-0.000 000 152 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 152(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
-0.000 000 152(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:

|-0.000 000 152| = 0.000 000 152


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: 0.000 000 152.

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) 0.000 000 152 × 2 = 0 + 0.000 000 304;
  • 2) 0.000 000 304 × 2 = 0 + 0.000 000 608;
  • 3) 0.000 000 608 × 2 = 0 + 0.000 001 216;
  • 4) 0.000 001 216 × 2 = 0 + 0.000 002 432;
  • 5) 0.000 002 432 × 2 = 0 + 0.000 004 864;
  • 6) 0.000 004 864 × 2 = 0 + 0.000 009 728;
  • 7) 0.000 009 728 × 2 = 0 + 0.000 019 456;
  • 8) 0.000 019 456 × 2 = 0 + 0.000 038 912;
  • 9) 0.000 038 912 × 2 = 0 + 0.000 077 824;
  • 10) 0.000 077 824 × 2 = 0 + 0.000 155 648;
  • 11) 0.000 155 648 × 2 = 0 + 0.000 311 296;
  • 12) 0.000 311 296 × 2 = 0 + 0.000 622 592;
  • 13) 0.000 622 592 × 2 = 0 + 0.001 245 184;
  • 14) 0.001 245 184 × 2 = 0 + 0.002 490 368;
  • 15) 0.002 490 368 × 2 = 0 + 0.004 980 736;
  • 16) 0.004 980 736 × 2 = 0 + 0.009 961 472;
  • 17) 0.009 961 472 × 2 = 0 + 0.019 922 944;
  • 18) 0.019 922 944 × 2 = 0 + 0.039 845 888;
  • 19) 0.039 845 888 × 2 = 0 + 0.079 691 776;
  • 20) 0.079 691 776 × 2 = 0 + 0.159 383 552;
  • 21) 0.159 383 552 × 2 = 0 + 0.318 767 104;
  • 22) 0.318 767 104 × 2 = 0 + 0.637 534 208;
  • 23) 0.637 534 208 × 2 = 1 + 0.275 068 416;
  • 24) 0.275 068 416 × 2 = 0 + 0.550 136 832;
  • 25) 0.550 136 832 × 2 = 1 + 0.100 273 664;
  • 26) 0.100 273 664 × 2 = 0 + 0.200 547 328;
  • 27) 0.200 547 328 × 2 = 0 + 0.401 094 656;
  • 28) 0.401 094 656 × 2 = 0 + 0.802 189 312;
  • 29) 0.802 189 312 × 2 = 1 + 0.604 378 624;
  • 30) 0.604 378 624 × 2 = 1 + 0.208 757 248;
  • 31) 0.208 757 248 × 2 = 0 + 0.417 514 496;
  • 32) 0.417 514 496 × 2 = 0 + 0.835 028 992;
  • 33) 0.835 028 992 × 2 = 1 + 0.670 057 984;
  • 34) 0.670 057 984 × 2 = 1 + 0.340 115 968;
  • 35) 0.340 115 968 × 2 = 0 + 0.680 231 936;
  • 36) 0.680 231 936 × 2 = 1 + 0.360 463 872;
  • 37) 0.360 463 872 × 2 = 0 + 0.720 927 744;
  • 38) 0.720 927 744 × 2 = 1 + 0.441 855 488;
  • 39) 0.441 855 488 × 2 = 0 + 0.883 710 976;
  • 40) 0.883 710 976 × 2 = 1 + 0.767 421 952;
  • 41) 0.767 421 952 × 2 = 1 + 0.534 843 904;
  • 42) 0.534 843 904 × 2 = 1 + 0.069 687 808;
  • 43) 0.069 687 808 × 2 = 0 + 0.139 375 616;
  • 44) 0.139 375 616 × 2 = 0 + 0.278 751 232;
  • 45) 0.278 751 232 × 2 = 0 + 0.557 502 464;
  • 46) 0.557 502 464 × 2 = 1 + 0.115 004 928;

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:


0.000 000 152(10) =

0.0000 0000 0000 0000 0000 0010 1000 1100 1101 0101 1100 01(2)

6. Positive number before normalization:

0.000 000 152(10) =

0.0000 0000 0000 0000 0000 0010 1000 1100 1101 0101 1100 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 152(10) =

0.0000 0000 0000 0000 0000 0010 1000 1100 1101 0101 1100 01(2) =

0.0000 0000 0000 0000 0000 0010 1000 1100 1101 0101 1100 01(2) × 20 =

1.0100 0110 0110 1010 1110 001(2) × 2-23


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): -23

Mantissa (not normalized):
1.0100 0110 0110 1010 1110 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(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;
  • 104 ÷ 2 = 52 + 0;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

104(10) =

0110 1000(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. 010 0011 0011 0101 0111 0001 =

010 0011 0011 0101 0111 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) =
0110 1000

Mantissa (23 bits) =
010 0011 0011 0101 0111 0001


Decimal number -0.000 000 152 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1000 - 010 0011 0011 0101 0111 0001

Share

» Convert decimal -0.000 000 085 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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