-0.000 000 000 742 143 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 143(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
-0.000 000 000 742 143(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:

|-0.000 000 000 742 143| = 0.000 000 000 742 143


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: 0.000 000 000 742 143.

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) 0.000 000 000 742 143 × 2 = 0 + 0.000 000 001 484 286;
  • 2) 0.000 000 001 484 286 × 2 = 0 + 0.000 000 002 968 572;
  • 3) 0.000 000 002 968 572 × 2 = 0 + 0.000 000 005 937 144;
  • 4) 0.000 000 005 937 144 × 2 = 0 + 0.000 000 011 874 288;
  • 5) 0.000 000 011 874 288 × 2 = 0 + 0.000 000 023 748 576;
  • 6) 0.000 000 023 748 576 × 2 = 0 + 0.000 000 047 497 152;
  • 7) 0.000 000 047 497 152 × 2 = 0 + 0.000 000 094 994 304;
  • 8) 0.000 000 094 994 304 × 2 = 0 + 0.000 000 189 988 608;
  • 9) 0.000 000 189 988 608 × 2 = 0 + 0.000 000 379 977 216;
  • 10) 0.000 000 379 977 216 × 2 = 0 + 0.000 000 759 954 432;
  • 11) 0.000 000 759 954 432 × 2 = 0 + 0.000 001 519 908 864;
  • 12) 0.000 001 519 908 864 × 2 = 0 + 0.000 003 039 817 728;
  • 13) 0.000 003 039 817 728 × 2 = 0 + 0.000 006 079 635 456;
  • 14) 0.000 006 079 635 456 × 2 = 0 + 0.000 012 159 270 912;
  • 15) 0.000 012 159 270 912 × 2 = 0 + 0.000 024 318 541 824;
  • 16) 0.000 024 318 541 824 × 2 = 0 + 0.000 048 637 083 648;
  • 17) 0.000 048 637 083 648 × 2 = 0 + 0.000 097 274 167 296;
  • 18) 0.000 097 274 167 296 × 2 = 0 + 0.000 194 548 334 592;
  • 19) 0.000 194 548 334 592 × 2 = 0 + 0.000 389 096 669 184;
  • 20) 0.000 389 096 669 184 × 2 = 0 + 0.000 778 193 338 368;
  • 21) 0.000 778 193 338 368 × 2 = 0 + 0.001 556 386 676 736;
  • 22) 0.001 556 386 676 736 × 2 = 0 + 0.003 112 773 353 472;
  • 23) 0.003 112 773 353 472 × 2 = 0 + 0.006 225 546 706 944;
  • 24) 0.006 225 546 706 944 × 2 = 0 + 0.012 451 093 413 888;
  • 25) 0.012 451 093 413 888 × 2 = 0 + 0.024 902 186 827 776;
  • 26) 0.024 902 186 827 776 × 2 = 0 + 0.049 804 373 655 552;
  • 27) 0.049 804 373 655 552 × 2 = 0 + 0.099 608 747 311 104;
  • 28) 0.099 608 747 311 104 × 2 = 0 + 0.199 217 494 622 208;
  • 29) 0.199 217 494 622 208 × 2 = 0 + 0.398 434 989 244 416;
  • 30) 0.398 434 989 244 416 × 2 = 0 + 0.796 869 978 488 832;
  • 31) 0.796 869 978 488 832 × 2 = 1 + 0.593 739 956 977 664;
  • 32) 0.593 739 956 977 664 × 2 = 1 + 0.187 479 913 955 328;
  • 33) 0.187 479 913 955 328 × 2 = 0 + 0.374 959 827 910 656;
  • 34) 0.374 959 827 910 656 × 2 = 0 + 0.749 919 655 821 312;
  • 35) 0.749 919 655 821 312 × 2 = 1 + 0.499 839 311 642 624;
  • 36) 0.499 839 311 642 624 × 2 = 0 + 0.999 678 623 285 248;
  • 37) 0.999 678 623 285 248 × 2 = 1 + 0.999 357 246 570 496;
  • 38) 0.999 357 246 570 496 × 2 = 1 + 0.998 714 493 140 992;
  • 39) 0.998 714 493 140 992 × 2 = 1 + 0.997 428 986 281 984;
  • 40) 0.997 428 986 281 984 × 2 = 1 + 0.994 857 972 563 968;
  • 41) 0.994 857 972 563 968 × 2 = 1 + 0.989 715 945 127 936;
  • 42) 0.989 715 945 127 936 × 2 = 1 + 0.979 431 890 255 872;
  • 43) 0.979 431 890 255 872 × 2 = 1 + 0.958 863 780 511 744;
  • 44) 0.958 863 780 511 744 × 2 = 1 + 0.917 727 561 023 488;
  • 45) 0.917 727 561 023 488 × 2 = 1 + 0.835 455 122 046 976;
  • 46) 0.835 455 122 046 976 × 2 = 1 + 0.670 910 244 093 952;
  • 47) 0.670 910 244 093 952 × 2 = 1 + 0.341 820 488 187 904;
  • 48) 0.341 820 488 187 904 × 2 = 0 + 0.683 640 976 375 808;
  • 49) 0.683 640 976 375 808 × 2 = 1 + 0.367 281 952 751 616;
  • 50) 0.367 281 952 751 616 × 2 = 0 + 0.734 563 905 503 232;
  • 51) 0.734 563 905 503 232 × 2 = 1 + 0.469 127 811 006 464;
  • 52) 0.469 127 811 006 464 × 2 = 0 + 0.938 255 622 012 928;
  • 53) 0.938 255 622 012 928 × 2 = 1 + 0.876 511 244 025 856;
  • 54) 0.876 511 244 025 856 × 2 = 1 + 0.753 022 488 051 712;

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:


0.000 000 000 742 143(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1010 11(2)

6. Positive number before normalization:

0.000 000 000 742 143(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1010 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:


0.000 000 000 742 143(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1010 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1010 11(2) × 20 =

1.1001 0111 1111 1111 0101 011(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 0101 011


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 1010 1011 =

100 1011 1111 1111 1010 1011


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 1010 1011


Decimal number -0.000 000 000 742 143 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1010 1011

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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