-0.000 000 000 742 114 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 114(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
-0.000 000 000 742 114(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:

|-0.000 000 000 742 114| = 0.000 000 000 742 114


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: 0.000 000 000 742 114.

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) 0.000 000 000 742 114 × 2 = 0 + 0.000 000 001 484 228;
  • 2) 0.000 000 001 484 228 × 2 = 0 + 0.000 000 002 968 456;
  • 3) 0.000 000 002 968 456 × 2 = 0 + 0.000 000 005 936 912;
  • 4) 0.000 000 005 936 912 × 2 = 0 + 0.000 000 011 873 824;
  • 5) 0.000 000 011 873 824 × 2 = 0 + 0.000 000 023 747 648;
  • 6) 0.000 000 023 747 648 × 2 = 0 + 0.000 000 047 495 296;
  • 7) 0.000 000 047 495 296 × 2 = 0 + 0.000 000 094 990 592;
  • 8) 0.000 000 094 990 592 × 2 = 0 + 0.000 000 189 981 184;
  • 9) 0.000 000 189 981 184 × 2 = 0 + 0.000 000 379 962 368;
  • 10) 0.000 000 379 962 368 × 2 = 0 + 0.000 000 759 924 736;
  • 11) 0.000 000 759 924 736 × 2 = 0 + 0.000 001 519 849 472;
  • 12) 0.000 001 519 849 472 × 2 = 0 + 0.000 003 039 698 944;
  • 13) 0.000 003 039 698 944 × 2 = 0 + 0.000 006 079 397 888;
  • 14) 0.000 006 079 397 888 × 2 = 0 + 0.000 012 158 795 776;
  • 15) 0.000 012 158 795 776 × 2 = 0 + 0.000 024 317 591 552;
  • 16) 0.000 024 317 591 552 × 2 = 0 + 0.000 048 635 183 104;
  • 17) 0.000 048 635 183 104 × 2 = 0 + 0.000 097 270 366 208;
  • 18) 0.000 097 270 366 208 × 2 = 0 + 0.000 194 540 732 416;
  • 19) 0.000 194 540 732 416 × 2 = 0 + 0.000 389 081 464 832;
  • 20) 0.000 389 081 464 832 × 2 = 0 + 0.000 778 162 929 664;
  • 21) 0.000 778 162 929 664 × 2 = 0 + 0.001 556 325 859 328;
  • 22) 0.001 556 325 859 328 × 2 = 0 + 0.003 112 651 718 656;
  • 23) 0.003 112 651 718 656 × 2 = 0 + 0.006 225 303 437 312;
  • 24) 0.006 225 303 437 312 × 2 = 0 + 0.012 450 606 874 624;
  • 25) 0.012 450 606 874 624 × 2 = 0 + 0.024 901 213 749 248;
  • 26) 0.024 901 213 749 248 × 2 = 0 + 0.049 802 427 498 496;
  • 27) 0.049 802 427 498 496 × 2 = 0 + 0.099 604 854 996 992;
  • 28) 0.099 604 854 996 992 × 2 = 0 + 0.199 209 709 993 984;
  • 29) 0.199 209 709 993 984 × 2 = 0 + 0.398 419 419 987 968;
  • 30) 0.398 419 419 987 968 × 2 = 0 + 0.796 838 839 975 936;
  • 31) 0.796 838 839 975 936 × 2 = 1 + 0.593 677 679 951 872;
  • 32) 0.593 677 679 951 872 × 2 = 1 + 0.187 355 359 903 744;
  • 33) 0.187 355 359 903 744 × 2 = 0 + 0.374 710 719 807 488;
  • 34) 0.374 710 719 807 488 × 2 = 0 + 0.749 421 439 614 976;
  • 35) 0.749 421 439 614 976 × 2 = 1 + 0.498 842 879 229 952;
  • 36) 0.498 842 879 229 952 × 2 = 0 + 0.997 685 758 459 904;
  • 37) 0.997 685 758 459 904 × 2 = 1 + 0.995 371 516 919 808;
  • 38) 0.995 371 516 919 808 × 2 = 1 + 0.990 743 033 839 616;
  • 39) 0.990 743 033 839 616 × 2 = 1 + 0.981 486 067 679 232;
  • 40) 0.981 486 067 679 232 × 2 = 1 + 0.962 972 135 358 464;
  • 41) 0.962 972 135 358 464 × 2 = 1 + 0.925 944 270 716 928;
  • 42) 0.925 944 270 716 928 × 2 = 1 + 0.851 888 541 433 856;
  • 43) 0.851 888 541 433 856 × 2 = 1 + 0.703 777 082 867 712;
  • 44) 0.703 777 082 867 712 × 2 = 1 + 0.407 554 165 735 424;
  • 45) 0.407 554 165 735 424 × 2 = 0 + 0.815 108 331 470 848;
  • 46) 0.815 108 331 470 848 × 2 = 1 + 0.630 216 662 941 696;
  • 47) 0.630 216 662 941 696 × 2 = 1 + 0.260 433 325 883 392;
  • 48) 0.260 433 325 883 392 × 2 = 0 + 0.520 866 651 766 784;
  • 49) 0.520 866 651 766 784 × 2 = 1 + 0.041 733 303 533 568;
  • 50) 0.041 733 303 533 568 × 2 = 0 + 0.083 466 607 067 136;
  • 51) 0.083 466 607 067 136 × 2 = 0 + 0.166 933 214 134 272;
  • 52) 0.166 933 214 134 272 × 2 = 0 + 0.333 866 428 268 544;
  • 53) 0.333 866 428 268 544 × 2 = 0 + 0.667 732 856 537 088;
  • 54) 0.667 732 856 537 088 × 2 = 1 + 0.335 465 713 074 176;

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:


0.000 000 000 742 114(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0110 1000 01(2)

6. Positive number before normalization:

0.000 000 000 742 114(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0110 1000 01(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:


0.000 000 000 742 114(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0110 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0110 1000 01(2) × 20 =

1.1001 0111 1111 1011 0100 001(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 1011 0100 001


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 1101 1010 0001 =

100 1011 1111 1101 1010 0001


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 1101 1010 0001


Decimal number -0.000 000 000 742 114 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1101 1010 0001

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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