-0.000 000 004 63 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 004 63(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
-0.000 000 004 63(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:

|-0.000 000 004 63| = 0.000 000 004 63


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: 0.000 000 004 63.

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) 0.000 000 004 63 × 2 = 0 + 0.000 000 009 26;
  • 2) 0.000 000 009 26 × 2 = 0 + 0.000 000 018 52;
  • 3) 0.000 000 018 52 × 2 = 0 + 0.000 000 037 04;
  • 4) 0.000 000 037 04 × 2 = 0 + 0.000 000 074 08;
  • 5) 0.000 000 074 08 × 2 = 0 + 0.000 000 148 16;
  • 6) 0.000 000 148 16 × 2 = 0 + 0.000 000 296 32;
  • 7) 0.000 000 296 32 × 2 = 0 + 0.000 000 592 64;
  • 8) 0.000 000 592 64 × 2 = 0 + 0.000 001 185 28;
  • 9) 0.000 001 185 28 × 2 = 0 + 0.000 002 370 56;
  • 10) 0.000 002 370 56 × 2 = 0 + 0.000 004 741 12;
  • 11) 0.000 004 741 12 × 2 = 0 + 0.000 009 482 24;
  • 12) 0.000 009 482 24 × 2 = 0 + 0.000 018 964 48;
  • 13) 0.000 018 964 48 × 2 = 0 + 0.000 037 928 96;
  • 14) 0.000 037 928 96 × 2 = 0 + 0.000 075 857 92;
  • 15) 0.000 075 857 92 × 2 = 0 + 0.000 151 715 84;
  • 16) 0.000 151 715 84 × 2 = 0 + 0.000 303 431 68;
  • 17) 0.000 303 431 68 × 2 = 0 + 0.000 606 863 36;
  • 18) 0.000 606 863 36 × 2 = 0 + 0.001 213 726 72;
  • 19) 0.001 213 726 72 × 2 = 0 + 0.002 427 453 44;
  • 20) 0.002 427 453 44 × 2 = 0 + 0.004 854 906 88;
  • 21) 0.004 854 906 88 × 2 = 0 + 0.009 709 813 76;
  • 22) 0.009 709 813 76 × 2 = 0 + 0.019 419 627 52;
  • 23) 0.019 419 627 52 × 2 = 0 + 0.038 839 255 04;
  • 24) 0.038 839 255 04 × 2 = 0 + 0.077 678 510 08;
  • 25) 0.077 678 510 08 × 2 = 0 + 0.155 357 020 16;
  • 26) 0.155 357 020 16 × 2 = 0 + 0.310 714 040 32;
  • 27) 0.310 714 040 32 × 2 = 0 + 0.621 428 080 64;
  • 28) 0.621 428 080 64 × 2 = 1 + 0.242 856 161 28;
  • 29) 0.242 856 161 28 × 2 = 0 + 0.485 712 322 56;
  • 30) 0.485 712 322 56 × 2 = 0 + 0.971 424 645 12;
  • 31) 0.971 424 645 12 × 2 = 1 + 0.942 849 290 24;
  • 32) 0.942 849 290 24 × 2 = 1 + 0.885 698 580 48;
  • 33) 0.885 698 580 48 × 2 = 1 + 0.771 397 160 96;
  • 34) 0.771 397 160 96 × 2 = 1 + 0.542 794 321 92;
  • 35) 0.542 794 321 92 × 2 = 1 + 0.085 588 643 84;
  • 36) 0.085 588 643 84 × 2 = 0 + 0.171 177 287 68;
  • 37) 0.171 177 287 68 × 2 = 0 + 0.342 354 575 36;
  • 38) 0.342 354 575 36 × 2 = 0 + 0.684 709 150 72;
  • 39) 0.684 709 150 72 × 2 = 1 + 0.369 418 301 44;
  • 40) 0.369 418 301 44 × 2 = 0 + 0.738 836 602 88;
  • 41) 0.738 836 602 88 × 2 = 1 + 0.477 673 205 76;
  • 42) 0.477 673 205 76 × 2 = 0 + 0.955 346 411 52;
  • 43) 0.955 346 411 52 × 2 = 1 + 0.910 692 823 04;
  • 44) 0.910 692 823 04 × 2 = 1 + 0.821 385 646 08;
  • 45) 0.821 385 646 08 × 2 = 1 + 0.642 771 292 16;
  • 46) 0.642 771 292 16 × 2 = 1 + 0.285 542 584 32;
  • 47) 0.285 542 584 32 × 2 = 0 + 0.571 085 168 64;
  • 48) 0.571 085 168 64 × 2 = 1 + 0.142 170 337 28;
  • 49) 0.142 170 337 28 × 2 = 0 + 0.284 340 674 56;
  • 50) 0.284 340 674 56 × 2 = 0 + 0.568 681 349 12;
  • 51) 0.568 681 349 12 × 2 = 1 + 0.137 362 698 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:


0.000 000 004 63(10) =

0.0000 0000 0000 0000 0000 0000 0001 0011 1110 0010 1011 1101 001(2)

6. Positive number before normalization:

0.000 000 004 63(10) =

0.0000 0000 0000 0000 0000 0000 0001 0011 1110 0010 1011 1101 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 004 63(10) =

0.0000 0000 0000 0000 0000 0000 0001 0011 1110 0010 1011 1101 001(2) =

0.0000 0000 0000 0000 0000 0000 0001 0011 1110 0010 1011 1101 001(2) × 20 =

1.0011 1110 0010 1011 1101 001(2) × 2-28


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): -28

Mantissa (not normalized):
1.0011 1110 0010 1011 1101 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 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) =

99(10) =

0110 0011(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. 001 1111 0001 0101 1110 1001 =

001 1111 0001 0101 1110 1001


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 0011

Mantissa (23 bits) =
001 1111 0001 0101 1110 1001


Decimal number -0.000 000 004 63 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0011 - 001 1111 0001 0101 1110 1001

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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