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

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


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.

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 × 2 = 0 + 0.000 000 001 484;
  • 2) 0.000 000 001 484 × 2 = 0 + 0.000 000 002 968;
  • 3) 0.000 000 002 968 × 2 = 0 + 0.000 000 005 936;
  • 4) 0.000 000 005 936 × 2 = 0 + 0.000 000 011 872;
  • 5) 0.000 000 011 872 × 2 = 0 + 0.000 000 023 744;
  • 6) 0.000 000 023 744 × 2 = 0 + 0.000 000 047 488;
  • 7) 0.000 000 047 488 × 2 = 0 + 0.000 000 094 976;
  • 8) 0.000 000 094 976 × 2 = 0 + 0.000 000 189 952;
  • 9) 0.000 000 189 952 × 2 = 0 + 0.000 000 379 904;
  • 10) 0.000 000 379 904 × 2 = 0 + 0.000 000 759 808;
  • 11) 0.000 000 759 808 × 2 = 0 + 0.000 001 519 616;
  • 12) 0.000 001 519 616 × 2 = 0 + 0.000 003 039 232;
  • 13) 0.000 003 039 232 × 2 = 0 + 0.000 006 078 464;
  • 14) 0.000 006 078 464 × 2 = 0 + 0.000 012 156 928;
  • 15) 0.000 012 156 928 × 2 = 0 + 0.000 024 313 856;
  • 16) 0.000 024 313 856 × 2 = 0 + 0.000 048 627 712;
  • 17) 0.000 048 627 712 × 2 = 0 + 0.000 097 255 424;
  • 18) 0.000 097 255 424 × 2 = 0 + 0.000 194 510 848;
  • 19) 0.000 194 510 848 × 2 = 0 + 0.000 389 021 696;
  • 20) 0.000 389 021 696 × 2 = 0 + 0.000 778 043 392;
  • 21) 0.000 778 043 392 × 2 = 0 + 0.001 556 086 784;
  • 22) 0.001 556 086 784 × 2 = 0 + 0.003 112 173 568;
  • 23) 0.003 112 173 568 × 2 = 0 + 0.006 224 347 136;
  • 24) 0.006 224 347 136 × 2 = 0 + 0.012 448 694 272;
  • 25) 0.012 448 694 272 × 2 = 0 + 0.024 897 388 544;
  • 26) 0.024 897 388 544 × 2 = 0 + 0.049 794 777 088;
  • 27) 0.049 794 777 088 × 2 = 0 + 0.099 589 554 176;
  • 28) 0.099 589 554 176 × 2 = 0 + 0.199 179 108 352;
  • 29) 0.199 179 108 352 × 2 = 0 + 0.398 358 216 704;
  • 30) 0.398 358 216 704 × 2 = 0 + 0.796 716 433 408;
  • 31) 0.796 716 433 408 × 2 = 1 + 0.593 432 866 816;
  • 32) 0.593 432 866 816 × 2 = 1 + 0.186 865 733 632;
  • 33) 0.186 865 733 632 × 2 = 0 + 0.373 731 467 264;
  • 34) 0.373 731 467 264 × 2 = 0 + 0.747 462 934 528;
  • 35) 0.747 462 934 528 × 2 = 1 + 0.494 925 869 056;
  • 36) 0.494 925 869 056 × 2 = 0 + 0.989 851 738 112;
  • 37) 0.989 851 738 112 × 2 = 1 + 0.979 703 476 224;
  • 38) 0.979 703 476 224 × 2 = 1 + 0.959 406 952 448;
  • 39) 0.959 406 952 448 × 2 = 1 + 0.918 813 904 896;
  • 40) 0.918 813 904 896 × 2 = 1 + 0.837 627 809 792;
  • 41) 0.837 627 809 792 × 2 = 1 + 0.675 255 619 584;
  • 42) 0.675 255 619 584 × 2 = 1 + 0.350 511 239 168;
  • 43) 0.350 511 239 168 × 2 = 0 + 0.701 022 478 336;
  • 44) 0.701 022 478 336 × 2 = 1 + 0.402 044 956 672;
  • 45) 0.402 044 956 672 × 2 = 0 + 0.804 089 913 344;
  • 46) 0.804 089 913 344 × 2 = 1 + 0.608 179 826 688;
  • 47) 0.608 179 826 688 × 2 = 1 + 0.216 359 653 376;
  • 48) 0.216 359 653 376 × 2 = 0 + 0.432 719 306 752;
  • 49) 0.432 719 306 752 × 2 = 0 + 0.865 438 613 504;
  • 50) 0.865 438 613 504 × 2 = 1 + 0.730 877 227 008;
  • 51) 0.730 877 227 008 × 2 = 1 + 0.461 754 454 016;
  • 52) 0.461 754 454 016 × 2 = 0 + 0.923 508 908 032;
  • 53) 0.923 508 908 032 × 2 = 1 + 0.847 017 816 064;
  • 54) 0.847 017 816 064 × 2 = 1 + 0.694 035 632 128;

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(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0110 0110 11(2)

6. Positive number before normalization:

0.000 000 000 742(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0110 0110 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(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0110 0110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0110 0110 11(2) × 20 =

1.1001 0111 1110 1011 0011 011(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 1110 1011 0011 011


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 0101 1001 1011 =

100 1011 1111 0101 1001 1011


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 0101 1001 1011


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

1 - 0110 0000 - 100 1011 1111 0101 1001 1011

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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