-0.000 000 151 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 151(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
-0.000 000 151(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:

|-0.000 000 151| = 0.000 000 151


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: 0.000 000 151.

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) 0.000 000 151 × 2 = 0 + 0.000 000 302;
  • 2) 0.000 000 302 × 2 = 0 + 0.000 000 604;
  • 3) 0.000 000 604 × 2 = 0 + 0.000 001 208;
  • 4) 0.000 001 208 × 2 = 0 + 0.000 002 416;
  • 5) 0.000 002 416 × 2 = 0 + 0.000 004 832;
  • 6) 0.000 004 832 × 2 = 0 + 0.000 009 664;
  • 7) 0.000 009 664 × 2 = 0 + 0.000 019 328;
  • 8) 0.000 019 328 × 2 = 0 + 0.000 038 656;
  • 9) 0.000 038 656 × 2 = 0 + 0.000 077 312;
  • 10) 0.000 077 312 × 2 = 0 + 0.000 154 624;
  • 11) 0.000 154 624 × 2 = 0 + 0.000 309 248;
  • 12) 0.000 309 248 × 2 = 0 + 0.000 618 496;
  • 13) 0.000 618 496 × 2 = 0 + 0.001 236 992;
  • 14) 0.001 236 992 × 2 = 0 + 0.002 473 984;
  • 15) 0.002 473 984 × 2 = 0 + 0.004 947 968;
  • 16) 0.004 947 968 × 2 = 0 + 0.009 895 936;
  • 17) 0.009 895 936 × 2 = 0 + 0.019 791 872;
  • 18) 0.019 791 872 × 2 = 0 + 0.039 583 744;
  • 19) 0.039 583 744 × 2 = 0 + 0.079 167 488;
  • 20) 0.079 167 488 × 2 = 0 + 0.158 334 976;
  • 21) 0.158 334 976 × 2 = 0 + 0.316 669 952;
  • 22) 0.316 669 952 × 2 = 0 + 0.633 339 904;
  • 23) 0.633 339 904 × 2 = 1 + 0.266 679 808;
  • 24) 0.266 679 808 × 2 = 0 + 0.533 359 616;
  • 25) 0.533 359 616 × 2 = 1 + 0.066 719 232;
  • 26) 0.066 719 232 × 2 = 0 + 0.133 438 464;
  • 27) 0.133 438 464 × 2 = 0 + 0.266 876 928;
  • 28) 0.266 876 928 × 2 = 0 + 0.533 753 856;
  • 29) 0.533 753 856 × 2 = 1 + 0.067 507 712;
  • 30) 0.067 507 712 × 2 = 0 + 0.135 015 424;
  • 31) 0.135 015 424 × 2 = 0 + 0.270 030 848;
  • 32) 0.270 030 848 × 2 = 0 + 0.540 061 696;
  • 33) 0.540 061 696 × 2 = 1 + 0.080 123 392;
  • 34) 0.080 123 392 × 2 = 0 + 0.160 246 784;
  • 35) 0.160 246 784 × 2 = 0 + 0.320 493 568;
  • 36) 0.320 493 568 × 2 = 0 + 0.640 987 136;
  • 37) 0.640 987 136 × 2 = 1 + 0.281 974 272;
  • 38) 0.281 974 272 × 2 = 0 + 0.563 948 544;
  • 39) 0.563 948 544 × 2 = 1 + 0.127 897 088;
  • 40) 0.127 897 088 × 2 = 0 + 0.255 794 176;
  • 41) 0.255 794 176 × 2 = 0 + 0.511 588 352;
  • 42) 0.511 588 352 × 2 = 1 + 0.023 176 704;
  • 43) 0.023 176 704 × 2 = 0 + 0.046 353 408;
  • 44) 0.046 353 408 × 2 = 0 + 0.092 706 816;
  • 45) 0.092 706 816 × 2 = 0 + 0.185 413 632;
  • 46) 0.185 413 632 × 2 = 0 + 0.370 827 264;

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:


0.000 000 151(10) =

0.0000 0000 0000 0000 0000 0010 1000 1000 1000 1010 0100 00(2)

6. Positive number before normalization:

0.000 000 151(10) =

0.0000 0000 0000 0000 0000 0010 1000 1000 1000 1010 0100 00(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:


0.000 000 151(10) =

0.0000 0000 0000 0000 0000 0010 1000 1000 1000 1010 0100 00(2) =

0.0000 0000 0000 0000 0000 0010 1000 1000 1000 1010 0100 00(2) × 20 =

1.0100 0100 0100 0101 0010 000(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 0100 0100 0101 0010 000


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 0010 0010 0010 1001 0000 =

010 0010 0010 0010 1001 0000


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 0010 0010 0010 1001 0000


Decimal number -0.000 000 151 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1000 - 010 0010 0010 0010 1001 0000

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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