0.000 000 017 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 017(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
0.000 000 017(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 017.

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) 0.000 000 017 × 2 = 0 + 0.000 000 034;
  • 2) 0.000 000 034 × 2 = 0 + 0.000 000 068;
  • 3) 0.000 000 068 × 2 = 0 + 0.000 000 136;
  • 4) 0.000 000 136 × 2 = 0 + 0.000 000 272;
  • 5) 0.000 000 272 × 2 = 0 + 0.000 000 544;
  • 6) 0.000 000 544 × 2 = 0 + 0.000 001 088;
  • 7) 0.000 001 088 × 2 = 0 + 0.000 002 176;
  • 8) 0.000 002 176 × 2 = 0 + 0.000 004 352;
  • 9) 0.000 004 352 × 2 = 0 + 0.000 008 704;
  • 10) 0.000 008 704 × 2 = 0 + 0.000 017 408;
  • 11) 0.000 017 408 × 2 = 0 + 0.000 034 816;
  • 12) 0.000 034 816 × 2 = 0 + 0.000 069 632;
  • 13) 0.000 069 632 × 2 = 0 + 0.000 139 264;
  • 14) 0.000 139 264 × 2 = 0 + 0.000 278 528;
  • 15) 0.000 278 528 × 2 = 0 + 0.000 557 056;
  • 16) 0.000 557 056 × 2 = 0 + 0.001 114 112;
  • 17) 0.001 114 112 × 2 = 0 + 0.002 228 224;
  • 18) 0.002 228 224 × 2 = 0 + 0.004 456 448;
  • 19) 0.004 456 448 × 2 = 0 + 0.008 912 896;
  • 20) 0.008 912 896 × 2 = 0 + 0.017 825 792;
  • 21) 0.017 825 792 × 2 = 0 + 0.035 651 584;
  • 22) 0.035 651 584 × 2 = 0 + 0.071 303 168;
  • 23) 0.071 303 168 × 2 = 0 + 0.142 606 336;
  • 24) 0.142 606 336 × 2 = 0 + 0.285 212 672;
  • 25) 0.285 212 672 × 2 = 0 + 0.570 425 344;
  • 26) 0.570 425 344 × 2 = 1 + 0.140 850 688;
  • 27) 0.140 850 688 × 2 = 0 + 0.281 701 376;
  • 28) 0.281 701 376 × 2 = 0 + 0.563 402 752;
  • 29) 0.563 402 752 × 2 = 1 + 0.126 805 504;
  • 30) 0.126 805 504 × 2 = 0 + 0.253 611 008;
  • 31) 0.253 611 008 × 2 = 0 + 0.507 222 016;
  • 32) 0.507 222 016 × 2 = 1 + 0.014 444 032;
  • 33) 0.014 444 032 × 2 = 0 + 0.028 888 064;
  • 34) 0.028 888 064 × 2 = 0 + 0.057 776 128;
  • 35) 0.057 776 128 × 2 = 0 + 0.115 552 256;
  • 36) 0.115 552 256 × 2 = 0 + 0.231 104 512;
  • 37) 0.231 104 512 × 2 = 0 + 0.462 209 024;
  • 38) 0.462 209 024 × 2 = 0 + 0.924 418 048;
  • 39) 0.924 418 048 × 2 = 1 + 0.848 836 096;
  • 40) 0.848 836 096 × 2 = 1 + 0.697 672 192;
  • 41) 0.697 672 192 × 2 = 1 + 0.395 344 384;
  • 42) 0.395 344 384 × 2 = 0 + 0.790 688 768;
  • 43) 0.790 688 768 × 2 = 1 + 0.581 377 536;
  • 44) 0.581 377 536 × 2 = 1 + 0.162 755 072;
  • 45) 0.162 755 072 × 2 = 0 + 0.325 510 144;
  • 46) 0.325 510 144 × 2 = 0 + 0.651 020 288;
  • 47) 0.651 020 288 × 2 = 1 + 0.302 040 576;
  • 48) 0.302 040 576 × 2 = 0 + 0.604 081 152;
  • 49) 0.604 081 152 × 2 = 1 + 0.208 162 304;

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).


4. 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:


0.000 000 017(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 0000 0011 1011 0010 1(2)

5. Positive number before normalization:

0.000 000 017(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 0000 0011 1011 0010 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 017(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 0000 0011 1011 0010 1(2) =

0.0000 0000 0000 0000 0000 0000 0100 1001 0000 0011 1011 0010 1(2) × 20 =

1.0010 0100 0000 1110 1100 101(2) × 2-26


7. 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 0 (a positive number)

Exponent (unadjusted): -26

Mantissa (not normalized):
1.0010 0100 0000 1110 1100 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(10)


9. 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;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

101(10) =

0110 0101(2)


11. 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. 001 0010 0000 0111 0110 0101 =

001 0010 0000 0111 0110 0101


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0101

Mantissa (23 bits) =
001 0010 0000 0111 0110 0101


Decimal number 0.000 000 017 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 001 0010 0000 0111 0110 0101

Share

» Convert decimal -0.000 000 043 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