-0.000 000 000 742 089 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 089(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
-0.000 000 000 742 089(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:

|-0.000 000 000 742 089| = 0.000 000 000 742 089


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: 0.000 000 000 742 089.

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) 0.000 000 000 742 089 × 2 = 0 + 0.000 000 001 484 178;
  • 2) 0.000 000 001 484 178 × 2 = 0 + 0.000 000 002 968 356;
  • 3) 0.000 000 002 968 356 × 2 = 0 + 0.000 000 005 936 712;
  • 4) 0.000 000 005 936 712 × 2 = 0 + 0.000 000 011 873 424;
  • 5) 0.000 000 011 873 424 × 2 = 0 + 0.000 000 023 746 848;
  • 6) 0.000 000 023 746 848 × 2 = 0 + 0.000 000 047 493 696;
  • 7) 0.000 000 047 493 696 × 2 = 0 + 0.000 000 094 987 392;
  • 8) 0.000 000 094 987 392 × 2 = 0 + 0.000 000 189 974 784;
  • 9) 0.000 000 189 974 784 × 2 = 0 + 0.000 000 379 949 568;
  • 10) 0.000 000 379 949 568 × 2 = 0 + 0.000 000 759 899 136;
  • 11) 0.000 000 759 899 136 × 2 = 0 + 0.000 001 519 798 272;
  • 12) 0.000 001 519 798 272 × 2 = 0 + 0.000 003 039 596 544;
  • 13) 0.000 003 039 596 544 × 2 = 0 + 0.000 006 079 193 088;
  • 14) 0.000 006 079 193 088 × 2 = 0 + 0.000 012 158 386 176;
  • 15) 0.000 012 158 386 176 × 2 = 0 + 0.000 024 316 772 352;
  • 16) 0.000 024 316 772 352 × 2 = 0 + 0.000 048 633 544 704;
  • 17) 0.000 048 633 544 704 × 2 = 0 + 0.000 097 267 089 408;
  • 18) 0.000 097 267 089 408 × 2 = 0 + 0.000 194 534 178 816;
  • 19) 0.000 194 534 178 816 × 2 = 0 + 0.000 389 068 357 632;
  • 20) 0.000 389 068 357 632 × 2 = 0 + 0.000 778 136 715 264;
  • 21) 0.000 778 136 715 264 × 2 = 0 + 0.001 556 273 430 528;
  • 22) 0.001 556 273 430 528 × 2 = 0 + 0.003 112 546 861 056;
  • 23) 0.003 112 546 861 056 × 2 = 0 + 0.006 225 093 722 112;
  • 24) 0.006 225 093 722 112 × 2 = 0 + 0.012 450 187 444 224;
  • 25) 0.012 450 187 444 224 × 2 = 0 + 0.024 900 374 888 448;
  • 26) 0.024 900 374 888 448 × 2 = 0 + 0.049 800 749 776 896;
  • 27) 0.049 800 749 776 896 × 2 = 0 + 0.099 601 499 553 792;
  • 28) 0.099 601 499 553 792 × 2 = 0 + 0.199 202 999 107 584;
  • 29) 0.199 202 999 107 584 × 2 = 0 + 0.398 405 998 215 168;
  • 30) 0.398 405 998 215 168 × 2 = 0 + 0.796 811 996 430 336;
  • 31) 0.796 811 996 430 336 × 2 = 1 + 0.593 623 992 860 672;
  • 32) 0.593 623 992 860 672 × 2 = 1 + 0.187 247 985 721 344;
  • 33) 0.187 247 985 721 344 × 2 = 0 + 0.374 495 971 442 688;
  • 34) 0.374 495 971 442 688 × 2 = 0 + 0.748 991 942 885 376;
  • 35) 0.748 991 942 885 376 × 2 = 1 + 0.497 983 885 770 752;
  • 36) 0.497 983 885 770 752 × 2 = 0 + 0.995 967 771 541 504;
  • 37) 0.995 967 771 541 504 × 2 = 1 + 0.991 935 543 083 008;
  • 38) 0.991 935 543 083 008 × 2 = 1 + 0.983 871 086 166 016;
  • 39) 0.983 871 086 166 016 × 2 = 1 + 0.967 742 172 332 032;
  • 40) 0.967 742 172 332 032 × 2 = 1 + 0.935 484 344 664 064;
  • 41) 0.935 484 344 664 064 × 2 = 1 + 0.870 968 689 328 128;
  • 42) 0.870 968 689 328 128 × 2 = 1 + 0.741 937 378 656 256;
  • 43) 0.741 937 378 656 256 × 2 = 1 + 0.483 874 757 312 512;
  • 44) 0.483 874 757 312 512 × 2 = 0 + 0.967 749 514 625 024;
  • 45) 0.967 749 514 625 024 × 2 = 1 + 0.935 499 029 250 048;
  • 46) 0.935 499 029 250 048 × 2 = 1 + 0.870 998 058 500 096;
  • 47) 0.870 998 058 500 096 × 2 = 1 + 0.741 996 117 000 192;
  • 48) 0.741 996 117 000 192 × 2 = 1 + 0.483 992 234 000 384;
  • 49) 0.483 992 234 000 384 × 2 = 0 + 0.967 984 468 000 768;
  • 50) 0.967 984 468 000 768 × 2 = 1 + 0.935 968 936 001 536;
  • 51) 0.935 968 936 001 536 × 2 = 1 + 0.871 937 872 003 072;
  • 52) 0.871 937 872 003 072 × 2 = 1 + 0.743 875 744 006 144;
  • 53) 0.743 875 744 006 144 × 2 = 1 + 0.487 751 488 012 288;
  • 54) 0.487 751 488 012 288 × 2 = 0 + 0.975 502 976 024 576;

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:


0.000 000 000 742 089(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1111 0111 10(2)

6. Positive number before normalization:

0.000 000 000 742 089(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1111 0111 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:


0.000 000 000 742 089(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1111 0111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1111 0111 10(2) × 20 =

1.1001 0111 1111 0111 1011 110(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 0111 1011 110


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 1011 1101 1110 =

100 1011 1111 1011 1101 1110


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 1011 1101 1110


Decimal number -0.000 000 000 742 089 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1011 1101 1110

Share

» Convert decimal -0.000 000 000 742 082 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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