-0.000 000 000 742 031 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 031(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
-0.000 000 000 742 031(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:

|-0.000 000 000 742 031| = 0.000 000 000 742 031


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: 0.000 000 000 742 031.

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) 0.000 000 000 742 031 × 2 = 0 + 0.000 000 001 484 062;
  • 2) 0.000 000 001 484 062 × 2 = 0 + 0.000 000 002 968 124;
  • 3) 0.000 000 002 968 124 × 2 = 0 + 0.000 000 005 936 248;
  • 4) 0.000 000 005 936 248 × 2 = 0 + 0.000 000 011 872 496;
  • 5) 0.000 000 011 872 496 × 2 = 0 + 0.000 000 023 744 992;
  • 6) 0.000 000 023 744 992 × 2 = 0 + 0.000 000 047 489 984;
  • 7) 0.000 000 047 489 984 × 2 = 0 + 0.000 000 094 979 968;
  • 8) 0.000 000 094 979 968 × 2 = 0 + 0.000 000 189 959 936;
  • 9) 0.000 000 189 959 936 × 2 = 0 + 0.000 000 379 919 872;
  • 10) 0.000 000 379 919 872 × 2 = 0 + 0.000 000 759 839 744;
  • 11) 0.000 000 759 839 744 × 2 = 0 + 0.000 001 519 679 488;
  • 12) 0.000 001 519 679 488 × 2 = 0 + 0.000 003 039 358 976;
  • 13) 0.000 003 039 358 976 × 2 = 0 + 0.000 006 078 717 952;
  • 14) 0.000 006 078 717 952 × 2 = 0 + 0.000 012 157 435 904;
  • 15) 0.000 012 157 435 904 × 2 = 0 + 0.000 024 314 871 808;
  • 16) 0.000 024 314 871 808 × 2 = 0 + 0.000 048 629 743 616;
  • 17) 0.000 048 629 743 616 × 2 = 0 + 0.000 097 259 487 232;
  • 18) 0.000 097 259 487 232 × 2 = 0 + 0.000 194 518 974 464;
  • 19) 0.000 194 518 974 464 × 2 = 0 + 0.000 389 037 948 928;
  • 20) 0.000 389 037 948 928 × 2 = 0 + 0.000 778 075 897 856;
  • 21) 0.000 778 075 897 856 × 2 = 0 + 0.001 556 151 795 712;
  • 22) 0.001 556 151 795 712 × 2 = 0 + 0.003 112 303 591 424;
  • 23) 0.003 112 303 591 424 × 2 = 0 + 0.006 224 607 182 848;
  • 24) 0.006 224 607 182 848 × 2 = 0 + 0.012 449 214 365 696;
  • 25) 0.012 449 214 365 696 × 2 = 0 + 0.024 898 428 731 392;
  • 26) 0.024 898 428 731 392 × 2 = 0 + 0.049 796 857 462 784;
  • 27) 0.049 796 857 462 784 × 2 = 0 + 0.099 593 714 925 568;
  • 28) 0.099 593 714 925 568 × 2 = 0 + 0.199 187 429 851 136;
  • 29) 0.199 187 429 851 136 × 2 = 0 + 0.398 374 859 702 272;
  • 30) 0.398 374 859 702 272 × 2 = 0 + 0.796 749 719 404 544;
  • 31) 0.796 749 719 404 544 × 2 = 1 + 0.593 499 438 809 088;
  • 32) 0.593 499 438 809 088 × 2 = 1 + 0.186 998 877 618 176;
  • 33) 0.186 998 877 618 176 × 2 = 0 + 0.373 997 755 236 352;
  • 34) 0.373 997 755 236 352 × 2 = 0 + 0.747 995 510 472 704;
  • 35) 0.747 995 510 472 704 × 2 = 1 + 0.495 991 020 945 408;
  • 36) 0.495 991 020 945 408 × 2 = 0 + 0.991 982 041 890 816;
  • 37) 0.991 982 041 890 816 × 2 = 1 + 0.983 964 083 781 632;
  • 38) 0.983 964 083 781 632 × 2 = 1 + 0.967 928 167 563 264;
  • 39) 0.967 928 167 563 264 × 2 = 1 + 0.935 856 335 126 528;
  • 40) 0.935 856 335 126 528 × 2 = 1 + 0.871 712 670 253 056;
  • 41) 0.871 712 670 253 056 × 2 = 1 + 0.743 425 340 506 112;
  • 42) 0.743 425 340 506 112 × 2 = 1 + 0.486 850 681 012 224;
  • 43) 0.486 850 681 012 224 × 2 = 0 + 0.973 701 362 024 448;
  • 44) 0.973 701 362 024 448 × 2 = 1 + 0.947 402 724 048 896;
  • 45) 0.947 402 724 048 896 × 2 = 1 + 0.894 805 448 097 792;
  • 46) 0.894 805 448 097 792 × 2 = 1 + 0.789 610 896 195 584;
  • 47) 0.789 610 896 195 584 × 2 = 1 + 0.579 221 792 391 168;
  • 48) 0.579 221 792 391 168 × 2 = 1 + 0.158 443 584 782 336;
  • 49) 0.158 443 584 782 336 × 2 = 0 + 0.316 887 169 564 672;
  • 50) 0.316 887 169 564 672 × 2 = 0 + 0.633 774 339 129 344;
  • 51) 0.633 774 339 129 344 × 2 = 1 + 0.267 548 678 258 688;
  • 52) 0.267 548 678 258 688 × 2 = 0 + 0.535 097 356 517 376;
  • 53) 0.535 097 356 517 376 × 2 = 1 + 0.070 194 713 034 752;
  • 54) 0.070 194 713 034 752 × 2 = 0 + 0.140 389 426 069 504;

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:


0.000 000 000 742 031(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1111 0010 10(2)

6. Positive number before normalization:

0.000 000 000 742 031(10) =

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


0.000 000 000 742 031(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1111 0010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1111 0010 10(2) × 20 =

1.1001 0111 1110 1111 1001 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 1110 1111 1001 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 0111 1100 1010 =

100 1011 1111 0111 1100 1010


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 0111 1100 1010


Decimal number -0.000 000 000 742 031 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0111 1100 1010

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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