-0.000 000 000 742 147 661 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 661 7(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
-0.000 000 000 742 147 661 7(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:

|-0.000 000 000 742 147 661 7| = 0.000 000 000 742 147 661 7


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: 0.000 000 000 742 147 661 7.

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) 0.000 000 000 742 147 661 7 × 2 = 0 + 0.000 000 001 484 295 323 4;
  • 2) 0.000 000 001 484 295 323 4 × 2 = 0 + 0.000 000 002 968 590 646 8;
  • 3) 0.000 000 002 968 590 646 8 × 2 = 0 + 0.000 000 005 937 181 293 6;
  • 4) 0.000 000 005 937 181 293 6 × 2 = 0 + 0.000 000 011 874 362 587 2;
  • 5) 0.000 000 011 874 362 587 2 × 2 = 0 + 0.000 000 023 748 725 174 4;
  • 6) 0.000 000 023 748 725 174 4 × 2 = 0 + 0.000 000 047 497 450 348 8;
  • 7) 0.000 000 047 497 450 348 8 × 2 = 0 + 0.000 000 094 994 900 697 6;
  • 8) 0.000 000 094 994 900 697 6 × 2 = 0 + 0.000 000 189 989 801 395 2;
  • 9) 0.000 000 189 989 801 395 2 × 2 = 0 + 0.000 000 379 979 602 790 4;
  • 10) 0.000 000 379 979 602 790 4 × 2 = 0 + 0.000 000 759 959 205 580 8;
  • 11) 0.000 000 759 959 205 580 8 × 2 = 0 + 0.000 001 519 918 411 161 6;
  • 12) 0.000 001 519 918 411 161 6 × 2 = 0 + 0.000 003 039 836 822 323 2;
  • 13) 0.000 003 039 836 822 323 2 × 2 = 0 + 0.000 006 079 673 644 646 4;
  • 14) 0.000 006 079 673 644 646 4 × 2 = 0 + 0.000 012 159 347 289 292 8;
  • 15) 0.000 012 159 347 289 292 8 × 2 = 0 + 0.000 024 318 694 578 585 6;
  • 16) 0.000 024 318 694 578 585 6 × 2 = 0 + 0.000 048 637 389 157 171 2;
  • 17) 0.000 048 637 389 157 171 2 × 2 = 0 + 0.000 097 274 778 314 342 4;
  • 18) 0.000 097 274 778 314 342 4 × 2 = 0 + 0.000 194 549 556 628 684 8;
  • 19) 0.000 194 549 556 628 684 8 × 2 = 0 + 0.000 389 099 113 257 369 6;
  • 20) 0.000 389 099 113 257 369 6 × 2 = 0 + 0.000 778 198 226 514 739 2;
  • 21) 0.000 778 198 226 514 739 2 × 2 = 0 + 0.001 556 396 453 029 478 4;
  • 22) 0.001 556 396 453 029 478 4 × 2 = 0 + 0.003 112 792 906 058 956 8;
  • 23) 0.003 112 792 906 058 956 8 × 2 = 0 + 0.006 225 585 812 117 913 6;
  • 24) 0.006 225 585 812 117 913 6 × 2 = 0 + 0.012 451 171 624 235 827 2;
  • 25) 0.012 451 171 624 235 827 2 × 2 = 0 + 0.024 902 343 248 471 654 4;
  • 26) 0.024 902 343 248 471 654 4 × 2 = 0 + 0.049 804 686 496 943 308 8;
  • 27) 0.049 804 686 496 943 308 8 × 2 = 0 + 0.099 609 372 993 886 617 6;
  • 28) 0.099 609 372 993 886 617 6 × 2 = 0 + 0.199 218 745 987 773 235 2;
  • 29) 0.199 218 745 987 773 235 2 × 2 = 0 + 0.398 437 491 975 546 470 4;
  • 30) 0.398 437 491 975 546 470 4 × 2 = 0 + 0.796 874 983 951 092 940 8;
  • 31) 0.796 874 983 951 092 940 8 × 2 = 1 + 0.593 749 967 902 185 881 6;
  • 32) 0.593 749 967 902 185 881 6 × 2 = 1 + 0.187 499 935 804 371 763 2;
  • 33) 0.187 499 935 804 371 763 2 × 2 = 0 + 0.374 999 871 608 743 526 4;
  • 34) 0.374 999 871 608 743 526 4 × 2 = 0 + 0.749 999 743 217 487 052 8;
  • 35) 0.749 999 743 217 487 052 8 × 2 = 1 + 0.499 999 486 434 974 105 6;
  • 36) 0.499 999 486 434 974 105 6 × 2 = 0 + 0.999 998 972 869 948 211 2;
  • 37) 0.999 998 972 869 948 211 2 × 2 = 1 + 0.999 997 945 739 896 422 4;
  • 38) 0.999 997 945 739 896 422 4 × 2 = 1 + 0.999 995 891 479 792 844 8;
  • 39) 0.999 995 891 479 792 844 8 × 2 = 1 + 0.999 991 782 959 585 689 6;
  • 40) 0.999 991 782 959 585 689 6 × 2 = 1 + 0.999 983 565 919 171 379 2;
  • 41) 0.999 983 565 919 171 379 2 × 2 = 1 + 0.999 967 131 838 342 758 4;
  • 42) 0.999 967 131 838 342 758 4 × 2 = 1 + 0.999 934 263 676 685 516 8;
  • 43) 0.999 934 263 676 685 516 8 × 2 = 1 + 0.999 868 527 353 371 033 6;
  • 44) 0.999 868 527 353 371 033 6 × 2 = 1 + 0.999 737 054 706 742 067 2;
  • 45) 0.999 737 054 706 742 067 2 × 2 = 1 + 0.999 474 109 413 484 134 4;
  • 46) 0.999 474 109 413 484 134 4 × 2 = 1 + 0.998 948 218 826 968 268 8;
  • 47) 0.998 948 218 826 968 268 8 × 2 = 1 + 0.997 896 437 653 936 537 6;
  • 48) 0.997 896 437 653 936 537 6 × 2 = 1 + 0.995 792 875 307 873 075 2;
  • 49) 0.995 792 875 307 873 075 2 × 2 = 1 + 0.991 585 750 615 746 150 4;
  • 50) 0.991 585 750 615 746 150 4 × 2 = 1 + 0.983 171 501 231 492 300 8;
  • 51) 0.983 171 501 231 492 300 8 × 2 = 1 + 0.966 343 002 462 984 601 6;
  • 52) 0.966 343 002 462 984 601 6 × 2 = 1 + 0.932 686 004 925 969 203 2;
  • 53) 0.932 686 004 925 969 203 2 × 2 = 1 + 0.865 372 009 851 938 406 4;
  • 54) 0.865 372 009 851 938 406 4 × 2 = 1 + 0.730 744 019 703 876 812 8;

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:


0.000 000 000 742 147 661 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 661 7(10) =

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


0.000 000 000 742 147 661 7(10) =

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

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

1.1001 0111 1111 1111 1111 111(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 1111 111


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 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111 1111


Decimal number -0.000 000 000 742 147 661 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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