-0.000 000 000 742 14 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 14(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 14(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 14| = 0.000 000 000 742 14


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 14.

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 14 × 2 = 0 + 0.000 000 001 484 28;
  • 2) 0.000 000 001 484 28 × 2 = 0 + 0.000 000 002 968 56;
  • 3) 0.000 000 002 968 56 × 2 = 0 + 0.000 000 005 937 12;
  • 4) 0.000 000 005 937 12 × 2 = 0 + 0.000 000 011 874 24;
  • 5) 0.000 000 011 874 24 × 2 = 0 + 0.000 000 023 748 48;
  • 6) 0.000 000 023 748 48 × 2 = 0 + 0.000 000 047 496 96;
  • 7) 0.000 000 047 496 96 × 2 = 0 + 0.000 000 094 993 92;
  • 8) 0.000 000 094 993 92 × 2 = 0 + 0.000 000 189 987 84;
  • 9) 0.000 000 189 987 84 × 2 = 0 + 0.000 000 379 975 68;
  • 10) 0.000 000 379 975 68 × 2 = 0 + 0.000 000 759 951 36;
  • 11) 0.000 000 759 951 36 × 2 = 0 + 0.000 001 519 902 72;
  • 12) 0.000 001 519 902 72 × 2 = 0 + 0.000 003 039 805 44;
  • 13) 0.000 003 039 805 44 × 2 = 0 + 0.000 006 079 610 88;
  • 14) 0.000 006 079 610 88 × 2 = 0 + 0.000 012 159 221 76;
  • 15) 0.000 012 159 221 76 × 2 = 0 + 0.000 024 318 443 52;
  • 16) 0.000 024 318 443 52 × 2 = 0 + 0.000 048 636 887 04;
  • 17) 0.000 048 636 887 04 × 2 = 0 + 0.000 097 273 774 08;
  • 18) 0.000 097 273 774 08 × 2 = 0 + 0.000 194 547 548 16;
  • 19) 0.000 194 547 548 16 × 2 = 0 + 0.000 389 095 096 32;
  • 20) 0.000 389 095 096 32 × 2 = 0 + 0.000 778 190 192 64;
  • 21) 0.000 778 190 192 64 × 2 = 0 + 0.001 556 380 385 28;
  • 22) 0.001 556 380 385 28 × 2 = 0 + 0.003 112 760 770 56;
  • 23) 0.003 112 760 770 56 × 2 = 0 + 0.006 225 521 541 12;
  • 24) 0.006 225 521 541 12 × 2 = 0 + 0.012 451 043 082 24;
  • 25) 0.012 451 043 082 24 × 2 = 0 + 0.024 902 086 164 48;
  • 26) 0.024 902 086 164 48 × 2 = 0 + 0.049 804 172 328 96;
  • 27) 0.049 804 172 328 96 × 2 = 0 + 0.099 608 344 657 92;
  • 28) 0.099 608 344 657 92 × 2 = 0 + 0.199 216 689 315 84;
  • 29) 0.199 216 689 315 84 × 2 = 0 + 0.398 433 378 631 68;
  • 30) 0.398 433 378 631 68 × 2 = 0 + 0.796 866 757 263 36;
  • 31) 0.796 866 757 263 36 × 2 = 1 + 0.593 733 514 526 72;
  • 32) 0.593 733 514 526 72 × 2 = 1 + 0.187 467 029 053 44;
  • 33) 0.187 467 029 053 44 × 2 = 0 + 0.374 934 058 106 88;
  • 34) 0.374 934 058 106 88 × 2 = 0 + 0.749 868 116 213 76;
  • 35) 0.749 868 116 213 76 × 2 = 1 + 0.499 736 232 427 52;
  • 36) 0.499 736 232 427 52 × 2 = 0 + 0.999 472 464 855 04;
  • 37) 0.999 472 464 855 04 × 2 = 1 + 0.998 944 929 710 08;
  • 38) 0.998 944 929 710 08 × 2 = 1 + 0.997 889 859 420 16;
  • 39) 0.997 889 859 420 16 × 2 = 1 + 0.995 779 718 840 32;
  • 40) 0.995 779 718 840 32 × 2 = 1 + 0.991 559 437 680 64;
  • 41) 0.991 559 437 680 64 × 2 = 1 + 0.983 118 875 361 28;
  • 42) 0.983 118 875 361 28 × 2 = 1 + 0.966 237 750 722 56;
  • 43) 0.966 237 750 722 56 × 2 = 1 + 0.932 475 501 445 12;
  • 44) 0.932 475 501 445 12 × 2 = 1 + 0.864 951 002 890 24;
  • 45) 0.864 951 002 890 24 × 2 = 1 + 0.729 902 005 780 48;
  • 46) 0.729 902 005 780 48 × 2 = 1 + 0.459 804 011 560 96;
  • 47) 0.459 804 011 560 96 × 2 = 0 + 0.919 608 023 121 92;
  • 48) 0.919 608 023 121 92 × 2 = 1 + 0.839 216 046 243 84;
  • 49) 0.839 216 046 243 84 × 2 = 1 + 0.678 432 092 487 68;
  • 50) 0.678 432 092 487 68 × 2 = 1 + 0.356 864 184 975 36;
  • 51) 0.356 864 184 975 36 × 2 = 0 + 0.713 728 369 950 72;
  • 52) 0.713 728 369 950 72 × 2 = 1 + 0.427 456 739 901 44;
  • 53) 0.427 456 739 901 44 × 2 = 0 + 0.854 913 479 802 88;
  • 54) 0.854 913 479 802 88 × 2 = 1 + 0.709 826 959 605 76;

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 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1101 01(2)

6. Positive number before normalization:

0.000 000 000 742 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1101 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:


0.000 000 000 742 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1101 01(2) × 20 =

1.1001 0111 1111 1110 1110 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 1111 1110 1110 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 1111 1111 0111 0101 =

100 1011 1111 1111 0111 0101


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 1111 0111 0101


Decimal number -0.000 000 000 742 14 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 0111 0101

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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