-0.000 000 000 742 182 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 182(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
-0.000 000 000 742 182(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:

|-0.000 000 000 742 182| = 0.000 000 000 742 182


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: 0.000 000 000 742 182.

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) 0.000 000 000 742 182 × 2 = 0 + 0.000 000 001 484 364;
  • 2) 0.000 000 001 484 364 × 2 = 0 + 0.000 000 002 968 728;
  • 3) 0.000 000 002 968 728 × 2 = 0 + 0.000 000 005 937 456;
  • 4) 0.000 000 005 937 456 × 2 = 0 + 0.000 000 011 874 912;
  • 5) 0.000 000 011 874 912 × 2 = 0 + 0.000 000 023 749 824;
  • 6) 0.000 000 023 749 824 × 2 = 0 + 0.000 000 047 499 648;
  • 7) 0.000 000 047 499 648 × 2 = 0 + 0.000 000 094 999 296;
  • 8) 0.000 000 094 999 296 × 2 = 0 + 0.000 000 189 998 592;
  • 9) 0.000 000 189 998 592 × 2 = 0 + 0.000 000 379 997 184;
  • 10) 0.000 000 379 997 184 × 2 = 0 + 0.000 000 759 994 368;
  • 11) 0.000 000 759 994 368 × 2 = 0 + 0.000 001 519 988 736;
  • 12) 0.000 001 519 988 736 × 2 = 0 + 0.000 003 039 977 472;
  • 13) 0.000 003 039 977 472 × 2 = 0 + 0.000 006 079 954 944;
  • 14) 0.000 006 079 954 944 × 2 = 0 + 0.000 012 159 909 888;
  • 15) 0.000 012 159 909 888 × 2 = 0 + 0.000 024 319 819 776;
  • 16) 0.000 024 319 819 776 × 2 = 0 + 0.000 048 639 639 552;
  • 17) 0.000 048 639 639 552 × 2 = 0 + 0.000 097 279 279 104;
  • 18) 0.000 097 279 279 104 × 2 = 0 + 0.000 194 558 558 208;
  • 19) 0.000 194 558 558 208 × 2 = 0 + 0.000 389 117 116 416;
  • 20) 0.000 389 117 116 416 × 2 = 0 + 0.000 778 234 232 832;
  • 21) 0.000 778 234 232 832 × 2 = 0 + 0.001 556 468 465 664;
  • 22) 0.001 556 468 465 664 × 2 = 0 + 0.003 112 936 931 328;
  • 23) 0.003 112 936 931 328 × 2 = 0 + 0.006 225 873 862 656;
  • 24) 0.006 225 873 862 656 × 2 = 0 + 0.012 451 747 725 312;
  • 25) 0.012 451 747 725 312 × 2 = 0 + 0.024 903 495 450 624;
  • 26) 0.024 903 495 450 624 × 2 = 0 + 0.049 806 990 901 248;
  • 27) 0.049 806 990 901 248 × 2 = 0 + 0.099 613 981 802 496;
  • 28) 0.099 613 981 802 496 × 2 = 0 + 0.199 227 963 604 992;
  • 29) 0.199 227 963 604 992 × 2 = 0 + 0.398 455 927 209 984;
  • 30) 0.398 455 927 209 984 × 2 = 0 + 0.796 911 854 419 968;
  • 31) 0.796 911 854 419 968 × 2 = 1 + 0.593 823 708 839 936;
  • 32) 0.593 823 708 839 936 × 2 = 1 + 0.187 647 417 679 872;
  • 33) 0.187 647 417 679 872 × 2 = 0 + 0.375 294 835 359 744;
  • 34) 0.375 294 835 359 744 × 2 = 0 + 0.750 589 670 719 488;
  • 35) 0.750 589 670 719 488 × 2 = 1 + 0.501 179 341 438 976;
  • 36) 0.501 179 341 438 976 × 2 = 1 + 0.002 358 682 877 952;
  • 37) 0.002 358 682 877 952 × 2 = 0 + 0.004 717 365 755 904;
  • 38) 0.004 717 365 755 904 × 2 = 0 + 0.009 434 731 511 808;
  • 39) 0.009 434 731 511 808 × 2 = 0 + 0.018 869 463 023 616;
  • 40) 0.018 869 463 023 616 × 2 = 0 + 0.037 738 926 047 232;
  • 41) 0.037 738 926 047 232 × 2 = 0 + 0.075 477 852 094 464;
  • 42) 0.075 477 852 094 464 × 2 = 0 + 0.150 955 704 188 928;
  • 43) 0.150 955 704 188 928 × 2 = 0 + 0.301 911 408 377 856;
  • 44) 0.301 911 408 377 856 × 2 = 0 + 0.603 822 816 755 712;
  • 45) 0.603 822 816 755 712 × 2 = 1 + 0.207 645 633 511 424;
  • 46) 0.207 645 633 511 424 × 2 = 0 + 0.415 291 267 022 848;
  • 47) 0.415 291 267 022 848 × 2 = 0 + 0.830 582 534 045 696;
  • 48) 0.830 582 534 045 696 × 2 = 1 + 0.661 165 068 091 392;
  • 49) 0.661 165 068 091 392 × 2 = 1 + 0.322 330 136 182 784;
  • 50) 0.322 330 136 182 784 × 2 = 0 + 0.644 660 272 365 568;
  • 51) 0.644 660 272 365 568 × 2 = 1 + 0.289 320 544 731 136;
  • 52) 0.289 320 544 731 136 × 2 = 0 + 0.578 641 089 462 272;
  • 53) 0.578 641 089 462 272 × 2 = 1 + 0.157 282 178 924 544;
  • 54) 0.157 282 178 924 544 × 2 = 0 + 0.314 564 357 849 088;

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:


0.000 000 000 742 182(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 1001 1010 10(2)

6. Positive number before normalization:

0.000 000 000 742 182(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 1001 1010 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:


0.000 000 000 742 182(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 1001 1010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 1001 1010 10(2) × 20 =

1.1001 1000 0000 0100 1101 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 0100 1101 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 0010 0110 1010 =

100 1100 0000 0010 0110 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 0010 0110 1010


Decimal number -0.000 000 000 742 182 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0010 0110 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