-0.000 000 000 742 142 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 142 5(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
-0.000 000 000 742 142 5(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:

|-0.000 000 000 742 142 5| = 0.000 000 000 742 142 5


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: 0.000 000 000 742 142 5.

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) 0.000 000 000 742 142 5 × 2 = 0 + 0.000 000 001 484 285;
  • 2) 0.000 000 001 484 285 × 2 = 0 + 0.000 000 002 968 57;
  • 3) 0.000 000 002 968 57 × 2 = 0 + 0.000 000 005 937 14;
  • 4) 0.000 000 005 937 14 × 2 = 0 + 0.000 000 011 874 28;
  • 5) 0.000 000 011 874 28 × 2 = 0 + 0.000 000 023 748 56;
  • 6) 0.000 000 023 748 56 × 2 = 0 + 0.000 000 047 497 12;
  • 7) 0.000 000 047 497 12 × 2 = 0 + 0.000 000 094 994 24;
  • 8) 0.000 000 094 994 24 × 2 = 0 + 0.000 000 189 988 48;
  • 9) 0.000 000 189 988 48 × 2 = 0 + 0.000 000 379 976 96;
  • 10) 0.000 000 379 976 96 × 2 = 0 + 0.000 000 759 953 92;
  • 11) 0.000 000 759 953 92 × 2 = 0 + 0.000 001 519 907 84;
  • 12) 0.000 001 519 907 84 × 2 = 0 + 0.000 003 039 815 68;
  • 13) 0.000 003 039 815 68 × 2 = 0 + 0.000 006 079 631 36;
  • 14) 0.000 006 079 631 36 × 2 = 0 + 0.000 012 159 262 72;
  • 15) 0.000 012 159 262 72 × 2 = 0 + 0.000 024 318 525 44;
  • 16) 0.000 024 318 525 44 × 2 = 0 + 0.000 048 637 050 88;
  • 17) 0.000 048 637 050 88 × 2 = 0 + 0.000 097 274 101 76;
  • 18) 0.000 097 274 101 76 × 2 = 0 + 0.000 194 548 203 52;
  • 19) 0.000 194 548 203 52 × 2 = 0 + 0.000 389 096 407 04;
  • 20) 0.000 389 096 407 04 × 2 = 0 + 0.000 778 192 814 08;
  • 21) 0.000 778 192 814 08 × 2 = 0 + 0.001 556 385 628 16;
  • 22) 0.001 556 385 628 16 × 2 = 0 + 0.003 112 771 256 32;
  • 23) 0.003 112 771 256 32 × 2 = 0 + 0.006 225 542 512 64;
  • 24) 0.006 225 542 512 64 × 2 = 0 + 0.012 451 085 025 28;
  • 25) 0.012 451 085 025 28 × 2 = 0 + 0.024 902 170 050 56;
  • 26) 0.024 902 170 050 56 × 2 = 0 + 0.049 804 340 101 12;
  • 27) 0.049 804 340 101 12 × 2 = 0 + 0.099 608 680 202 24;
  • 28) 0.099 608 680 202 24 × 2 = 0 + 0.199 217 360 404 48;
  • 29) 0.199 217 360 404 48 × 2 = 0 + 0.398 434 720 808 96;
  • 30) 0.398 434 720 808 96 × 2 = 0 + 0.796 869 441 617 92;
  • 31) 0.796 869 441 617 92 × 2 = 1 + 0.593 738 883 235 84;
  • 32) 0.593 738 883 235 84 × 2 = 1 + 0.187 477 766 471 68;
  • 33) 0.187 477 766 471 68 × 2 = 0 + 0.374 955 532 943 36;
  • 34) 0.374 955 532 943 36 × 2 = 0 + 0.749 911 065 886 72;
  • 35) 0.749 911 065 886 72 × 2 = 1 + 0.499 822 131 773 44;
  • 36) 0.499 822 131 773 44 × 2 = 0 + 0.999 644 263 546 88;
  • 37) 0.999 644 263 546 88 × 2 = 1 + 0.999 288 527 093 76;
  • 38) 0.999 288 527 093 76 × 2 = 1 + 0.998 577 054 187 52;
  • 39) 0.998 577 054 187 52 × 2 = 1 + 0.997 154 108 375 04;
  • 40) 0.997 154 108 375 04 × 2 = 1 + 0.994 308 216 750 08;
  • 41) 0.994 308 216 750 08 × 2 = 1 + 0.988 616 433 500 16;
  • 42) 0.988 616 433 500 16 × 2 = 1 + 0.977 232 867 000 32;
  • 43) 0.977 232 867 000 32 × 2 = 1 + 0.954 465 734 000 64;
  • 44) 0.954 465 734 000 64 × 2 = 1 + 0.908 931 468 001 28;
  • 45) 0.908 931 468 001 28 × 2 = 1 + 0.817 862 936 002 56;
  • 46) 0.817 862 936 002 56 × 2 = 1 + 0.635 725 872 005 12;
  • 47) 0.635 725 872 005 12 × 2 = 1 + 0.271 451 744 010 24;
  • 48) 0.271 451 744 010 24 × 2 = 0 + 0.542 903 488 020 48;
  • 49) 0.542 903 488 020 48 × 2 = 1 + 0.085 806 976 040 96;
  • 50) 0.085 806 976 040 96 × 2 = 0 + 0.171 613 952 081 92;
  • 51) 0.171 613 952 081 92 × 2 = 0 + 0.343 227 904 163 84;
  • 52) 0.343 227 904 163 84 × 2 = 0 + 0.686 455 808 327 68;
  • 53) 0.686 455 808 327 68 × 2 = 1 + 0.372 911 616 655 36;
  • 54) 0.372 911 616 655 36 × 2 = 0 + 0.745 823 233 310 72;

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:


0.000 000 000 742 142 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1000 10(2)

6. Positive number before normalization:

0.000 000 000 742 142 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 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:


0.000 000 000 742 142 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1000 10(2) × 20 =

1.1001 0111 1111 1111 0100 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 1111 1111 0100 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 1111 1111 1010 0010 =

100 1011 1111 1111 1010 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 1111 1111 1010 0010


Decimal number -0.000 000 000 742 142 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1010 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