-0.000 000 000 742 157 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 157 9(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
-0.000 000 000 742 157 9(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:

|-0.000 000 000 742 157 9| = 0.000 000 000 742 157 9


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: 0.000 000 000 742 157 9.

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) 0.000 000 000 742 157 9 × 2 = 0 + 0.000 000 001 484 315 8;
  • 2) 0.000 000 001 484 315 8 × 2 = 0 + 0.000 000 002 968 631 6;
  • 3) 0.000 000 002 968 631 6 × 2 = 0 + 0.000 000 005 937 263 2;
  • 4) 0.000 000 005 937 263 2 × 2 = 0 + 0.000 000 011 874 526 4;
  • 5) 0.000 000 011 874 526 4 × 2 = 0 + 0.000 000 023 749 052 8;
  • 6) 0.000 000 023 749 052 8 × 2 = 0 + 0.000 000 047 498 105 6;
  • 7) 0.000 000 047 498 105 6 × 2 = 0 + 0.000 000 094 996 211 2;
  • 8) 0.000 000 094 996 211 2 × 2 = 0 + 0.000 000 189 992 422 4;
  • 9) 0.000 000 189 992 422 4 × 2 = 0 + 0.000 000 379 984 844 8;
  • 10) 0.000 000 379 984 844 8 × 2 = 0 + 0.000 000 759 969 689 6;
  • 11) 0.000 000 759 969 689 6 × 2 = 0 + 0.000 001 519 939 379 2;
  • 12) 0.000 001 519 939 379 2 × 2 = 0 + 0.000 003 039 878 758 4;
  • 13) 0.000 003 039 878 758 4 × 2 = 0 + 0.000 006 079 757 516 8;
  • 14) 0.000 006 079 757 516 8 × 2 = 0 + 0.000 012 159 515 033 6;
  • 15) 0.000 012 159 515 033 6 × 2 = 0 + 0.000 024 319 030 067 2;
  • 16) 0.000 024 319 030 067 2 × 2 = 0 + 0.000 048 638 060 134 4;
  • 17) 0.000 048 638 060 134 4 × 2 = 0 + 0.000 097 276 120 268 8;
  • 18) 0.000 097 276 120 268 8 × 2 = 0 + 0.000 194 552 240 537 6;
  • 19) 0.000 194 552 240 537 6 × 2 = 0 + 0.000 389 104 481 075 2;
  • 20) 0.000 389 104 481 075 2 × 2 = 0 + 0.000 778 208 962 150 4;
  • 21) 0.000 778 208 962 150 4 × 2 = 0 + 0.001 556 417 924 300 8;
  • 22) 0.001 556 417 924 300 8 × 2 = 0 + 0.003 112 835 848 601 6;
  • 23) 0.003 112 835 848 601 6 × 2 = 0 + 0.006 225 671 697 203 2;
  • 24) 0.006 225 671 697 203 2 × 2 = 0 + 0.012 451 343 394 406 4;
  • 25) 0.012 451 343 394 406 4 × 2 = 0 + 0.024 902 686 788 812 8;
  • 26) 0.024 902 686 788 812 8 × 2 = 0 + 0.049 805 373 577 625 6;
  • 27) 0.049 805 373 577 625 6 × 2 = 0 + 0.099 610 747 155 251 2;
  • 28) 0.099 610 747 155 251 2 × 2 = 0 + 0.199 221 494 310 502 4;
  • 29) 0.199 221 494 310 502 4 × 2 = 0 + 0.398 442 988 621 004 8;
  • 30) 0.398 442 988 621 004 8 × 2 = 0 + 0.796 885 977 242 009 6;
  • 31) 0.796 885 977 242 009 6 × 2 = 1 + 0.593 771 954 484 019 2;
  • 32) 0.593 771 954 484 019 2 × 2 = 1 + 0.187 543 908 968 038 4;
  • 33) 0.187 543 908 968 038 4 × 2 = 0 + 0.375 087 817 936 076 8;
  • 34) 0.375 087 817 936 076 8 × 2 = 0 + 0.750 175 635 872 153 6;
  • 35) 0.750 175 635 872 153 6 × 2 = 1 + 0.500 351 271 744 307 2;
  • 36) 0.500 351 271 744 307 2 × 2 = 1 + 0.000 702 543 488 614 4;
  • 37) 0.000 702 543 488 614 4 × 2 = 0 + 0.001 405 086 977 228 8;
  • 38) 0.001 405 086 977 228 8 × 2 = 0 + 0.002 810 173 954 457 6;
  • 39) 0.002 810 173 954 457 6 × 2 = 0 + 0.005 620 347 908 915 2;
  • 40) 0.005 620 347 908 915 2 × 2 = 0 + 0.011 240 695 817 830 4;
  • 41) 0.011 240 695 817 830 4 × 2 = 0 + 0.022 481 391 635 660 8;
  • 42) 0.022 481 391 635 660 8 × 2 = 0 + 0.044 962 783 271 321 6;
  • 43) 0.044 962 783 271 321 6 × 2 = 0 + 0.089 925 566 542 643 2;
  • 44) 0.089 925 566 542 643 2 × 2 = 0 + 0.179 851 133 085 286 4;
  • 45) 0.179 851 133 085 286 4 × 2 = 0 + 0.359 702 266 170 572 8;
  • 46) 0.359 702 266 170 572 8 × 2 = 0 + 0.719 404 532 341 145 6;
  • 47) 0.719 404 532 341 145 6 × 2 = 1 + 0.438 809 064 682 291 2;
  • 48) 0.438 809 064 682 291 2 × 2 = 0 + 0.877 618 129 364 582 4;
  • 49) 0.877 618 129 364 582 4 × 2 = 1 + 0.755 236 258 729 164 8;
  • 50) 0.755 236 258 729 164 8 × 2 = 1 + 0.510 472 517 458 329 6;
  • 51) 0.510 472 517 458 329 6 × 2 = 1 + 0.020 945 034 916 659 2;
  • 52) 0.020 945 034 916 659 2 × 2 = 0 + 0.041 890 069 833 318 4;
  • 53) 0.041 890 069 833 318 4 × 2 = 0 + 0.083 780 139 666 636 8;
  • 54) 0.083 780 139 666 636 8 × 2 = 0 + 0.167 560 279 333 273 6;

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:


0.000 000 000 742 157 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0010 1110 00(2)

6. Positive number before normalization:

0.000 000 000 742 157 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0010 1110 00(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:


0.000 000 000 742 157 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0010 1110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0010 1110 00(2) × 20 =

1.1001 1000 0000 0001 0111 000(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 0001 0111 000


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 1011 1000 =

100 1100 0000 0000 1011 1000


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 1011 1000


Decimal number -0.000 000 000 742 157 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 1011 1000

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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