-0.000 000 000 742 147 681 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 681 4(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
-0.000 000 000 742 147 681 4(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:

|-0.000 000 000 742 147 681 4| = 0.000 000 000 742 147 681 4


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: 0.000 000 000 742 147 681 4.

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) 0.000 000 000 742 147 681 4 × 2 = 0 + 0.000 000 001 484 295 362 8;
  • 2) 0.000 000 001 484 295 362 8 × 2 = 0 + 0.000 000 002 968 590 725 6;
  • 3) 0.000 000 002 968 590 725 6 × 2 = 0 + 0.000 000 005 937 181 451 2;
  • 4) 0.000 000 005 937 181 451 2 × 2 = 0 + 0.000 000 011 874 362 902 4;
  • 5) 0.000 000 011 874 362 902 4 × 2 = 0 + 0.000 000 023 748 725 804 8;
  • 6) 0.000 000 023 748 725 804 8 × 2 = 0 + 0.000 000 047 497 451 609 6;
  • 7) 0.000 000 047 497 451 609 6 × 2 = 0 + 0.000 000 094 994 903 219 2;
  • 8) 0.000 000 094 994 903 219 2 × 2 = 0 + 0.000 000 189 989 806 438 4;
  • 9) 0.000 000 189 989 806 438 4 × 2 = 0 + 0.000 000 379 979 612 876 8;
  • 10) 0.000 000 379 979 612 876 8 × 2 = 0 + 0.000 000 759 959 225 753 6;
  • 11) 0.000 000 759 959 225 753 6 × 2 = 0 + 0.000 001 519 918 451 507 2;
  • 12) 0.000 001 519 918 451 507 2 × 2 = 0 + 0.000 003 039 836 903 014 4;
  • 13) 0.000 003 039 836 903 014 4 × 2 = 0 + 0.000 006 079 673 806 028 8;
  • 14) 0.000 006 079 673 806 028 8 × 2 = 0 + 0.000 012 159 347 612 057 6;
  • 15) 0.000 012 159 347 612 057 6 × 2 = 0 + 0.000 024 318 695 224 115 2;
  • 16) 0.000 024 318 695 224 115 2 × 2 = 0 + 0.000 048 637 390 448 230 4;
  • 17) 0.000 048 637 390 448 230 4 × 2 = 0 + 0.000 097 274 780 896 460 8;
  • 18) 0.000 097 274 780 896 460 8 × 2 = 0 + 0.000 194 549 561 792 921 6;
  • 19) 0.000 194 549 561 792 921 6 × 2 = 0 + 0.000 389 099 123 585 843 2;
  • 20) 0.000 389 099 123 585 843 2 × 2 = 0 + 0.000 778 198 247 171 686 4;
  • 21) 0.000 778 198 247 171 686 4 × 2 = 0 + 0.001 556 396 494 343 372 8;
  • 22) 0.001 556 396 494 343 372 8 × 2 = 0 + 0.003 112 792 988 686 745 6;
  • 23) 0.003 112 792 988 686 745 6 × 2 = 0 + 0.006 225 585 977 373 491 2;
  • 24) 0.006 225 585 977 373 491 2 × 2 = 0 + 0.012 451 171 954 746 982 4;
  • 25) 0.012 451 171 954 746 982 4 × 2 = 0 + 0.024 902 343 909 493 964 8;
  • 26) 0.024 902 343 909 493 964 8 × 2 = 0 + 0.049 804 687 818 987 929 6;
  • 27) 0.049 804 687 818 987 929 6 × 2 = 0 + 0.099 609 375 637 975 859 2;
  • 28) 0.099 609 375 637 975 859 2 × 2 = 0 + 0.199 218 751 275 951 718 4;
  • 29) 0.199 218 751 275 951 718 4 × 2 = 0 + 0.398 437 502 551 903 436 8;
  • 30) 0.398 437 502 551 903 436 8 × 2 = 0 + 0.796 875 005 103 806 873 6;
  • 31) 0.796 875 005 103 806 873 6 × 2 = 1 + 0.593 750 010 207 613 747 2;
  • 32) 0.593 750 010 207 613 747 2 × 2 = 1 + 0.187 500 020 415 227 494 4;
  • 33) 0.187 500 020 415 227 494 4 × 2 = 0 + 0.375 000 040 830 454 988 8;
  • 34) 0.375 000 040 830 454 988 8 × 2 = 0 + 0.750 000 081 660 909 977 6;
  • 35) 0.750 000 081 660 909 977 6 × 2 = 1 + 0.500 000 163 321 819 955 2;
  • 36) 0.500 000 163 321 819 955 2 × 2 = 1 + 0.000 000 326 643 639 910 4;
  • 37) 0.000 000 326 643 639 910 4 × 2 = 0 + 0.000 000 653 287 279 820 8;
  • 38) 0.000 000 653 287 279 820 8 × 2 = 0 + 0.000 001 306 574 559 641 6;
  • 39) 0.000 001 306 574 559 641 6 × 2 = 0 + 0.000 002 613 149 119 283 2;
  • 40) 0.000 002 613 149 119 283 2 × 2 = 0 + 0.000 005 226 298 238 566 4;
  • 41) 0.000 005 226 298 238 566 4 × 2 = 0 + 0.000 010 452 596 477 132 8;
  • 42) 0.000 010 452 596 477 132 8 × 2 = 0 + 0.000 020 905 192 954 265 6;
  • 43) 0.000 020 905 192 954 265 6 × 2 = 0 + 0.000 041 810 385 908 531 2;
  • 44) 0.000 041 810 385 908 531 2 × 2 = 0 + 0.000 083 620 771 817 062 4;
  • 45) 0.000 083 620 771 817 062 4 × 2 = 0 + 0.000 167 241 543 634 124 8;
  • 46) 0.000 167 241 543 634 124 8 × 2 = 0 + 0.000 334 483 087 268 249 6;
  • 47) 0.000 334 483 087 268 249 6 × 2 = 0 + 0.000 668 966 174 536 499 2;
  • 48) 0.000 668 966 174 536 499 2 × 2 = 0 + 0.001 337 932 349 072 998 4;
  • 49) 0.001 337 932 349 072 998 4 × 2 = 0 + 0.002 675 864 698 145 996 8;
  • 50) 0.002 675 864 698 145 996 8 × 2 = 0 + 0.005 351 729 396 291 993 6;
  • 51) 0.005 351 729 396 291 993 6 × 2 = 0 + 0.010 703 458 792 583 987 2;
  • 52) 0.010 703 458 792 583 987 2 × 2 = 0 + 0.021 406 917 585 167 974 4;
  • 53) 0.021 406 917 585 167 974 4 × 2 = 0 + 0.042 813 835 170 335 948 8;
  • 54) 0.042 813 835 170 335 948 8 × 2 = 0 + 0.085 627 670 340 671 897 6;

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:


0.000 000 000 742 147 681 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 681 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 681 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 681 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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