-0.000 000 000 742 147 496 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 496(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
-0.000 000 000 742 147 496(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:

|-0.000 000 000 742 147 496| = 0.000 000 000 742 147 496


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: 0.000 000 000 742 147 496.

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) 0.000 000 000 742 147 496 × 2 = 0 + 0.000 000 001 484 294 992;
  • 2) 0.000 000 001 484 294 992 × 2 = 0 + 0.000 000 002 968 589 984;
  • 3) 0.000 000 002 968 589 984 × 2 = 0 + 0.000 000 005 937 179 968;
  • 4) 0.000 000 005 937 179 968 × 2 = 0 + 0.000 000 011 874 359 936;
  • 5) 0.000 000 011 874 359 936 × 2 = 0 + 0.000 000 023 748 719 872;
  • 6) 0.000 000 023 748 719 872 × 2 = 0 + 0.000 000 047 497 439 744;
  • 7) 0.000 000 047 497 439 744 × 2 = 0 + 0.000 000 094 994 879 488;
  • 8) 0.000 000 094 994 879 488 × 2 = 0 + 0.000 000 189 989 758 976;
  • 9) 0.000 000 189 989 758 976 × 2 = 0 + 0.000 000 379 979 517 952;
  • 10) 0.000 000 379 979 517 952 × 2 = 0 + 0.000 000 759 959 035 904;
  • 11) 0.000 000 759 959 035 904 × 2 = 0 + 0.000 001 519 918 071 808;
  • 12) 0.000 001 519 918 071 808 × 2 = 0 + 0.000 003 039 836 143 616;
  • 13) 0.000 003 039 836 143 616 × 2 = 0 + 0.000 006 079 672 287 232;
  • 14) 0.000 006 079 672 287 232 × 2 = 0 + 0.000 012 159 344 574 464;
  • 15) 0.000 012 159 344 574 464 × 2 = 0 + 0.000 024 318 689 148 928;
  • 16) 0.000 024 318 689 148 928 × 2 = 0 + 0.000 048 637 378 297 856;
  • 17) 0.000 048 637 378 297 856 × 2 = 0 + 0.000 097 274 756 595 712;
  • 18) 0.000 097 274 756 595 712 × 2 = 0 + 0.000 194 549 513 191 424;
  • 19) 0.000 194 549 513 191 424 × 2 = 0 + 0.000 389 099 026 382 848;
  • 20) 0.000 389 099 026 382 848 × 2 = 0 + 0.000 778 198 052 765 696;
  • 21) 0.000 778 198 052 765 696 × 2 = 0 + 0.001 556 396 105 531 392;
  • 22) 0.001 556 396 105 531 392 × 2 = 0 + 0.003 112 792 211 062 784;
  • 23) 0.003 112 792 211 062 784 × 2 = 0 + 0.006 225 584 422 125 568;
  • 24) 0.006 225 584 422 125 568 × 2 = 0 + 0.012 451 168 844 251 136;
  • 25) 0.012 451 168 844 251 136 × 2 = 0 + 0.024 902 337 688 502 272;
  • 26) 0.024 902 337 688 502 272 × 2 = 0 + 0.049 804 675 377 004 544;
  • 27) 0.049 804 675 377 004 544 × 2 = 0 + 0.099 609 350 754 009 088;
  • 28) 0.099 609 350 754 009 088 × 2 = 0 + 0.199 218 701 508 018 176;
  • 29) 0.199 218 701 508 018 176 × 2 = 0 + 0.398 437 403 016 036 352;
  • 30) 0.398 437 403 016 036 352 × 2 = 0 + 0.796 874 806 032 072 704;
  • 31) 0.796 874 806 032 072 704 × 2 = 1 + 0.593 749 612 064 145 408;
  • 32) 0.593 749 612 064 145 408 × 2 = 1 + 0.187 499 224 128 290 816;
  • 33) 0.187 499 224 128 290 816 × 2 = 0 + 0.374 998 448 256 581 632;
  • 34) 0.374 998 448 256 581 632 × 2 = 0 + 0.749 996 896 513 163 264;
  • 35) 0.749 996 896 513 163 264 × 2 = 1 + 0.499 993 793 026 326 528;
  • 36) 0.499 993 793 026 326 528 × 2 = 0 + 0.999 987 586 052 653 056;
  • 37) 0.999 987 586 052 653 056 × 2 = 1 + 0.999 975 172 105 306 112;
  • 38) 0.999 975 172 105 306 112 × 2 = 1 + 0.999 950 344 210 612 224;
  • 39) 0.999 950 344 210 612 224 × 2 = 1 + 0.999 900 688 421 224 448;
  • 40) 0.999 900 688 421 224 448 × 2 = 1 + 0.999 801 376 842 448 896;
  • 41) 0.999 801 376 842 448 896 × 2 = 1 + 0.999 602 753 684 897 792;
  • 42) 0.999 602 753 684 897 792 × 2 = 1 + 0.999 205 507 369 795 584;
  • 43) 0.999 205 507 369 795 584 × 2 = 1 + 0.998 411 014 739 591 168;
  • 44) 0.998 411 014 739 591 168 × 2 = 1 + 0.996 822 029 479 182 336;
  • 45) 0.996 822 029 479 182 336 × 2 = 1 + 0.993 644 058 958 364 672;
  • 46) 0.993 644 058 958 364 672 × 2 = 1 + 0.987 288 117 916 729 344;
  • 47) 0.987 288 117 916 729 344 × 2 = 1 + 0.974 576 235 833 458 688;
  • 48) 0.974 576 235 833 458 688 × 2 = 1 + 0.949 152 471 666 917 376;
  • 49) 0.949 152 471 666 917 376 × 2 = 1 + 0.898 304 943 333 834 752;
  • 50) 0.898 304 943 333 834 752 × 2 = 1 + 0.796 609 886 667 669 504;
  • 51) 0.796 609 886 667 669 504 × 2 = 1 + 0.593 219 773 335 339 008;
  • 52) 0.593 219 773 335 339 008 × 2 = 1 + 0.186 439 546 670 678 016;
  • 53) 0.186 439 546 670 678 016 × 2 = 0 + 0.372 879 093 341 356 032;
  • 54) 0.372 879 093 341 356 032 × 2 = 0 + 0.745 758 186 682 712 064;

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:


0.000 000 000 742 147 496(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 496(10) =

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


0.000 000 000 742 147 496(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 100


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 1100 =

100 1011 1111 1111 1111 1100


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 1100


Decimal number -0.000 000 000 742 147 496 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

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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