-0.000 000 000 741 67 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 67(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 67(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 67| = 0.000 000 000 741 67


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 67.

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 67 × 2 = 0 + 0.000 000 001 483 34;
  • 2) 0.000 000 001 483 34 × 2 = 0 + 0.000 000 002 966 68;
  • 3) 0.000 000 002 966 68 × 2 = 0 + 0.000 000 005 933 36;
  • 4) 0.000 000 005 933 36 × 2 = 0 + 0.000 000 011 866 72;
  • 5) 0.000 000 011 866 72 × 2 = 0 + 0.000 000 023 733 44;
  • 6) 0.000 000 023 733 44 × 2 = 0 + 0.000 000 047 466 88;
  • 7) 0.000 000 047 466 88 × 2 = 0 + 0.000 000 094 933 76;
  • 8) 0.000 000 094 933 76 × 2 = 0 + 0.000 000 189 867 52;
  • 9) 0.000 000 189 867 52 × 2 = 0 + 0.000 000 379 735 04;
  • 10) 0.000 000 379 735 04 × 2 = 0 + 0.000 000 759 470 08;
  • 11) 0.000 000 759 470 08 × 2 = 0 + 0.000 001 518 940 16;
  • 12) 0.000 001 518 940 16 × 2 = 0 + 0.000 003 037 880 32;
  • 13) 0.000 003 037 880 32 × 2 = 0 + 0.000 006 075 760 64;
  • 14) 0.000 006 075 760 64 × 2 = 0 + 0.000 012 151 521 28;
  • 15) 0.000 012 151 521 28 × 2 = 0 + 0.000 024 303 042 56;
  • 16) 0.000 024 303 042 56 × 2 = 0 + 0.000 048 606 085 12;
  • 17) 0.000 048 606 085 12 × 2 = 0 + 0.000 097 212 170 24;
  • 18) 0.000 097 212 170 24 × 2 = 0 + 0.000 194 424 340 48;
  • 19) 0.000 194 424 340 48 × 2 = 0 + 0.000 388 848 680 96;
  • 20) 0.000 388 848 680 96 × 2 = 0 + 0.000 777 697 361 92;
  • 21) 0.000 777 697 361 92 × 2 = 0 + 0.001 555 394 723 84;
  • 22) 0.001 555 394 723 84 × 2 = 0 + 0.003 110 789 447 68;
  • 23) 0.003 110 789 447 68 × 2 = 0 + 0.006 221 578 895 36;
  • 24) 0.006 221 578 895 36 × 2 = 0 + 0.012 443 157 790 72;
  • 25) 0.012 443 157 790 72 × 2 = 0 + 0.024 886 315 581 44;
  • 26) 0.024 886 315 581 44 × 2 = 0 + 0.049 772 631 162 88;
  • 27) 0.049 772 631 162 88 × 2 = 0 + 0.099 545 262 325 76;
  • 28) 0.099 545 262 325 76 × 2 = 0 + 0.199 090 524 651 52;
  • 29) 0.199 090 524 651 52 × 2 = 0 + 0.398 181 049 303 04;
  • 30) 0.398 181 049 303 04 × 2 = 0 + 0.796 362 098 606 08;
  • 31) 0.796 362 098 606 08 × 2 = 1 + 0.592 724 197 212 16;
  • 32) 0.592 724 197 212 16 × 2 = 1 + 0.185 448 394 424 32;
  • 33) 0.185 448 394 424 32 × 2 = 0 + 0.370 896 788 848 64;
  • 34) 0.370 896 788 848 64 × 2 = 0 + 0.741 793 577 697 28;
  • 35) 0.741 793 577 697 28 × 2 = 1 + 0.483 587 155 394 56;
  • 36) 0.483 587 155 394 56 × 2 = 0 + 0.967 174 310 789 12;
  • 37) 0.967 174 310 789 12 × 2 = 1 + 0.934 348 621 578 24;
  • 38) 0.934 348 621 578 24 × 2 = 1 + 0.868 697 243 156 48;
  • 39) 0.868 697 243 156 48 × 2 = 1 + 0.737 394 486 312 96;
  • 40) 0.737 394 486 312 96 × 2 = 1 + 0.474 788 972 625 92;
  • 41) 0.474 788 972 625 92 × 2 = 0 + 0.949 577 945 251 84;
  • 42) 0.949 577 945 251 84 × 2 = 1 + 0.899 155 890 503 68;
  • 43) 0.899 155 890 503 68 × 2 = 1 + 0.798 311 781 007 36;
  • 44) 0.798 311 781 007 36 × 2 = 1 + 0.596 623 562 014 72;
  • 45) 0.596 623 562 014 72 × 2 = 1 + 0.193 247 124 029 44;
  • 46) 0.193 247 124 029 44 × 2 = 0 + 0.386 494 248 058 88;
  • 47) 0.386 494 248 058 88 × 2 = 0 + 0.772 988 496 117 76;
  • 48) 0.772 988 496 117 76 × 2 = 1 + 0.545 976 992 235 52;
  • 49) 0.545 976 992 235 52 × 2 = 1 + 0.091 953 984 471 04;
  • 50) 0.091 953 984 471 04 × 2 = 0 + 0.183 907 968 942 08;
  • 51) 0.183 907 968 942 08 × 2 = 0 + 0.367 815 937 884 16;
  • 52) 0.367 815 937 884 16 × 2 = 0 + 0.735 631 875 768 32;
  • 53) 0.735 631 875 768 32 × 2 = 1 + 0.471 263 751 536 64;
  • 54) 0.471 263 751 536 64 × 2 = 0 + 0.942 527 503 073 28;

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 67(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0111 1001 1000 10(2)

6. Positive number before normalization:

0.000 000 000 741 67(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0111 1001 1000 10(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 67(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0111 1001 1000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0111 1001 1000 10(2) × 20 =

1.1001 0111 1011 1100 1100 010(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 1011 1100 1100 010


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 1101 1110 0110 0010 =

100 1011 1101 1110 0110 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 0000

Mantissa (23 bits) =
100 1011 1101 1110 0110 0010


Decimal number -0.000 000 000 741 67 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1101 1110 0110 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