-0.000 000 000 742 147 543 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 543(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
-0.000 000 000 742 147 543(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:

|-0.000 000 000 742 147 543| = 0.000 000 000 742 147 543


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: 0.000 000 000 742 147 543.

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) 0.000 000 000 742 147 543 × 2 = 0 + 0.000 000 001 484 295 086;
  • 2) 0.000 000 001 484 295 086 × 2 = 0 + 0.000 000 002 968 590 172;
  • 3) 0.000 000 002 968 590 172 × 2 = 0 + 0.000 000 005 937 180 344;
  • 4) 0.000 000 005 937 180 344 × 2 = 0 + 0.000 000 011 874 360 688;
  • 5) 0.000 000 011 874 360 688 × 2 = 0 + 0.000 000 023 748 721 376;
  • 6) 0.000 000 023 748 721 376 × 2 = 0 + 0.000 000 047 497 442 752;
  • 7) 0.000 000 047 497 442 752 × 2 = 0 + 0.000 000 094 994 885 504;
  • 8) 0.000 000 094 994 885 504 × 2 = 0 + 0.000 000 189 989 771 008;
  • 9) 0.000 000 189 989 771 008 × 2 = 0 + 0.000 000 379 979 542 016;
  • 10) 0.000 000 379 979 542 016 × 2 = 0 + 0.000 000 759 959 084 032;
  • 11) 0.000 000 759 959 084 032 × 2 = 0 + 0.000 001 519 918 168 064;
  • 12) 0.000 001 519 918 168 064 × 2 = 0 + 0.000 003 039 836 336 128;
  • 13) 0.000 003 039 836 336 128 × 2 = 0 + 0.000 006 079 672 672 256;
  • 14) 0.000 006 079 672 672 256 × 2 = 0 + 0.000 012 159 345 344 512;
  • 15) 0.000 012 159 345 344 512 × 2 = 0 + 0.000 024 318 690 689 024;
  • 16) 0.000 024 318 690 689 024 × 2 = 0 + 0.000 048 637 381 378 048;
  • 17) 0.000 048 637 381 378 048 × 2 = 0 + 0.000 097 274 762 756 096;
  • 18) 0.000 097 274 762 756 096 × 2 = 0 + 0.000 194 549 525 512 192;
  • 19) 0.000 194 549 525 512 192 × 2 = 0 + 0.000 389 099 051 024 384;
  • 20) 0.000 389 099 051 024 384 × 2 = 0 + 0.000 778 198 102 048 768;
  • 21) 0.000 778 198 102 048 768 × 2 = 0 + 0.001 556 396 204 097 536;
  • 22) 0.001 556 396 204 097 536 × 2 = 0 + 0.003 112 792 408 195 072;
  • 23) 0.003 112 792 408 195 072 × 2 = 0 + 0.006 225 584 816 390 144;
  • 24) 0.006 225 584 816 390 144 × 2 = 0 + 0.012 451 169 632 780 288;
  • 25) 0.012 451 169 632 780 288 × 2 = 0 + 0.024 902 339 265 560 576;
  • 26) 0.024 902 339 265 560 576 × 2 = 0 + 0.049 804 678 531 121 152;
  • 27) 0.049 804 678 531 121 152 × 2 = 0 + 0.099 609 357 062 242 304;
  • 28) 0.099 609 357 062 242 304 × 2 = 0 + 0.199 218 714 124 484 608;
  • 29) 0.199 218 714 124 484 608 × 2 = 0 + 0.398 437 428 248 969 216;
  • 30) 0.398 437 428 248 969 216 × 2 = 0 + 0.796 874 856 497 938 432;
  • 31) 0.796 874 856 497 938 432 × 2 = 1 + 0.593 749 712 995 876 864;
  • 32) 0.593 749 712 995 876 864 × 2 = 1 + 0.187 499 425 991 753 728;
  • 33) 0.187 499 425 991 753 728 × 2 = 0 + 0.374 998 851 983 507 456;
  • 34) 0.374 998 851 983 507 456 × 2 = 0 + 0.749 997 703 967 014 912;
  • 35) 0.749 997 703 967 014 912 × 2 = 1 + 0.499 995 407 934 029 824;
  • 36) 0.499 995 407 934 029 824 × 2 = 0 + 0.999 990 815 868 059 648;
  • 37) 0.999 990 815 868 059 648 × 2 = 1 + 0.999 981 631 736 119 296;
  • 38) 0.999 981 631 736 119 296 × 2 = 1 + 0.999 963 263 472 238 592;
  • 39) 0.999 963 263 472 238 592 × 2 = 1 + 0.999 926 526 944 477 184;
  • 40) 0.999 926 526 944 477 184 × 2 = 1 + 0.999 853 053 888 954 368;
  • 41) 0.999 853 053 888 954 368 × 2 = 1 + 0.999 706 107 777 908 736;
  • 42) 0.999 706 107 777 908 736 × 2 = 1 + 0.999 412 215 555 817 472;
  • 43) 0.999 412 215 555 817 472 × 2 = 1 + 0.998 824 431 111 634 944;
  • 44) 0.998 824 431 111 634 944 × 2 = 1 + 0.997 648 862 223 269 888;
  • 45) 0.997 648 862 223 269 888 × 2 = 1 + 0.995 297 724 446 539 776;
  • 46) 0.995 297 724 446 539 776 × 2 = 1 + 0.990 595 448 893 079 552;
  • 47) 0.990 595 448 893 079 552 × 2 = 1 + 0.981 190 897 786 159 104;
  • 48) 0.981 190 897 786 159 104 × 2 = 1 + 0.962 381 795 572 318 208;
  • 49) 0.962 381 795 572 318 208 × 2 = 1 + 0.924 763 591 144 636 416;
  • 50) 0.924 763 591 144 636 416 × 2 = 1 + 0.849 527 182 289 272 832;
  • 51) 0.849 527 182 289 272 832 × 2 = 1 + 0.699 054 364 578 545 664;
  • 52) 0.699 054 364 578 545 664 × 2 = 1 + 0.398 108 729 157 091 328;
  • 53) 0.398 108 729 157 091 328 × 2 = 0 + 0.796 217 458 314 182 656;
  • 54) 0.796 217 458 314 182 656 × 2 = 1 + 0.592 434 916 628 365 312;

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:


0.000 000 000 742 147 543(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 543(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(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:


0.000 000 000 742 147 543(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) × 20 =

1.1001 0111 1111 1111 1111 101(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 101


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 1101 =

100 1011 1111 1111 1111 1101


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 1101


Decimal number -0.000 000 000 742 147 543 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1101

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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