0.000 000 060 508 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 060 508(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
0.000 000 060 508(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 060 508.

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) 0.000 000 060 508 × 2 = 0 + 0.000 000 121 016;
  • 2) 0.000 000 121 016 × 2 = 0 + 0.000 000 242 032;
  • 3) 0.000 000 242 032 × 2 = 0 + 0.000 000 484 064;
  • 4) 0.000 000 484 064 × 2 = 0 + 0.000 000 968 128;
  • 5) 0.000 000 968 128 × 2 = 0 + 0.000 001 936 256;
  • 6) 0.000 001 936 256 × 2 = 0 + 0.000 003 872 512;
  • 7) 0.000 003 872 512 × 2 = 0 + 0.000 007 745 024;
  • 8) 0.000 007 745 024 × 2 = 0 + 0.000 015 490 048;
  • 9) 0.000 015 490 048 × 2 = 0 + 0.000 030 980 096;
  • 10) 0.000 030 980 096 × 2 = 0 + 0.000 061 960 192;
  • 11) 0.000 061 960 192 × 2 = 0 + 0.000 123 920 384;
  • 12) 0.000 123 920 384 × 2 = 0 + 0.000 247 840 768;
  • 13) 0.000 247 840 768 × 2 = 0 + 0.000 495 681 536;
  • 14) 0.000 495 681 536 × 2 = 0 + 0.000 991 363 072;
  • 15) 0.000 991 363 072 × 2 = 0 + 0.001 982 726 144;
  • 16) 0.001 982 726 144 × 2 = 0 + 0.003 965 452 288;
  • 17) 0.003 965 452 288 × 2 = 0 + 0.007 930 904 576;
  • 18) 0.007 930 904 576 × 2 = 0 + 0.015 861 809 152;
  • 19) 0.015 861 809 152 × 2 = 0 + 0.031 723 618 304;
  • 20) 0.031 723 618 304 × 2 = 0 + 0.063 447 236 608;
  • 21) 0.063 447 236 608 × 2 = 0 + 0.126 894 473 216;
  • 22) 0.126 894 473 216 × 2 = 0 + 0.253 788 946 432;
  • 23) 0.253 788 946 432 × 2 = 0 + 0.507 577 892 864;
  • 24) 0.507 577 892 864 × 2 = 1 + 0.015 155 785 728;
  • 25) 0.015 155 785 728 × 2 = 0 + 0.030 311 571 456;
  • 26) 0.030 311 571 456 × 2 = 0 + 0.060 623 142 912;
  • 27) 0.060 623 142 912 × 2 = 0 + 0.121 246 285 824;
  • 28) 0.121 246 285 824 × 2 = 0 + 0.242 492 571 648;
  • 29) 0.242 492 571 648 × 2 = 0 + 0.484 985 143 296;
  • 30) 0.484 985 143 296 × 2 = 0 + 0.969 970 286 592;
  • 31) 0.969 970 286 592 × 2 = 1 + 0.939 940 573 184;
  • 32) 0.939 940 573 184 × 2 = 1 + 0.879 881 146 368;
  • 33) 0.879 881 146 368 × 2 = 1 + 0.759 762 292 736;
  • 34) 0.759 762 292 736 × 2 = 1 + 0.519 524 585 472;
  • 35) 0.519 524 585 472 × 2 = 1 + 0.039 049 170 944;
  • 36) 0.039 049 170 944 × 2 = 0 + 0.078 098 341 888;
  • 37) 0.078 098 341 888 × 2 = 0 + 0.156 196 683 776;
  • 38) 0.156 196 683 776 × 2 = 0 + 0.312 393 367 552;
  • 39) 0.312 393 367 552 × 2 = 0 + 0.624 786 735 104;
  • 40) 0.624 786 735 104 × 2 = 1 + 0.249 573 470 208;
  • 41) 0.249 573 470 208 × 2 = 0 + 0.499 146 940 416;
  • 42) 0.499 146 940 416 × 2 = 0 + 0.998 293 880 832;
  • 43) 0.998 293 880 832 × 2 = 1 + 0.996 587 761 664;
  • 44) 0.996 587 761 664 × 2 = 1 + 0.993 175 523 328;
  • 45) 0.993 175 523 328 × 2 = 1 + 0.986 351 046 656;
  • 46) 0.986 351 046 656 × 2 = 1 + 0.972 702 093 312;
  • 47) 0.972 702 093 312 × 2 = 1 + 0.945 404 186 624;

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).


4. 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:


0.000 000 060 508(10) =

0.0000 0000 0000 0000 0000 0001 0000 0011 1110 0001 0011 111(2)

5. Positive number before normalization:

0.000 000 060 508(10) =

0.0000 0000 0000 0000 0000 0001 0000 0011 1110 0001 0011 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 060 508(10) =

0.0000 0000 0000 0000 0000 0001 0000 0011 1110 0001 0011 111(2) =

0.0000 0000 0000 0000 0000 0001 0000 0011 1110 0001 0011 111(2) × 20 =

1.0000 0011 1110 0001 0011 111(2) × 2-24


7. 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 0 (a positive number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.0000 0011 1110 0001 0011 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(10)


9. 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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

103(10) =

0110 0111(2)


11. 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 0001 1111 0000 1001 1111 =

000 0001 1111 0000 1001 1111


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0111

Mantissa (23 bits) =
000 0001 1111 0000 1001 1111


Decimal number 0.000 000 060 508 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 000 0001 1111 0000 1001 1111

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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