-0.000 000 000 98 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 98(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 98(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 98| = 0.000 000 000 98


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 98.

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 98 × 2 = 0 + 0.000 000 001 96;
  • 2) 0.000 000 001 96 × 2 = 0 + 0.000 000 003 92;
  • 3) 0.000 000 003 92 × 2 = 0 + 0.000 000 007 84;
  • 4) 0.000 000 007 84 × 2 = 0 + 0.000 000 015 68;
  • 5) 0.000 000 015 68 × 2 = 0 + 0.000 000 031 36;
  • 6) 0.000 000 031 36 × 2 = 0 + 0.000 000 062 72;
  • 7) 0.000 000 062 72 × 2 = 0 + 0.000 000 125 44;
  • 8) 0.000 000 125 44 × 2 = 0 + 0.000 000 250 88;
  • 9) 0.000 000 250 88 × 2 = 0 + 0.000 000 501 76;
  • 10) 0.000 000 501 76 × 2 = 0 + 0.000 001 003 52;
  • 11) 0.000 001 003 52 × 2 = 0 + 0.000 002 007 04;
  • 12) 0.000 002 007 04 × 2 = 0 + 0.000 004 014 08;
  • 13) 0.000 004 014 08 × 2 = 0 + 0.000 008 028 16;
  • 14) 0.000 008 028 16 × 2 = 0 + 0.000 016 056 32;
  • 15) 0.000 016 056 32 × 2 = 0 + 0.000 032 112 64;
  • 16) 0.000 032 112 64 × 2 = 0 + 0.000 064 225 28;
  • 17) 0.000 064 225 28 × 2 = 0 + 0.000 128 450 56;
  • 18) 0.000 128 450 56 × 2 = 0 + 0.000 256 901 12;
  • 19) 0.000 256 901 12 × 2 = 0 + 0.000 513 802 24;
  • 20) 0.000 513 802 24 × 2 = 0 + 0.001 027 604 48;
  • 21) 0.001 027 604 48 × 2 = 0 + 0.002 055 208 96;
  • 22) 0.002 055 208 96 × 2 = 0 + 0.004 110 417 92;
  • 23) 0.004 110 417 92 × 2 = 0 + 0.008 220 835 84;
  • 24) 0.008 220 835 84 × 2 = 0 + 0.016 441 671 68;
  • 25) 0.016 441 671 68 × 2 = 0 + 0.032 883 343 36;
  • 26) 0.032 883 343 36 × 2 = 0 + 0.065 766 686 72;
  • 27) 0.065 766 686 72 × 2 = 0 + 0.131 533 373 44;
  • 28) 0.131 533 373 44 × 2 = 0 + 0.263 066 746 88;
  • 29) 0.263 066 746 88 × 2 = 0 + 0.526 133 493 76;
  • 30) 0.526 133 493 76 × 2 = 1 + 0.052 266 987 52;
  • 31) 0.052 266 987 52 × 2 = 0 + 0.104 533 975 04;
  • 32) 0.104 533 975 04 × 2 = 0 + 0.209 067 950 08;
  • 33) 0.209 067 950 08 × 2 = 0 + 0.418 135 900 16;
  • 34) 0.418 135 900 16 × 2 = 0 + 0.836 271 800 32;
  • 35) 0.836 271 800 32 × 2 = 1 + 0.672 543 600 64;
  • 36) 0.672 543 600 64 × 2 = 1 + 0.345 087 201 28;
  • 37) 0.345 087 201 28 × 2 = 0 + 0.690 174 402 56;
  • 38) 0.690 174 402 56 × 2 = 1 + 0.380 348 805 12;
  • 39) 0.380 348 805 12 × 2 = 0 + 0.760 697 610 24;
  • 40) 0.760 697 610 24 × 2 = 1 + 0.521 395 220 48;
  • 41) 0.521 395 220 48 × 2 = 1 + 0.042 790 440 96;
  • 42) 0.042 790 440 96 × 2 = 0 + 0.085 580 881 92;
  • 43) 0.085 580 881 92 × 2 = 0 + 0.171 161 763 84;
  • 44) 0.171 161 763 84 × 2 = 0 + 0.342 323 527 68;
  • 45) 0.342 323 527 68 × 2 = 0 + 0.684 647 055 36;
  • 46) 0.684 647 055 36 × 2 = 1 + 0.369 294 110 72;
  • 47) 0.369 294 110 72 × 2 = 0 + 0.738 588 221 44;
  • 48) 0.738 588 221 44 × 2 = 1 + 0.477 176 442 88;
  • 49) 0.477 176 442 88 × 2 = 0 + 0.954 352 885 76;
  • 50) 0.954 352 885 76 × 2 = 1 + 0.908 705 771 52;
  • 51) 0.908 705 771 52 × 2 = 1 + 0.817 411 543 04;
  • 52) 0.817 411 543 04 × 2 = 1 + 0.634 823 086 08;
  • 53) 0.634 823 086 08 × 2 = 1 + 0.269 646 172 16;

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 98(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0011 0101 1000 0101 0111 1(2)

6. Positive number before normalization:

0.000 000 000 98(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0011 0101 1000 0101 0111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 98(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0011 0101 1000 0101 0111 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0011 0101 1000 0101 0111 1(2) × 20 =

1.0000 1101 0110 0001 0101 111(2) × 2-30


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): -30

Mantissa (not normalized):
1.0000 1101 0110 0001 0101 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-30 + 2(8-1) - 1 =

(-30 + 127)(10) =

97(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;
  • 97 ÷ 2 = 48 + 1;
  • 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) =

97(10) =

0110 0001(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. 000 0110 1011 0000 1010 1111 =

000 0110 1011 0000 1010 1111


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 0001

Mantissa (23 bits) =
000 0110 1011 0000 1010 1111


Decimal number -0.000 000 000 98 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0001 - 000 0110 1011 0000 1010 1111

Share

» Convert decimal -0.000 000 000 59 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