000 094 899 2;
  • 8) 0.000 000 094 899 2 × 2 = 0 + 0.000 000 189 798 4;
  • 9) 0.000 000 189 798 4 × 2 = 0 + 0.000 000 379 596 8;
  • 10) 0.000 000 379 596 8 × 2 = 0 + 0.000 000 759 193 6;
  • 11) 0.000 000 759 193 6 × 2 = 0 + 0.000 001 518 387 2;
  • 12) 0.000 001 518 387 2 × 2 = 0 + 0.000 003 036 774 4;
  • 13) 0.000 003 036 774 4 × 2 = 0 + 0.000 006 073 548 8;
  • 14) 0.000 006 073 548 8 × 2 = 0 + 0.000 012 147 097 6;
  • 15) 0.000 012 147 097 6 × 2 = 0 + 0.000 024 294 195 2;
  • 16) 0.000 024 294 195 2 × 2 = 0 + 0.000 048 588 390 4;
  • 17) 0.000 048 588 390 4 × 2 = 0 + 0.000 097 176 780 8;
  • 18) 0.000 097 176 780 8 × 2 = 0 + 0.000 194 353 561 6;
  • 19) 0.000 194 353 561 6 × 2 = 0 + 0.000 388 707 123 2;
  • 20) 0.000 388 707 123 2 × 2 = 0 + 0.000 777 414 246 4;
  • 21) 0.000 777 414 246 4 × 2 = 0 + 0.001 554 828 492 8;
  • 22) 0.001 554 828 492 8 × 2 = 0 + 0.003 109 656 985 6;
  • 23) 0.003 109 656 985 6 × 2 = 0 + 0.006 219 313 971 2;
  • 24) 0.006 219 313 971 2 × 2 = 0 + 0.012 438 627 942 4;
  • 25) 0.012 438 627 942 4 × 2 = 0 + 0.024 877 255 884 8;
  • 26) 0.024 877 255 884 8 × 2 = 0 + 0.049 754 511 769 6;
  • 27) 0.049 754 511 769 6 × 2 = 0 + 0.099 509 023 539 2;
  • 28) 0.099 509 023 539 2 × 2 = 0 + 0.199 018 047 078 4;
  • 29) 0.199 018 047 078 4 × 2 = 0 + 0.398 036 094 156 8;
  • 30) 0.398 036 094 156 8 × 2 = 0 + 0.796 072 188 313 6;
  • 31) 0.796 072 188 313 6 × 2 = 1 + 0.592 144 376 627 2;
  • 32) 0.592 144 376 627 2 × 2 = 1 + 0.184 288 753 254 4;
  • 33) 0.184 288 753 254 4 × 2 = 0 + 0.368 577 506 508 8;
  • 34) 0.368 577 506 508 8 × 2 = 0 + 0.737 155 013 017 6;
  • 35) 0.737 155 013 017 6 × 2 = 1 + 0.474 310 026 035 2;
  • 36) 0.474 310 026 035 2 × 2 = 0 + 0.948 620 052 070 4;
  • 37) 0.948 620 052 070 4 × 2 = 1 + 0.897 240 104 140 8;
  • 38) 0.897 240 104 140 8 × 2 = 1 + 0.794 480 208 281 6;
  • 39) 0.794 480 208 281 6 × 2 = 1 + 0.588 960 416 563 2;
  • 40) 0.588 960 416 563 2 × 2 = 1 + 0.177 920 833 126 4;
  • 41) 0.177 920 833 126 4 × 2 = 0 + 0.355 841 666 252 8;
  • 42) 0.355 841 666 252 8 × 2 = 0 + 0.711 683 332 505 6;
  • 43) 0.711 683 332 505 6 × 2 = 1 + 0.423 366 665 011 2;
  • 44) 0.423 366 665 011 2 × 2 = 0 + 0.846 733 330 022 4;
  • 45) 0.846 733 330 022 4 × 2 = 1 + 0.693 466 660 044 8;
  • 46) 0.693 466 660 044 8 × 2 = 1 + 0.386 933 320 089 6;
  • 47) 0.386 933 320 089 6 × 2 = 0 + 0.773 866 640 179 2;
  • 48) 0.773 866 640 179 2 × 2 = 1 + 0.547 733 280 358 4;
  • 49) 0.547 733 280 358 4 × 2 = 1 + 0.095 466 560 716 8;
  • 50) 0.095 466 560 716 8 × 2 = 0 + 0.190 933 121 433 6;
  • 51) 0.190 933 121 433 6 × 2 = 0 + 0.381 866 242 867 2;
  • 52) 0.381 866 242 867 2 × 2 = 0 + 0.763 732 485 734 4;
  • 53) 0.763 732 485 734 4 × 2 = 1 + 0.527 464 971 468 8;
  • 54) 0.527 464 971 468 8 × 2 = 1 + 0.054 929 942 937 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:


0.000 000 000 741 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0010 1101 1000 11(2)

6. Positive number before normalization:

0.000 000 000 741 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0010 1101 1000 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 741 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0010 1101 1000 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0010 1101 1000 11(2) × 20 =

1.1001 0111 1001 0110 1100 011(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0111 1001 0110 1100 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 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) =

96(10) =

0110 0000(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. 100 1011 1100 1011 0110 0011 =

100 1011 1100 1011 0110 0011


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 0000

Mantissa (23 bits) =
100 1011 1100 1011 0110 0011


Decimal number -0.000 000 000 741 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1100 1011 0110 0011

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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