-0.000 000 153 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 153(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
-0.000 000 153(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:

|-0.000 000 153| = 0.000 000 153


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: 0.000 000 153.

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) 0.000 000 153 × 2 = 0 + 0.000 000 306;
  • 2) 0.000 000 306 × 2 = 0 + 0.000 000 612;
  • 3) 0.000 000 612 × 2 = 0 + 0.000 001 224;
  • 4) 0.000 001 224 × 2 = 0 + 0.000 002 448;
  • 5) 0.000 002 448 × 2 = 0 + 0.000 004 896;
  • 6) 0.000 004 896 × 2 = 0 + 0.000 009 792;
  • 7) 0.000 009 792 × 2 = 0 + 0.000 019 584;
  • 8) 0.000 019 584 × 2 = 0 + 0.000 039 168;
  • 9) 0.000 039 168 × 2 = 0 + 0.000 078 336;
  • 10) 0.000 078 336 × 2 = 0 + 0.000 156 672;
  • 11) 0.000 156 672 × 2 = 0 + 0.000 313 344;
  • 12) 0.000 313 344 × 2 = 0 + 0.000 626 688;
  • 13) 0.000 626 688 × 2 = 0 + 0.001 253 376;
  • 14) 0.001 253 376 × 2 = 0 + 0.002 506 752;
  • 15) 0.002 506 752 × 2 = 0 + 0.005 013 504;
  • 16) 0.005 013 504 × 2 = 0 + 0.010 027 008;
  • 17) 0.010 027 008 × 2 = 0 + 0.020 054 016;
  • 18) 0.020 054 016 × 2 = 0 + 0.040 108 032;
  • 19) 0.040 108 032 × 2 = 0 + 0.080 216 064;
  • 20) 0.080 216 064 × 2 = 0 + 0.160 432 128;
  • 21) 0.160 432 128 × 2 = 0 + 0.320 864 256;
  • 22) 0.320 864 256 × 2 = 0 + 0.641 728 512;
  • 23) 0.641 728 512 × 2 = 1 + 0.283 457 024;
  • 24) 0.283 457 024 × 2 = 0 + 0.566 914 048;
  • 25) 0.566 914 048 × 2 = 1 + 0.133 828 096;
  • 26) 0.133 828 096 × 2 = 0 + 0.267 656 192;
  • 27) 0.267 656 192 × 2 = 0 + 0.535 312 384;
  • 28) 0.535 312 384 × 2 = 1 + 0.070 624 768;
  • 29) 0.070 624 768 × 2 = 0 + 0.141 249 536;
  • 30) 0.141 249 536 × 2 = 0 + 0.282 499 072;
  • 31) 0.282 499 072 × 2 = 0 + 0.564 998 144;
  • 32) 0.564 998 144 × 2 = 1 + 0.129 996 288;
  • 33) 0.129 996 288 × 2 = 0 + 0.259 992 576;
  • 34) 0.259 992 576 × 2 = 0 + 0.519 985 152;
  • 35) 0.519 985 152 × 2 = 1 + 0.039 970 304;
  • 36) 0.039 970 304 × 2 = 0 + 0.079 940 608;
  • 37) 0.079 940 608 × 2 = 0 + 0.159 881 216;
  • 38) 0.159 881 216 × 2 = 0 + 0.319 762 432;
  • 39) 0.319 762 432 × 2 = 0 + 0.639 524 864;
  • 40) 0.639 524 864 × 2 = 1 + 0.279 049 728;
  • 41) 0.279 049 728 × 2 = 0 + 0.558 099 456;
  • 42) 0.558 099 456 × 2 = 1 + 0.116 198 912;
  • 43) 0.116 198 912 × 2 = 0 + 0.232 397 824;
  • 44) 0.232 397 824 × 2 = 0 + 0.464 795 648;
  • 45) 0.464 795 648 × 2 = 0 + 0.929 591 296;
  • 46) 0.929 591 296 × 2 = 1 + 0.859 182 592;

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:


0.000 000 153(10) =

0.0000 0000 0000 0000 0000 0010 1001 0001 0010 0001 0100 01(2)

6. Positive number before normalization:

0.000 000 153(10) =

0.0000 0000 0000 0000 0000 0010 1001 0001 0010 0001 0100 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:


0.000 000 153(10) =

0.0000 0000 0000 0000 0000 0010 1001 0001 0010 0001 0100 01(2) =

0.0000 0000 0000 0000 0000 0010 1001 0001 0010 0001 0100 01(2) × 20 =

1.0100 1000 1001 0000 1010 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 1000 1001 0000 1010 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 0100 0100 1000 0101 0001 =

010 0100 0100 1000 0101 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 0100 0100 1000 0101 0001


Decimal number -0.000 000 153 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1000 - 010 0100 0100 1000 0101 0001

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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