-0.000 000 000 742 132 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 132 5(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
-0.000 000 000 742 132 5(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:

|-0.000 000 000 742 132 5| = 0.000 000 000 742 132 5


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: 0.000 000 000 742 132 5.

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) 0.000 000 000 742 132 5 × 2 = 0 + 0.000 000 001 484 265;
  • 2) 0.000 000 001 484 265 × 2 = 0 + 0.000 000 002 968 53;
  • 3) 0.000 000 002 968 53 × 2 = 0 + 0.000 000 005 937 06;
  • 4) 0.000 000 005 937 06 × 2 = 0 + 0.000 000 011 874 12;
  • 5) 0.000 000 011 874 12 × 2 = 0 + 0.000 000 023 748 24;
  • 6) 0.000 000 023 748 24 × 2 = 0 + 0.000 000 047 496 48;
  • 7) 0.000 000 047 496 48 × 2 = 0 + 0.000 000 094 992 96;
  • 8) 0.000 000 094 992 96 × 2 = 0 + 0.000 000 189 985 92;
  • 9) 0.000 000 189 985 92 × 2 = 0 + 0.000 000 379 971 84;
  • 10) 0.000 000 379 971 84 × 2 = 0 + 0.000 000 759 943 68;
  • 11) 0.000 000 759 943 68 × 2 = 0 + 0.000 001 519 887 36;
  • 12) 0.000 001 519 887 36 × 2 = 0 + 0.000 003 039 774 72;
  • 13) 0.000 003 039 774 72 × 2 = 0 + 0.000 006 079 549 44;
  • 14) 0.000 006 079 549 44 × 2 = 0 + 0.000 012 159 098 88;
  • 15) 0.000 012 159 098 88 × 2 = 0 + 0.000 024 318 197 76;
  • 16) 0.000 024 318 197 76 × 2 = 0 + 0.000 048 636 395 52;
  • 17) 0.000 048 636 395 52 × 2 = 0 + 0.000 097 272 791 04;
  • 18) 0.000 097 272 791 04 × 2 = 0 + 0.000 194 545 582 08;
  • 19) 0.000 194 545 582 08 × 2 = 0 + 0.000 389 091 164 16;
  • 20) 0.000 389 091 164 16 × 2 = 0 + 0.000 778 182 328 32;
  • 21) 0.000 778 182 328 32 × 2 = 0 + 0.001 556 364 656 64;
  • 22) 0.001 556 364 656 64 × 2 = 0 + 0.003 112 729 313 28;
  • 23) 0.003 112 729 313 28 × 2 = 0 + 0.006 225 458 626 56;
  • 24) 0.006 225 458 626 56 × 2 = 0 + 0.012 450 917 253 12;
  • 25) 0.012 450 917 253 12 × 2 = 0 + 0.024 901 834 506 24;
  • 26) 0.024 901 834 506 24 × 2 = 0 + 0.049 803 669 012 48;
  • 27) 0.049 803 669 012 48 × 2 = 0 + 0.099 607 338 024 96;
  • 28) 0.099 607 338 024 96 × 2 = 0 + 0.199 214 676 049 92;
  • 29) 0.199 214 676 049 92 × 2 = 0 + 0.398 429 352 099 84;
  • 30) 0.398 429 352 099 84 × 2 = 0 + 0.796 858 704 199 68;
  • 31) 0.796 858 704 199 68 × 2 = 1 + 0.593 717 408 399 36;
  • 32) 0.593 717 408 399 36 × 2 = 1 + 0.187 434 816 798 72;
  • 33) 0.187 434 816 798 72 × 2 = 0 + 0.374 869 633 597 44;
  • 34) 0.374 869 633 597 44 × 2 = 0 + 0.749 739 267 194 88;
  • 35) 0.749 739 267 194 88 × 2 = 1 + 0.499 478 534 389 76;
  • 36) 0.499 478 534 389 76 × 2 = 0 + 0.998 957 068 779 52;
  • 37) 0.998 957 068 779 52 × 2 = 1 + 0.997 914 137 559 04;
  • 38) 0.997 914 137 559 04 × 2 = 1 + 0.995 828 275 118 08;
  • 39) 0.995 828 275 118 08 × 2 = 1 + 0.991 656 550 236 16;
  • 40) 0.991 656 550 236 16 × 2 = 1 + 0.983 313 100 472 32;
  • 41) 0.983 313 100 472 32 × 2 = 1 + 0.966 626 200 944 64;
  • 42) 0.966 626 200 944 64 × 2 = 1 + 0.933 252 401 889 28;
  • 43) 0.933 252 401 889 28 × 2 = 1 + 0.866 504 803 778 56;
  • 44) 0.866 504 803 778 56 × 2 = 1 + 0.733 009 607 557 12;
  • 45) 0.733 009 607 557 12 × 2 = 1 + 0.466 019 215 114 24;
  • 46) 0.466 019 215 114 24 × 2 = 0 + 0.932 038 430 228 48;
  • 47) 0.932 038 430 228 48 × 2 = 1 + 0.864 076 860 456 96;
  • 48) 0.864 076 860 456 96 × 2 = 1 + 0.728 153 720 913 92;
  • 49) 0.728 153 720 913 92 × 2 = 1 + 0.456 307 441 827 84;
  • 50) 0.456 307 441 827 84 × 2 = 0 + 0.912 614 883 655 68;
  • 51) 0.912 614 883 655 68 × 2 = 1 + 0.825 229 767 311 36;
  • 52) 0.825 229 767 311 36 × 2 = 1 + 0.650 459 534 622 72;
  • 53) 0.650 459 534 622 72 × 2 = 1 + 0.300 919 069 245 44;
  • 54) 0.300 919 069 245 44 × 2 = 0 + 0.601 838 138 490 88;

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:


0.000 000 000 742 132 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 1011 10(2)

6. Positive number before normalization:

0.000 000 000 742 132 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 1011 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:


0.000 000 000 742 132 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 1011 10(2) × 20 =

1.1001 0111 1111 1101 1101 110(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 1101 1101 110


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 1110 1110 1110 =

100 1011 1111 1110 1110 1110


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 1110 1110 1110


Decimal number -0.000 000 000 742 132 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1110 1110

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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