-0.000 000 000 742 148 28 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 28(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
-0.000 000 000 742 148 28(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:

|-0.000 000 000 742 148 28| = 0.000 000 000 742 148 28


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: 0.000 000 000 742 148 28.

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) 0.000 000 000 742 148 28 × 2 = 0 + 0.000 000 001 484 296 56;
  • 2) 0.000 000 001 484 296 56 × 2 = 0 + 0.000 000 002 968 593 12;
  • 3) 0.000 000 002 968 593 12 × 2 = 0 + 0.000 000 005 937 186 24;
  • 4) 0.000 000 005 937 186 24 × 2 = 0 + 0.000 000 011 874 372 48;
  • 5) 0.000 000 011 874 372 48 × 2 = 0 + 0.000 000 023 748 744 96;
  • 6) 0.000 000 023 748 744 96 × 2 = 0 + 0.000 000 047 497 489 92;
  • 7) 0.000 000 047 497 489 92 × 2 = 0 + 0.000 000 094 994 979 84;
  • 8) 0.000 000 094 994 979 84 × 2 = 0 + 0.000 000 189 989 959 68;
  • 9) 0.000 000 189 989 959 68 × 2 = 0 + 0.000 000 379 979 919 36;
  • 10) 0.000 000 379 979 919 36 × 2 = 0 + 0.000 000 759 959 838 72;
  • 11) 0.000 000 759 959 838 72 × 2 = 0 + 0.000 001 519 919 677 44;
  • 12) 0.000 001 519 919 677 44 × 2 = 0 + 0.000 003 039 839 354 88;
  • 13) 0.000 003 039 839 354 88 × 2 = 0 + 0.000 006 079 678 709 76;
  • 14) 0.000 006 079 678 709 76 × 2 = 0 + 0.000 012 159 357 419 52;
  • 15) 0.000 012 159 357 419 52 × 2 = 0 + 0.000 024 318 714 839 04;
  • 16) 0.000 024 318 714 839 04 × 2 = 0 + 0.000 048 637 429 678 08;
  • 17) 0.000 048 637 429 678 08 × 2 = 0 + 0.000 097 274 859 356 16;
  • 18) 0.000 097 274 859 356 16 × 2 = 0 + 0.000 194 549 718 712 32;
  • 19) 0.000 194 549 718 712 32 × 2 = 0 + 0.000 389 099 437 424 64;
  • 20) 0.000 389 099 437 424 64 × 2 = 0 + 0.000 778 198 874 849 28;
  • 21) 0.000 778 198 874 849 28 × 2 = 0 + 0.001 556 397 749 698 56;
  • 22) 0.001 556 397 749 698 56 × 2 = 0 + 0.003 112 795 499 397 12;
  • 23) 0.003 112 795 499 397 12 × 2 = 0 + 0.006 225 590 998 794 24;
  • 24) 0.006 225 590 998 794 24 × 2 = 0 + 0.012 451 181 997 588 48;
  • 25) 0.012 451 181 997 588 48 × 2 = 0 + 0.024 902 363 995 176 96;
  • 26) 0.024 902 363 995 176 96 × 2 = 0 + 0.049 804 727 990 353 92;
  • 27) 0.049 804 727 990 353 92 × 2 = 0 + 0.099 609 455 980 707 84;
  • 28) 0.099 609 455 980 707 84 × 2 = 0 + 0.199 218 911 961 415 68;
  • 29) 0.199 218 911 961 415 68 × 2 = 0 + 0.398 437 823 922 831 36;
  • 30) 0.398 437 823 922 831 36 × 2 = 0 + 0.796 875 647 845 662 72;
  • 31) 0.796 875 647 845 662 72 × 2 = 1 + 0.593 751 295 691 325 44;
  • 32) 0.593 751 295 691 325 44 × 2 = 1 + 0.187 502 591 382 650 88;
  • 33) 0.187 502 591 382 650 88 × 2 = 0 + 0.375 005 182 765 301 76;
  • 34) 0.375 005 182 765 301 76 × 2 = 0 + 0.750 010 365 530 603 52;
  • 35) 0.750 010 365 530 603 52 × 2 = 1 + 0.500 020 731 061 207 04;
  • 36) 0.500 020 731 061 207 04 × 2 = 1 + 0.000 041 462 122 414 08;
  • 37) 0.000 041 462 122 414 08 × 2 = 0 + 0.000 082 924 244 828 16;
  • 38) 0.000 082 924 244 828 16 × 2 = 0 + 0.000 165 848 489 656 32;
  • 39) 0.000 165 848 489 656 32 × 2 = 0 + 0.000 331 696 979 312 64;
  • 40) 0.000 331 696 979 312 64 × 2 = 0 + 0.000 663 393 958 625 28;
  • 41) 0.000 663 393 958 625 28 × 2 = 0 + 0.001 326 787 917 250 56;
  • 42) 0.001 326 787 917 250 56 × 2 = 0 + 0.002 653 575 834 501 12;
  • 43) 0.002 653 575 834 501 12 × 2 = 0 + 0.005 307 151 669 002 24;
  • 44) 0.005 307 151 669 002 24 × 2 = 0 + 0.010 614 303 338 004 48;
  • 45) 0.010 614 303 338 004 48 × 2 = 0 + 0.021 228 606 676 008 96;
  • 46) 0.021 228 606 676 008 96 × 2 = 0 + 0.042 457 213 352 017 92;
  • 47) 0.042 457 213 352 017 92 × 2 = 0 + 0.084 914 426 704 035 84;
  • 48) 0.084 914 426 704 035 84 × 2 = 0 + 0.169 828 853 408 071 68;
  • 49) 0.169 828 853 408 071 68 × 2 = 0 + 0.339 657 706 816 143 36;
  • 50) 0.339 657 706 816 143 36 × 2 = 0 + 0.679 315 413 632 286 72;
  • 51) 0.679 315 413 632 286 72 × 2 = 1 + 0.358 630 827 264 573 44;
  • 52) 0.358 630 827 264 573 44 × 2 = 0 + 0.717 261 654 529 146 88;
  • 53) 0.717 261 654 529 146 88 × 2 = 1 + 0.434 523 309 058 293 76;
  • 54) 0.434 523 309 058 293 76 × 2 = 0 + 0.869 046 618 116 587 52;

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:


0.000 000 000 742 148 28(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 10(2)

6. Positive number before normalization:

0.000 000 000 742 148 28(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 10(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:


0.000 000 000 742 148 28(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 10(2) × 20 =

1.1001 1000 0000 0000 0001 010(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 1000 0000 0000 0001 010


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 1100 0000 0000 0000 1010 =

100 1100 0000 0000 0000 1010


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 1100 0000 0000 0000 1010


Decimal number -0.000 000 000 742 148 28 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 1010

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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