-0.000 000 000 742 128 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 128(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
-0.000 000 000 742 128(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:

|-0.000 000 000 742 128| = 0.000 000 000 742 128


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: 0.000 000 000 742 128.

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) 0.000 000 000 742 128 × 2 = 0 + 0.000 000 001 484 256;
  • 2) 0.000 000 001 484 256 × 2 = 0 + 0.000 000 002 968 512;
  • 3) 0.000 000 002 968 512 × 2 = 0 + 0.000 000 005 937 024;
  • 4) 0.000 000 005 937 024 × 2 = 0 + 0.000 000 011 874 048;
  • 5) 0.000 000 011 874 048 × 2 = 0 + 0.000 000 023 748 096;
  • 6) 0.000 000 023 748 096 × 2 = 0 + 0.000 000 047 496 192;
  • 7) 0.000 000 047 496 192 × 2 = 0 + 0.000 000 094 992 384;
  • 8) 0.000 000 094 992 384 × 2 = 0 + 0.000 000 189 984 768;
  • 9) 0.000 000 189 984 768 × 2 = 0 + 0.000 000 379 969 536;
  • 10) 0.000 000 379 969 536 × 2 = 0 + 0.000 000 759 939 072;
  • 11) 0.000 000 759 939 072 × 2 = 0 + 0.000 001 519 878 144;
  • 12) 0.000 001 519 878 144 × 2 = 0 + 0.000 003 039 756 288;
  • 13) 0.000 003 039 756 288 × 2 = 0 + 0.000 006 079 512 576;
  • 14) 0.000 006 079 512 576 × 2 = 0 + 0.000 012 159 025 152;
  • 15) 0.000 012 159 025 152 × 2 = 0 + 0.000 024 318 050 304;
  • 16) 0.000 024 318 050 304 × 2 = 0 + 0.000 048 636 100 608;
  • 17) 0.000 048 636 100 608 × 2 = 0 + 0.000 097 272 201 216;
  • 18) 0.000 097 272 201 216 × 2 = 0 + 0.000 194 544 402 432;
  • 19) 0.000 194 544 402 432 × 2 = 0 + 0.000 389 088 804 864;
  • 20) 0.000 389 088 804 864 × 2 = 0 + 0.000 778 177 609 728;
  • 21) 0.000 778 177 609 728 × 2 = 0 + 0.001 556 355 219 456;
  • 22) 0.001 556 355 219 456 × 2 = 0 + 0.003 112 710 438 912;
  • 23) 0.003 112 710 438 912 × 2 = 0 + 0.006 225 420 877 824;
  • 24) 0.006 225 420 877 824 × 2 = 0 + 0.012 450 841 755 648;
  • 25) 0.012 450 841 755 648 × 2 = 0 + 0.024 901 683 511 296;
  • 26) 0.024 901 683 511 296 × 2 = 0 + 0.049 803 367 022 592;
  • 27) 0.049 803 367 022 592 × 2 = 0 + 0.099 606 734 045 184;
  • 28) 0.099 606 734 045 184 × 2 = 0 + 0.199 213 468 090 368;
  • 29) 0.199 213 468 090 368 × 2 = 0 + 0.398 426 936 180 736;
  • 30) 0.398 426 936 180 736 × 2 = 0 + 0.796 853 872 361 472;
  • 31) 0.796 853 872 361 472 × 2 = 1 + 0.593 707 744 722 944;
  • 32) 0.593 707 744 722 944 × 2 = 1 + 0.187 415 489 445 888;
  • 33) 0.187 415 489 445 888 × 2 = 0 + 0.374 830 978 891 776;
  • 34) 0.374 830 978 891 776 × 2 = 0 + 0.749 661 957 783 552;
  • 35) 0.749 661 957 783 552 × 2 = 1 + 0.499 323 915 567 104;
  • 36) 0.499 323 915 567 104 × 2 = 0 + 0.998 647 831 134 208;
  • 37) 0.998 647 831 134 208 × 2 = 1 + 0.997 295 662 268 416;
  • 38) 0.997 295 662 268 416 × 2 = 1 + 0.994 591 324 536 832;
  • 39) 0.994 591 324 536 832 × 2 = 1 + 0.989 182 649 073 664;
  • 40) 0.989 182 649 073 664 × 2 = 1 + 0.978 365 298 147 328;
  • 41) 0.978 365 298 147 328 × 2 = 1 + 0.956 730 596 294 656;
  • 42) 0.956 730 596 294 656 × 2 = 1 + 0.913 461 192 589 312;
  • 43) 0.913 461 192 589 312 × 2 = 1 + 0.826 922 385 178 624;
  • 44) 0.826 922 385 178 624 × 2 = 1 + 0.653 844 770 357 248;
  • 45) 0.653 844 770 357 248 × 2 = 1 + 0.307 689 540 714 496;
  • 46) 0.307 689 540 714 496 × 2 = 0 + 0.615 379 081 428 992;
  • 47) 0.615 379 081 428 992 × 2 = 1 + 0.230 758 162 857 984;
  • 48) 0.230 758 162 857 984 × 2 = 0 + 0.461 516 325 715 968;
  • 49) 0.461 516 325 715 968 × 2 = 0 + 0.923 032 651 431 936;
  • 50) 0.923 032 651 431 936 × 2 = 1 + 0.846 065 302 863 872;
  • 51) 0.846 065 302 863 872 × 2 = 1 + 0.692 130 605 727 744;
  • 52) 0.692 130 605 727 744 × 2 = 1 + 0.384 261 211 455 488;
  • 53) 0.384 261 211 455 488 × 2 = 0 + 0.768 522 422 910 976;
  • 54) 0.768 522 422 910 976 × 2 = 1 + 0.537 044 845 821 952;

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:


0.000 000 000 742 128(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 0111 01(2)

6. Positive number before normalization:

0.000 000 000 742 128(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 0111 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:


0.000 000 000 742 128(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 0111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 0111 01(2) × 20 =

1.1001 0111 1111 1101 0011 101(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 0011 101


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 1001 1101 =

100 1011 1111 1110 1001 1101


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 1001 1101


Decimal number -0.000 000 000 742 128 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1001 1101

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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