-0.000 000 000 741 85 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 85(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 85(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 85| = 0.000 000 000 741 85


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 85.

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 85 × 2 = 0 + 0.000 000 001 483 7;
  • 2) 0.000 000 001 483 7 × 2 = 0 + 0.000 000 002 967 4;
  • 3) 0.000 000 002 967 4 × 2 = 0 + 0.000 000 005 934 8;
  • 4) 0.000 000 005 934 8 × 2 = 0 + 0.000 000 011 869 6;
  • 5) 0.000 000 011 869 6 × 2 = 0 + 0.000 000 023 739 2;
  • 6) 0.000 000 023 739 2 × 2 = 0 + 0.000 000 047 478 4;
  • 7) 0.000 000 047 478 4 × 2 = 0 + 0.000 000 094 956 8;
  • 8) 0.000 000 094 956 8 × 2 = 0 + 0.000 000 189 913 6;
  • 9) 0.000 000 189 913 6 × 2 = 0 + 0.000 000 379 827 2;
  • 10) 0.000 000 379 827 2 × 2 = 0 + 0.000 000 759 654 4;
  • 11) 0.000 000 759 654 4 × 2 = 0 + 0.000 001 519 308 8;
  • 12) 0.000 001 519 308 8 × 2 = 0 + 0.000 003 038 617 6;
  • 13) 0.000 003 038 617 6 × 2 = 0 + 0.000 006 077 235 2;
  • 14) 0.000 006 077 235 2 × 2 = 0 + 0.000 012 154 470 4;
  • 15) 0.000 012 154 470 4 × 2 = 0 + 0.000 024 308 940 8;
  • 16) 0.000 024 308 940 8 × 2 = 0 + 0.000 048 617 881 6;
  • 17) 0.000 048 617 881 6 × 2 = 0 + 0.000 097 235 763 2;
  • 18) 0.000 097 235 763 2 × 2 = 0 + 0.000 194 471 526 4;
  • 19) 0.000 194 471 526 4 × 2 = 0 + 0.000 388 943 052 8;
  • 20) 0.000 388 943 052 8 × 2 = 0 + 0.000 777 886 105 6;
  • 21) 0.000 777 886 105 6 × 2 = 0 + 0.001 555 772 211 2;
  • 22) 0.001 555 772 211 2 × 2 = 0 + 0.003 111 544 422 4;
  • 23) 0.003 111 544 422 4 × 2 = 0 + 0.006 223 088 844 8;
  • 24) 0.006 223 088 844 8 × 2 = 0 + 0.012 446 177 689 6;
  • 25) 0.012 446 177 689 6 × 2 = 0 + 0.024 892 355 379 2;
  • 26) 0.024 892 355 379 2 × 2 = 0 + 0.049 784 710 758 4;
  • 27) 0.049 784 710 758 4 × 2 = 0 + 0.099 569 421 516 8;
  • 28) 0.099 569 421 516 8 × 2 = 0 + 0.199 138 843 033 6;
  • 29) 0.199 138 843 033 6 × 2 = 0 + 0.398 277 686 067 2;
  • 30) 0.398 277 686 067 2 × 2 = 0 + 0.796 555 372 134 4;
  • 31) 0.796 555 372 134 4 × 2 = 1 + 0.593 110 744 268 8;
  • 32) 0.593 110 744 268 8 × 2 = 1 + 0.186 221 488 537 6;
  • 33) 0.186 221 488 537 6 × 2 = 0 + 0.372 442 977 075 2;
  • 34) 0.372 442 977 075 2 × 2 = 0 + 0.744 885 954 150 4;
  • 35) 0.744 885 954 150 4 × 2 = 1 + 0.489 771 908 300 8;
  • 36) 0.489 771 908 300 8 × 2 = 0 + 0.979 543 816 601 6;
  • 37) 0.979 543 816 601 6 × 2 = 1 + 0.959 087 633 203 2;
  • 38) 0.959 087 633 203 2 × 2 = 1 + 0.918 175 266 406 4;
  • 39) 0.918 175 266 406 4 × 2 = 1 + 0.836 350 532 812 8;
  • 40) 0.836 350 532 812 8 × 2 = 1 + 0.672 701 065 625 6;
  • 41) 0.672 701 065 625 6 × 2 = 1 + 0.345 402 131 251 2;
  • 42) 0.345 402 131 251 2 × 2 = 0 + 0.690 804 262 502 4;
  • 43) 0.690 804 262 502 4 × 2 = 1 + 0.381 608 525 004 8;
  • 44) 0.381 608 525 004 8 × 2 = 0 + 0.763 217 050 009 6;
  • 45) 0.763 217 050 009 6 × 2 = 1 + 0.526 434 100 019 2;
  • 46) 0.526 434 100 019 2 × 2 = 1 + 0.052 868 200 038 4;
  • 47) 0.052 868 200 038 4 × 2 = 0 + 0.105 736 400 076 8;
  • 48) 0.105 736 400 076 8 × 2 = 0 + 0.211 472 800 153 6;
  • 49) 0.211 472 800 153 6 × 2 = 0 + 0.422 945 600 307 2;
  • 50) 0.422 945 600 307 2 × 2 = 0 + 0.845 891 200 614 4;
  • 51) 0.845 891 200 614 4 × 2 = 1 + 0.691 782 401 228 8;
  • 52) 0.691 782 401 228 8 × 2 = 1 + 0.383 564 802 457 6;
  • 53) 0.383 564 802 457 6 × 2 = 0 + 0.767 129 604 915 2;
  • 54) 0.767 129 604 915 2 × 2 = 1 + 0.534 259 209 830 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:


0.000 000 000 741 85(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1010 1100 0011 01(2)

6. Positive number before normalization:

0.000 000 000 741 85(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1010 1100 0011 01(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 85(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1010 1100 0011 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1010 1100 0011 01(2) × 20 =

1.1001 0111 1101 0110 0001 101(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 1101 0110 0001 101


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 1110 1011 0000 1101 =

100 1011 1110 1011 0000 1101


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 1110 1011 0000 1101


Decimal number -0.000 000 000 741 85 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1110 1011 0000 1101

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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