-0.000 000 000 202 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 202(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
-0.000 000 000 202(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:

|-0.000 000 000 202| = 0.000 000 000 202


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: 0.000 000 000 202.

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) 0.000 000 000 202 × 2 = 0 + 0.000 000 000 404;
  • 2) 0.000 000 000 404 × 2 = 0 + 0.000 000 000 808;
  • 3) 0.000 000 000 808 × 2 = 0 + 0.000 000 001 616;
  • 4) 0.000 000 001 616 × 2 = 0 + 0.000 000 003 232;
  • 5) 0.000 000 003 232 × 2 = 0 + 0.000 000 006 464;
  • 6) 0.000 000 006 464 × 2 = 0 + 0.000 000 012 928;
  • 7) 0.000 000 012 928 × 2 = 0 + 0.000 000 025 856;
  • 8) 0.000 000 025 856 × 2 = 0 + 0.000 000 051 712;
  • 9) 0.000 000 051 712 × 2 = 0 + 0.000 000 103 424;
  • 10) 0.000 000 103 424 × 2 = 0 + 0.000 000 206 848;
  • 11) 0.000 000 206 848 × 2 = 0 + 0.000 000 413 696;
  • 12) 0.000 000 413 696 × 2 = 0 + 0.000 000 827 392;
  • 13) 0.000 000 827 392 × 2 = 0 + 0.000 001 654 784;
  • 14) 0.000 001 654 784 × 2 = 0 + 0.000 003 309 568;
  • 15) 0.000 003 309 568 × 2 = 0 + 0.000 006 619 136;
  • 16) 0.000 006 619 136 × 2 = 0 + 0.000 013 238 272;
  • 17) 0.000 013 238 272 × 2 = 0 + 0.000 026 476 544;
  • 18) 0.000 026 476 544 × 2 = 0 + 0.000 052 953 088;
  • 19) 0.000 052 953 088 × 2 = 0 + 0.000 105 906 176;
  • 20) 0.000 105 906 176 × 2 = 0 + 0.000 211 812 352;
  • 21) 0.000 211 812 352 × 2 = 0 + 0.000 423 624 704;
  • 22) 0.000 423 624 704 × 2 = 0 + 0.000 847 249 408;
  • 23) 0.000 847 249 408 × 2 = 0 + 0.001 694 498 816;
  • 24) 0.001 694 498 816 × 2 = 0 + 0.003 388 997 632;
  • 25) 0.003 388 997 632 × 2 = 0 + 0.006 777 995 264;
  • 26) 0.006 777 995 264 × 2 = 0 + 0.013 555 990 528;
  • 27) 0.013 555 990 528 × 2 = 0 + 0.027 111 981 056;
  • 28) 0.027 111 981 056 × 2 = 0 + 0.054 223 962 112;
  • 29) 0.054 223 962 112 × 2 = 0 + 0.108 447 924 224;
  • 30) 0.108 447 924 224 × 2 = 0 + 0.216 895 848 448;
  • 31) 0.216 895 848 448 × 2 = 0 + 0.433 791 696 896;
  • 32) 0.433 791 696 896 × 2 = 0 + 0.867 583 393 792;
  • 33) 0.867 583 393 792 × 2 = 1 + 0.735 166 787 584;
  • 34) 0.735 166 787 584 × 2 = 1 + 0.470 333 575 168;
  • 35) 0.470 333 575 168 × 2 = 0 + 0.940 667 150 336;
  • 36) 0.940 667 150 336 × 2 = 1 + 0.881 334 300 672;
  • 37) 0.881 334 300 672 × 2 = 1 + 0.762 668 601 344;
  • 38) 0.762 668 601 344 × 2 = 1 + 0.525 337 202 688;
  • 39) 0.525 337 202 688 × 2 = 1 + 0.050 674 405 376;
  • 40) 0.050 674 405 376 × 2 = 0 + 0.101 348 810 752;
  • 41) 0.101 348 810 752 × 2 = 0 + 0.202 697 621 504;
  • 42) 0.202 697 621 504 × 2 = 0 + 0.405 395 243 008;
  • 43) 0.405 395 243 008 × 2 = 0 + 0.810 790 486 016;
  • 44) 0.810 790 486 016 × 2 = 1 + 0.621 580 972 032;
  • 45) 0.621 580 972 032 × 2 = 1 + 0.243 161 944 064;
  • 46) 0.243 161 944 064 × 2 = 0 + 0.486 323 888 128;
  • 47) 0.486 323 888 128 × 2 = 0 + 0.972 647 776 256;
  • 48) 0.972 647 776 256 × 2 = 1 + 0.945 295 552 512;
  • 49) 0.945 295 552 512 × 2 = 1 + 0.890 591 105 024;
  • 50) 0.890 591 105 024 × 2 = 1 + 0.781 182 210 048;
  • 51) 0.781 182 210 048 × 2 = 1 + 0.562 364 420 096;
  • 52) 0.562 364 420 096 × 2 = 1 + 0.124 728 840 192;
  • 53) 0.124 728 840 192 × 2 = 0 + 0.249 457 680 384;
  • 54) 0.249 457 680 384 × 2 = 0 + 0.498 915 360 768;
  • 55) 0.498 915 360 768 × 2 = 0 + 0.997 830 721 536;
  • 56) 0.997 830 721 536 × 2 = 1 + 0.995 661 443 072;

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:


0.000 000 000 202(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1101 1110 0001 1001 1111 0001(2)

6. Positive number before normalization:

0.000 000 000 202(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1101 1110 0001 1001 1111 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 33 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 202(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1101 1110 0001 1001 1111 0001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1101 1110 0001 1001 1111 0001(2) × 20 =

1.1011 1100 0011 0011 1110 001(2) × 2-33


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): -33

Mantissa (not normalized):
1.1011 1100 0011 0011 1110 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-33 + 2(8-1) - 1 =

(-33 + 127)(10) =

94(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;
  • 94 ÷ 2 = 47 + 0;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

94(10) =

0101 1110(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. 101 1110 0001 1001 1111 0001 =

101 1110 0001 1001 1111 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) =
0101 1110

Mantissa (23 bits) =
101 1110 0001 1001 1111 0001


Decimal number -0.000 000 000 202 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1110 - 101 1110 0001 1001 1111 0001

Share

» Convert decimal -0.000 000 000 25 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: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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