-0.000 000 000 742 147 676 646 686 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 686 1(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
-0.000 000 000 742 147 676 646 686 1(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:

|-0.000 000 000 742 147 676 646 686 1| = 0.000 000 000 742 147 676 646 686 1


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: 0.000 000 000 742 147 676 646 686 1.

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) 0.000 000 000 742 147 676 646 686 1 × 2 = 0 + 0.000 000 001 484 295 353 293 372 2;
  • 2) 0.000 000 001 484 295 353 293 372 2 × 2 = 0 + 0.000 000 002 968 590 706 586 744 4;
  • 3) 0.000 000 002 968 590 706 586 744 4 × 2 = 0 + 0.000 000 005 937 181 413 173 488 8;
  • 4) 0.000 000 005 937 181 413 173 488 8 × 2 = 0 + 0.000 000 011 874 362 826 346 977 6;
  • 5) 0.000 000 011 874 362 826 346 977 6 × 2 = 0 + 0.000 000 023 748 725 652 693 955 2;
  • 6) 0.000 000 023 748 725 652 693 955 2 × 2 = 0 + 0.000 000 047 497 451 305 387 910 4;
  • 7) 0.000 000 047 497 451 305 387 910 4 × 2 = 0 + 0.000 000 094 994 902 610 775 820 8;
  • 8) 0.000 000 094 994 902 610 775 820 8 × 2 = 0 + 0.000 000 189 989 805 221 551 641 6;
  • 9) 0.000 000 189 989 805 221 551 641 6 × 2 = 0 + 0.000 000 379 979 610 443 103 283 2;
  • 10) 0.000 000 379 979 610 443 103 283 2 × 2 = 0 + 0.000 000 759 959 220 886 206 566 4;
  • 11) 0.000 000 759 959 220 886 206 566 4 × 2 = 0 + 0.000 001 519 918 441 772 413 132 8;
  • 12) 0.000 001 519 918 441 772 413 132 8 × 2 = 0 + 0.000 003 039 836 883 544 826 265 6;
  • 13) 0.000 003 039 836 883 544 826 265 6 × 2 = 0 + 0.000 006 079 673 767 089 652 531 2;
  • 14) 0.000 006 079 673 767 089 652 531 2 × 2 = 0 + 0.000 012 159 347 534 179 305 062 4;
  • 15) 0.000 012 159 347 534 179 305 062 4 × 2 = 0 + 0.000 024 318 695 068 358 610 124 8;
  • 16) 0.000 024 318 695 068 358 610 124 8 × 2 = 0 + 0.000 048 637 390 136 717 220 249 6;
  • 17) 0.000 048 637 390 136 717 220 249 6 × 2 = 0 + 0.000 097 274 780 273 434 440 499 2;
  • 18) 0.000 097 274 780 273 434 440 499 2 × 2 = 0 + 0.000 194 549 560 546 868 880 998 4;
  • 19) 0.000 194 549 560 546 868 880 998 4 × 2 = 0 + 0.000 389 099 121 093 737 761 996 8;
  • 20) 0.000 389 099 121 093 737 761 996 8 × 2 = 0 + 0.000 778 198 242 187 475 523 993 6;
  • 21) 0.000 778 198 242 187 475 523 993 6 × 2 = 0 + 0.001 556 396 484 374 951 047 987 2;
  • 22) 0.001 556 396 484 374 951 047 987 2 × 2 = 0 + 0.003 112 792 968 749 902 095 974 4;
  • 23) 0.003 112 792 968 749 902 095 974 4 × 2 = 0 + 0.006 225 585 937 499 804 191 948 8;
  • 24) 0.006 225 585 937 499 804 191 948 8 × 2 = 0 + 0.012 451 171 874 999 608 383 897 6;
  • 25) 0.012 451 171 874 999 608 383 897 6 × 2 = 0 + 0.024 902 343 749 999 216 767 795 2;
  • 26) 0.024 902 343 749 999 216 767 795 2 × 2 = 0 + 0.049 804 687 499 998 433 535 590 4;
  • 27) 0.049 804 687 499 998 433 535 590 4 × 2 = 0 + 0.099 609 374 999 996 867 071 180 8;
  • 28) 0.099 609 374 999 996 867 071 180 8 × 2 = 0 + 0.199 218 749 999 993 734 142 361 6;
  • 29) 0.199 218 749 999 993 734 142 361 6 × 2 = 0 + 0.398 437 499 999 987 468 284 723 2;
  • 30) 0.398 437 499 999 987 468 284 723 2 × 2 = 0 + 0.796 874 999 999 974 936 569 446 4;
  • 31) 0.796 874 999 999 974 936 569 446 4 × 2 = 1 + 0.593 749 999 999 949 873 138 892 8;
  • 32) 0.593 749 999 999 949 873 138 892 8 × 2 = 1 + 0.187 499 999 999 899 746 277 785 6;
  • 33) 0.187 499 999 999 899 746 277 785 6 × 2 = 0 + 0.374 999 999 999 799 492 555 571 2;
  • 34) 0.374 999 999 999 799 492 555 571 2 × 2 = 0 + 0.749 999 999 999 598 985 111 142 4;
  • 35) 0.749 999 999 999 598 985 111 142 4 × 2 = 1 + 0.499 999 999 999 197 970 222 284 8;
  • 36) 0.499 999 999 999 197 970 222 284 8 × 2 = 0 + 0.999 999 999 998 395 940 444 569 6;
  • 37) 0.999 999 999 998 395 940 444 569 6 × 2 = 1 + 0.999 999 999 996 791 880 889 139 2;
  • 38) 0.999 999 999 996 791 880 889 139 2 × 2 = 1 + 0.999 999 999 993 583 761 778 278 4;
  • 39) 0.999 999 999 993 583 761 778 278 4 × 2 = 1 + 0.999 999 999 987 167 523 556 556 8;
  • 40) 0.999 999 999 987 167 523 556 556 8 × 2 = 1 + 0.999 999 999 974 335 047 113 113 6;
  • 41) 0.999 999 999 974 335 047 113 113 6 × 2 = 1 + 0.999 999 999 948 670 094 226 227 2;
  • 42) 0.999 999 999 948 670 094 226 227 2 × 2 = 1 + 0.999 999 999 897 340 188 452 454 4;
  • 43) 0.999 999 999 897 340 188 452 454 4 × 2 = 1 + 0.999 999 999 794 680 376 904 908 8;
  • 44) 0.999 999 999 794 680 376 904 908 8 × 2 = 1 + 0.999 999 999 589 360 753 809 817 6;
  • 45) 0.999 999 999 589 360 753 809 817 6 × 2 = 1 + 0.999 999 999 178 721 507 619 635 2;
  • 46) 0.999 999 999 178 721 507 619 635 2 × 2 = 1 + 0.999 999 998 357 443 015 239 270 4;
  • 47) 0.999 999 998 357 443 015 239 270 4 × 2 = 1 + 0.999 999 996 714 886 030 478 540 8;
  • 48) 0.999 999 996 714 886 030 478 540 8 × 2 = 1 + 0.999 999 993 429 772 060 957 081 6;
  • 49) 0.999 999 993 429 772 060 957 081 6 × 2 = 1 + 0.999 999 986 859 544 121 914 163 2;
  • 50) 0.999 999 986 859 544 121 914 163 2 × 2 = 1 + 0.999 999 973 719 088 243 828 326 4;
  • 51) 0.999 999 973 719 088 243 828 326 4 × 2 = 1 + 0.999 999 947 438 176 487 656 652 8;
  • 52) 0.999 999 947 438 176 487 656 652 8 × 2 = 1 + 0.999 999 894 876 352 975 313 305 6;
  • 53) 0.999 999 894 876 352 975 313 305 6 × 2 = 1 + 0.999 999 789 752 705 950 626 611 2;
  • 54) 0.999 999 789 752 705 950 626 611 2 × 2 = 1 + 0.999 999 579 505 411 901 253 222 4;

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:


0.000 000 000 742 147 676 646 686 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 686 1(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:


0.000 000 000 742 147 676 646 686 1(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 -0.000 000 000 742 147 676 646 686 1 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