-0.000 000 000 742 131 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 131 2(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
-0.000 000 000 742 131 2(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:

|-0.000 000 000 742 131 2| = 0.000 000 000 742 131 2


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: 0.000 000 000 742 131 2.

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) 0.000 000 000 742 131 2 × 2 = 0 + 0.000 000 001 484 262 4;
  • 2) 0.000 000 001 484 262 4 × 2 = 0 + 0.000 000 002 968 524 8;
  • 3) 0.000 000 002 968 524 8 × 2 = 0 + 0.000 000 005 937 049 6;
  • 4) 0.000 000 005 937 049 6 × 2 = 0 + 0.000 000 011 874 099 2;
  • 5) 0.000 000 011 874 099 2 × 2 = 0 + 0.000 000 023 748 198 4;
  • 6) 0.000 000 023 748 198 4 × 2 = 0 + 0.000 000 047 496 396 8;
  • 7) 0.000 000 047 496 396 8 × 2 = 0 + 0.000 000 094 992 793 6;
  • 8) 0.000 000 094 992 793 6 × 2 = 0 + 0.000 000 189 985 587 2;
  • 9) 0.000 000 189 985 587 2 × 2 = 0 + 0.000 000 379 971 174 4;
  • 10) 0.000 000 379 971 174 4 × 2 = 0 + 0.000 000 759 942 348 8;
  • 11) 0.000 000 759 942 348 8 × 2 = 0 + 0.000 001 519 884 697 6;
  • 12) 0.000 001 519 884 697 6 × 2 = 0 + 0.000 003 039 769 395 2;
  • 13) 0.000 003 039 769 395 2 × 2 = 0 + 0.000 006 079 538 790 4;
  • 14) 0.000 006 079 538 790 4 × 2 = 0 + 0.000 012 159 077 580 8;
  • 15) 0.000 012 159 077 580 8 × 2 = 0 + 0.000 024 318 155 161 6;
  • 16) 0.000 024 318 155 161 6 × 2 = 0 + 0.000 048 636 310 323 2;
  • 17) 0.000 048 636 310 323 2 × 2 = 0 + 0.000 097 272 620 646 4;
  • 18) 0.000 097 272 620 646 4 × 2 = 0 + 0.000 194 545 241 292 8;
  • 19) 0.000 194 545 241 292 8 × 2 = 0 + 0.000 389 090 482 585 6;
  • 20) 0.000 389 090 482 585 6 × 2 = 0 + 0.000 778 180 965 171 2;
  • 21) 0.000 778 180 965 171 2 × 2 = 0 + 0.001 556 361 930 342 4;
  • 22) 0.001 556 361 930 342 4 × 2 = 0 + 0.003 112 723 860 684 8;
  • 23) 0.003 112 723 860 684 8 × 2 = 0 + 0.006 225 447 721 369 6;
  • 24) 0.006 225 447 721 369 6 × 2 = 0 + 0.012 450 895 442 739 2;
  • 25) 0.012 450 895 442 739 2 × 2 = 0 + 0.024 901 790 885 478 4;
  • 26) 0.024 901 790 885 478 4 × 2 = 0 + 0.049 803 581 770 956 8;
  • 27) 0.049 803 581 770 956 8 × 2 = 0 + 0.099 607 163 541 913 6;
  • 28) 0.099 607 163 541 913 6 × 2 = 0 + 0.199 214 327 083 827 2;
  • 29) 0.199 214 327 083 827 2 × 2 = 0 + 0.398 428 654 167 654 4;
  • 30) 0.398 428 654 167 654 4 × 2 = 0 + 0.796 857 308 335 308 8;
  • 31) 0.796 857 308 335 308 8 × 2 = 1 + 0.593 714 616 670 617 6;
  • 32) 0.593 714 616 670 617 6 × 2 = 1 + 0.187 429 233 341 235 2;
  • 33) 0.187 429 233 341 235 2 × 2 = 0 + 0.374 858 466 682 470 4;
  • 34) 0.374 858 466 682 470 4 × 2 = 0 + 0.749 716 933 364 940 8;
  • 35) 0.749 716 933 364 940 8 × 2 = 1 + 0.499 433 866 729 881 6;
  • 36) 0.499 433 866 729 881 6 × 2 = 0 + 0.998 867 733 459 763 2;
  • 37) 0.998 867 733 459 763 2 × 2 = 1 + 0.997 735 466 919 526 4;
  • 38) 0.997 735 466 919 526 4 × 2 = 1 + 0.995 470 933 839 052 8;
  • 39) 0.995 470 933 839 052 8 × 2 = 1 + 0.990 941 867 678 105 6;
  • 40) 0.990 941 867 678 105 6 × 2 = 1 + 0.981 883 735 356 211 2;
  • 41) 0.981 883 735 356 211 2 × 2 = 1 + 0.963 767 470 712 422 4;
  • 42) 0.963 767 470 712 422 4 × 2 = 1 + 0.927 534 941 424 844 8;
  • 43) 0.927 534 941 424 844 8 × 2 = 1 + 0.855 069 882 849 689 6;
  • 44) 0.855 069 882 849 689 6 × 2 = 1 + 0.710 139 765 699 379 2;
  • 45) 0.710 139 765 699 379 2 × 2 = 1 + 0.420 279 531 398 758 4;
  • 46) 0.420 279 531 398 758 4 × 2 = 0 + 0.840 559 062 797 516 8;
  • 47) 0.840 559 062 797 516 8 × 2 = 1 + 0.681 118 125 595 033 6;
  • 48) 0.681 118 125 595 033 6 × 2 = 1 + 0.362 236 251 190 067 2;
  • 49) 0.362 236 251 190 067 2 × 2 = 0 + 0.724 472 502 380 134 4;
  • 50) 0.724 472 502 380 134 4 × 2 = 1 + 0.448 945 004 760 268 8;
  • 51) 0.448 945 004 760 268 8 × 2 = 0 + 0.897 890 009 520 537 6;
  • 52) 0.897 890 009 520 537 6 × 2 = 1 + 0.795 780 019 041 075 2;
  • 53) 0.795 780 019 041 075 2 × 2 = 1 + 0.591 560 038 082 150 4;
  • 54) 0.591 560 038 082 150 4 × 2 = 1 + 0.183 120 076 164 300 8;

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:


0.000 000 000 742 131 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 0101 11(2)

6. Positive number before normalization:

0.000 000 000 742 131 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 0101 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:


0.000 000 000 742 131 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 0101 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 0101 11(2) × 20 =

1.1001 0111 1111 1101 1010 111(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 1111 1101 1010 111


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 1111 1110 1101 0111 =

100 1011 1111 1110 1101 0111


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 1111 1110 1101 0111


Decimal number -0.000 000 000 742 131 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1101 0111

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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