-0.000 000 000 742 147 676 646 708 55 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 708 55(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
-0.000 000 000 742 147 676 646 708 55(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:

|-0.000 000 000 742 147 676 646 708 55| = 0.000 000 000 742 147 676 646 708 55


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: 0.000 000 000 742 147 676 646 708 55.

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) 0.000 000 000 742 147 676 646 708 55 × 2 = 0 + 0.000 000 001 484 295 353 293 417 1;
  • 2) 0.000 000 001 484 295 353 293 417 1 × 2 = 0 + 0.000 000 002 968 590 706 586 834 2;
  • 3) 0.000 000 002 968 590 706 586 834 2 × 2 = 0 + 0.000 000 005 937 181 413 173 668 4;
  • 4) 0.000 000 005 937 181 413 173 668 4 × 2 = 0 + 0.000 000 011 874 362 826 347 336 8;
  • 5) 0.000 000 011 874 362 826 347 336 8 × 2 = 0 + 0.000 000 023 748 725 652 694 673 6;
  • 6) 0.000 000 023 748 725 652 694 673 6 × 2 = 0 + 0.000 000 047 497 451 305 389 347 2;
  • 7) 0.000 000 047 497 451 305 389 347 2 × 2 = 0 + 0.000 000 094 994 902 610 778 694 4;
  • 8) 0.000 000 094 994 902 610 778 694 4 × 2 = 0 + 0.000 000 189 989 805 221 557 388 8;
  • 9) 0.000 000 189 989 805 221 557 388 8 × 2 = 0 + 0.000 000 379 979 610 443 114 777 6;
  • 10) 0.000 000 379 979 610 443 114 777 6 × 2 = 0 + 0.000 000 759 959 220 886 229 555 2;
  • 11) 0.000 000 759 959 220 886 229 555 2 × 2 = 0 + 0.000 001 519 918 441 772 459 110 4;
  • 12) 0.000 001 519 918 441 772 459 110 4 × 2 = 0 + 0.000 003 039 836 883 544 918 220 8;
  • 13) 0.000 003 039 836 883 544 918 220 8 × 2 = 0 + 0.000 006 079 673 767 089 836 441 6;
  • 14) 0.000 006 079 673 767 089 836 441 6 × 2 = 0 + 0.000 012 159 347 534 179 672 883 2;
  • 15) 0.000 012 159 347 534 179 672 883 2 × 2 = 0 + 0.000 024 318 695 068 359 345 766 4;
  • 16) 0.000 024 318 695 068 359 345 766 4 × 2 = 0 + 0.000 048 637 390 136 718 691 532 8;
  • 17) 0.000 048 637 390 136 718 691 532 8 × 2 = 0 + 0.000 097 274 780 273 437 383 065 6;
  • 18) 0.000 097 274 780 273 437 383 065 6 × 2 = 0 + 0.000 194 549 560 546 874 766 131 2;
  • 19) 0.000 194 549 560 546 874 766 131 2 × 2 = 0 + 0.000 389 099 121 093 749 532 262 4;
  • 20) 0.000 389 099 121 093 749 532 262 4 × 2 = 0 + 0.000 778 198 242 187 499 064 524 8;
  • 21) 0.000 778 198 242 187 499 064 524 8 × 2 = 0 + 0.001 556 396 484 374 998 129 049 6;
  • 22) 0.001 556 396 484 374 998 129 049 6 × 2 = 0 + 0.003 112 792 968 749 996 258 099 2;
  • 23) 0.003 112 792 968 749 996 258 099 2 × 2 = 0 + 0.006 225 585 937 499 992 516 198 4;
  • 24) 0.006 225 585 937 499 992 516 198 4 × 2 = 0 + 0.012 451 171 874 999 985 032 396 8;
  • 25) 0.012 451 171 874 999 985 032 396 8 × 2 = 0 + 0.024 902 343 749 999 970 064 793 6;
  • 26) 0.024 902 343 749 999 970 064 793 6 × 2 = 0 + 0.049 804 687 499 999 940 129 587 2;
  • 27) 0.049 804 687 499 999 940 129 587 2 × 2 = 0 + 0.099 609 374 999 999 880 259 174 4;
  • 28) 0.099 609 374 999 999 880 259 174 4 × 2 = 0 + 0.199 218 749 999 999 760 518 348 8;
  • 29) 0.199 218 749 999 999 760 518 348 8 × 2 = 0 + 0.398 437 499 999 999 521 036 697 6;
  • 30) 0.398 437 499 999 999 521 036 697 6 × 2 = 0 + 0.796 874 999 999 999 042 073 395 2;
  • 31) 0.796 874 999 999 999 042 073 395 2 × 2 = 1 + 0.593 749 999 999 998 084 146 790 4;
  • 32) 0.593 749 999 999 998 084 146 790 4 × 2 = 1 + 0.187 499 999 999 996 168 293 580 8;
  • 33) 0.187 499 999 999 996 168 293 580 8 × 2 = 0 + 0.374 999 999 999 992 336 587 161 6;
  • 34) 0.374 999 999 999 992 336 587 161 6 × 2 = 0 + 0.749 999 999 999 984 673 174 323 2;
  • 35) 0.749 999 999 999 984 673 174 323 2 × 2 = 1 + 0.499 999 999 999 969 346 348 646 4;
  • 36) 0.499 999 999 999 969 346 348 646 4 × 2 = 0 + 0.999 999 999 999 938 692 697 292 8;
  • 37) 0.999 999 999 999 938 692 697 292 8 × 2 = 1 + 0.999 999 999 999 877 385 394 585 6;
  • 38) 0.999 999 999 999 877 385 394 585 6 × 2 = 1 + 0.999 999 999 999 754 770 789 171 2;
  • 39) 0.999 999 999 999 754 770 789 171 2 × 2 = 1 + 0.999 999 999 999 509 541 578 342 4;
  • 40) 0.999 999 999 999 509 541 578 342 4 × 2 = 1 + 0.999 999 999 999 019 083 156 684 8;
  • 41) 0.999 999 999 999 019 083 156 684 8 × 2 = 1 + 0.999 999 999 998 038 166 313 369 6;
  • 42) 0.999 999 999 998 038 166 313 369 6 × 2 = 1 + 0.999 999 999 996 076 332 626 739 2;
  • 43) 0.999 999 999 996 076 332 626 739 2 × 2 = 1 + 0.999 999 999 992 152 665 253 478 4;
  • 44) 0.999 999 999 992 152 665 253 478 4 × 2 = 1 + 0.999 999 999 984 305 330 506 956 8;
  • 45) 0.999 999 999 984 305 330 506 956 8 × 2 = 1 + 0.999 999 999 968 610 661 013 913 6;
  • 46) 0.999 999 999 968 610 661 013 913 6 × 2 = 1 + 0.999 999 999 937 221 322 027 827 2;
  • 47) 0.999 999 999 937 221 322 027 827 2 × 2 = 1 + 0.999 999 999 874 442 644 055 654 4;
  • 48) 0.999 999 999 874 442 644 055 654 4 × 2 = 1 + 0.999 999 999 748 885 288 111 308 8;
  • 49) 0.999 999 999 748 885 288 111 308 8 × 2 = 1 + 0.999 999 999 497 770 576 222 617 6;
  • 50) 0.999 999 999 497 770 576 222 617 6 × 2 = 1 + 0.999 999 998 995 541 152 445 235 2;
  • 51) 0.999 999 998 995 541 152 445 235 2 × 2 = 1 + 0.999 999 997 991 082 304 890 470 4;
  • 52) 0.999 999 997 991 082 304 890 470 4 × 2 = 1 + 0.999 999 995 982 164 609 780 940 8;
  • 53) 0.999 999 995 982 164 609 780 940 8 × 2 = 1 + 0.999 999 991 964 329 219 561 881 6;
  • 54) 0.999 999 991 964 329 219 561 881 6 × 2 = 1 + 0.999 999 983 928 658 439 123 763 2;

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:


0.000 000 000 742 147 676 646 708 55(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 708 55(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:


0.000 000 000 742 147 676 646 708 55(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 -0.000 000 000 742 147 676 646 708 55 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