-0.000 000 000 742 147 434 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 434(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
-0.000 000 000 742 147 434(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:

|-0.000 000 000 742 147 434| = 0.000 000 000 742 147 434


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: 0.000 000 000 742 147 434.

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) 0.000 000 000 742 147 434 × 2 = 0 + 0.000 000 001 484 294 868;
  • 2) 0.000 000 001 484 294 868 × 2 = 0 + 0.000 000 002 968 589 736;
  • 3) 0.000 000 002 968 589 736 × 2 = 0 + 0.000 000 005 937 179 472;
  • 4) 0.000 000 005 937 179 472 × 2 = 0 + 0.000 000 011 874 358 944;
  • 5) 0.000 000 011 874 358 944 × 2 = 0 + 0.000 000 023 748 717 888;
  • 6) 0.000 000 023 748 717 888 × 2 = 0 + 0.000 000 047 497 435 776;
  • 7) 0.000 000 047 497 435 776 × 2 = 0 + 0.000 000 094 994 871 552;
  • 8) 0.000 000 094 994 871 552 × 2 = 0 + 0.000 000 189 989 743 104;
  • 9) 0.000 000 189 989 743 104 × 2 = 0 + 0.000 000 379 979 486 208;
  • 10) 0.000 000 379 979 486 208 × 2 = 0 + 0.000 000 759 958 972 416;
  • 11) 0.000 000 759 958 972 416 × 2 = 0 + 0.000 001 519 917 944 832;
  • 12) 0.000 001 519 917 944 832 × 2 = 0 + 0.000 003 039 835 889 664;
  • 13) 0.000 003 039 835 889 664 × 2 = 0 + 0.000 006 079 671 779 328;
  • 14) 0.000 006 079 671 779 328 × 2 = 0 + 0.000 012 159 343 558 656;
  • 15) 0.000 012 159 343 558 656 × 2 = 0 + 0.000 024 318 687 117 312;
  • 16) 0.000 024 318 687 117 312 × 2 = 0 + 0.000 048 637 374 234 624;
  • 17) 0.000 048 637 374 234 624 × 2 = 0 + 0.000 097 274 748 469 248;
  • 18) 0.000 097 274 748 469 248 × 2 = 0 + 0.000 194 549 496 938 496;
  • 19) 0.000 194 549 496 938 496 × 2 = 0 + 0.000 389 098 993 876 992;
  • 20) 0.000 389 098 993 876 992 × 2 = 0 + 0.000 778 197 987 753 984;
  • 21) 0.000 778 197 987 753 984 × 2 = 0 + 0.001 556 395 975 507 968;
  • 22) 0.001 556 395 975 507 968 × 2 = 0 + 0.003 112 791 951 015 936;
  • 23) 0.003 112 791 951 015 936 × 2 = 0 + 0.006 225 583 902 031 872;
  • 24) 0.006 225 583 902 031 872 × 2 = 0 + 0.012 451 167 804 063 744;
  • 25) 0.012 451 167 804 063 744 × 2 = 0 + 0.024 902 335 608 127 488;
  • 26) 0.024 902 335 608 127 488 × 2 = 0 + 0.049 804 671 216 254 976;
  • 27) 0.049 804 671 216 254 976 × 2 = 0 + 0.099 609 342 432 509 952;
  • 28) 0.099 609 342 432 509 952 × 2 = 0 + 0.199 218 684 865 019 904;
  • 29) 0.199 218 684 865 019 904 × 2 = 0 + 0.398 437 369 730 039 808;
  • 30) 0.398 437 369 730 039 808 × 2 = 0 + 0.796 874 739 460 079 616;
  • 31) 0.796 874 739 460 079 616 × 2 = 1 + 0.593 749 478 920 159 232;
  • 32) 0.593 749 478 920 159 232 × 2 = 1 + 0.187 498 957 840 318 464;
  • 33) 0.187 498 957 840 318 464 × 2 = 0 + 0.374 997 915 680 636 928;
  • 34) 0.374 997 915 680 636 928 × 2 = 0 + 0.749 995 831 361 273 856;
  • 35) 0.749 995 831 361 273 856 × 2 = 1 + 0.499 991 662 722 547 712;
  • 36) 0.499 991 662 722 547 712 × 2 = 0 + 0.999 983 325 445 095 424;
  • 37) 0.999 983 325 445 095 424 × 2 = 1 + 0.999 966 650 890 190 848;
  • 38) 0.999 966 650 890 190 848 × 2 = 1 + 0.999 933 301 780 381 696;
  • 39) 0.999 933 301 780 381 696 × 2 = 1 + 0.999 866 603 560 763 392;
  • 40) 0.999 866 603 560 763 392 × 2 = 1 + 0.999 733 207 121 526 784;
  • 41) 0.999 733 207 121 526 784 × 2 = 1 + 0.999 466 414 243 053 568;
  • 42) 0.999 466 414 243 053 568 × 2 = 1 + 0.998 932 828 486 107 136;
  • 43) 0.998 932 828 486 107 136 × 2 = 1 + 0.997 865 656 972 214 272;
  • 44) 0.997 865 656 972 214 272 × 2 = 1 + 0.995 731 313 944 428 544;
  • 45) 0.995 731 313 944 428 544 × 2 = 1 + 0.991 462 627 888 857 088;
  • 46) 0.991 462 627 888 857 088 × 2 = 1 + 0.982 925 255 777 714 176;
  • 47) 0.982 925 255 777 714 176 × 2 = 1 + 0.965 850 511 555 428 352;
  • 48) 0.965 850 511 555 428 352 × 2 = 1 + 0.931 701 023 110 856 704;
  • 49) 0.931 701 023 110 856 704 × 2 = 1 + 0.863 402 046 221 713 408;
  • 50) 0.863 402 046 221 713 408 × 2 = 1 + 0.726 804 092 443 426 816;
  • 51) 0.726 804 092 443 426 816 × 2 = 1 + 0.453 608 184 886 853 632;
  • 52) 0.453 608 184 886 853 632 × 2 = 0 + 0.907 216 369 773 707 264;
  • 53) 0.907 216 369 773 707 264 × 2 = 1 + 0.814 432 739 547 414 528;
  • 54) 0.814 432 739 547 414 528 × 2 = 1 + 0.628 865 479 094 829 056;

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:


0.000 000 000 742 147 434(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 434(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 434(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 011


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 1011 =

100 1011 1111 1111 1111 1011


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 1011


Decimal number -0.000 000 000 742 147 434 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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