-0.000 000 000 97 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 97(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 97(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 97| = 0.000 000 000 97


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 97.

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 97 × 2 = 0 + 0.000 000 001 94;
  • 2) 0.000 000 001 94 × 2 = 0 + 0.000 000 003 88;
  • 3) 0.000 000 003 88 × 2 = 0 + 0.000 000 007 76;
  • 4) 0.000 000 007 76 × 2 = 0 + 0.000 000 015 52;
  • 5) 0.000 000 015 52 × 2 = 0 + 0.000 000 031 04;
  • 6) 0.000 000 031 04 × 2 = 0 + 0.000 000 062 08;
  • 7) 0.000 000 062 08 × 2 = 0 + 0.000 000 124 16;
  • 8) 0.000 000 124 16 × 2 = 0 + 0.000 000 248 32;
  • 9) 0.000 000 248 32 × 2 = 0 + 0.000 000 496 64;
  • 10) 0.000 000 496 64 × 2 = 0 + 0.000 000 993 28;
  • 11) 0.000 000 993 28 × 2 = 0 + 0.000 001 986 56;
  • 12) 0.000 001 986 56 × 2 = 0 + 0.000 003 973 12;
  • 13) 0.000 003 973 12 × 2 = 0 + 0.000 007 946 24;
  • 14) 0.000 007 946 24 × 2 = 0 + 0.000 015 892 48;
  • 15) 0.000 015 892 48 × 2 = 0 + 0.000 031 784 96;
  • 16) 0.000 031 784 96 × 2 = 0 + 0.000 063 569 92;
  • 17) 0.000 063 569 92 × 2 = 0 + 0.000 127 139 84;
  • 18) 0.000 127 139 84 × 2 = 0 + 0.000 254 279 68;
  • 19) 0.000 254 279 68 × 2 = 0 + 0.000 508 559 36;
  • 20) 0.000 508 559 36 × 2 = 0 + 0.001 017 118 72;
  • 21) 0.001 017 118 72 × 2 = 0 + 0.002 034 237 44;
  • 22) 0.002 034 237 44 × 2 = 0 + 0.004 068 474 88;
  • 23) 0.004 068 474 88 × 2 = 0 + 0.008 136 949 76;
  • 24) 0.008 136 949 76 × 2 = 0 + 0.016 273 899 52;
  • 25) 0.016 273 899 52 × 2 = 0 + 0.032 547 799 04;
  • 26) 0.032 547 799 04 × 2 = 0 + 0.065 095 598 08;
  • 27) 0.065 095 598 08 × 2 = 0 + 0.130 191 196 16;
  • 28) 0.130 191 196 16 × 2 = 0 + 0.260 382 392 32;
  • 29) 0.260 382 392 32 × 2 = 0 + 0.520 764 784 64;
  • 30) 0.520 764 784 64 × 2 = 1 + 0.041 529 569 28;
  • 31) 0.041 529 569 28 × 2 = 0 + 0.083 059 138 56;
  • 32) 0.083 059 138 56 × 2 = 0 + 0.166 118 277 12;
  • 33) 0.166 118 277 12 × 2 = 0 + 0.332 236 554 24;
  • 34) 0.332 236 554 24 × 2 = 0 + 0.664 473 108 48;
  • 35) 0.664 473 108 48 × 2 = 1 + 0.328 946 216 96;
  • 36) 0.328 946 216 96 × 2 = 0 + 0.657 892 433 92;
  • 37) 0.657 892 433 92 × 2 = 1 + 0.315 784 867 84;
  • 38) 0.315 784 867 84 × 2 = 0 + 0.631 569 735 68;
  • 39) 0.631 569 735 68 × 2 = 1 + 0.263 139 471 36;
  • 40) 0.263 139 471 36 × 2 = 0 + 0.526 278 942 72;
  • 41) 0.526 278 942 72 × 2 = 1 + 0.052 557 885 44;
  • 42) 0.052 557 885 44 × 2 = 0 + 0.105 115 770 88;
  • 43) 0.105 115 770 88 × 2 = 0 + 0.210 231 541 76;
  • 44) 0.210 231 541 76 × 2 = 0 + 0.420 463 083 52;
  • 45) 0.420 463 083 52 × 2 = 0 + 0.840 926 167 04;
  • 46) 0.840 926 167 04 × 2 = 1 + 0.681 852 334 08;
  • 47) 0.681 852 334 08 × 2 = 1 + 0.363 704 668 16;
  • 48) 0.363 704 668 16 × 2 = 0 + 0.727 409 336 32;
  • 49) 0.727 409 336 32 × 2 = 1 + 0.454 818 672 64;
  • 50) 0.454 818 672 64 × 2 = 0 + 0.909 637 345 28;
  • 51) 0.909 637 345 28 × 2 = 1 + 0.819 274 690 56;
  • 52) 0.819 274 690 56 × 2 = 1 + 0.638 549 381 12;
  • 53) 0.638 549 381 12 × 2 = 1 + 0.277 098 762 24;

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 97(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0010 1010 1000 0110 1011 1(2)

6. Positive number before normalization:

0.000 000 000 97(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0010 1010 1000 0110 1011 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 97(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0010 1010 1000 0110 1011 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0010 1010 1000 0110 1011 1(2) × 20 =

1.0000 1010 1010 0001 1010 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 1010 1010 0001 1010 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 0101 0101 0000 1101 0111 =

000 0101 0101 0000 1101 0111


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 0101 0101 0000 1101 0111


Decimal number -0.000 000 000 97 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0001 - 000 0101 0101 0000 1101 0111

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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