0.000 000 060 548 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 060 548(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
0.000 000 060 548(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: 0.000 000 060 548.

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) 0.000 000 060 548 × 2 = 0 + 0.000 000 121 096;
  • 2) 0.000 000 121 096 × 2 = 0 + 0.000 000 242 192;
  • 3) 0.000 000 242 192 × 2 = 0 + 0.000 000 484 384;
  • 4) 0.000 000 484 384 × 2 = 0 + 0.000 000 968 768;
  • 5) 0.000 000 968 768 × 2 = 0 + 0.000 001 937 536;
  • 6) 0.000 001 937 536 × 2 = 0 + 0.000 003 875 072;
  • 7) 0.000 003 875 072 × 2 = 0 + 0.000 007 750 144;
  • 8) 0.000 007 750 144 × 2 = 0 + 0.000 015 500 288;
  • 9) 0.000 015 500 288 × 2 = 0 + 0.000 031 000 576;
  • 10) 0.000 031 000 576 × 2 = 0 + 0.000 062 001 152;
  • 11) 0.000 062 001 152 × 2 = 0 + 0.000 124 002 304;
  • 12) 0.000 124 002 304 × 2 = 0 + 0.000 248 004 608;
  • 13) 0.000 248 004 608 × 2 = 0 + 0.000 496 009 216;
  • 14) 0.000 496 009 216 × 2 = 0 + 0.000 992 018 432;
  • 15) 0.000 992 018 432 × 2 = 0 + 0.001 984 036 864;
  • 16) 0.001 984 036 864 × 2 = 0 + 0.003 968 073 728;
  • 17) 0.003 968 073 728 × 2 = 0 + 0.007 936 147 456;
  • 18) 0.007 936 147 456 × 2 = 0 + 0.015 872 294 912;
  • 19) 0.015 872 294 912 × 2 = 0 + 0.031 744 589 824;
  • 20) 0.031 744 589 824 × 2 = 0 + 0.063 489 179 648;
  • 21) 0.063 489 179 648 × 2 = 0 + 0.126 978 359 296;
  • 22) 0.126 978 359 296 × 2 = 0 + 0.253 956 718 592;
  • 23) 0.253 956 718 592 × 2 = 0 + 0.507 913 437 184;
  • 24) 0.507 913 437 184 × 2 = 1 + 0.015 826 874 368;
  • 25) 0.015 826 874 368 × 2 = 0 + 0.031 653 748 736;
  • 26) 0.031 653 748 736 × 2 = 0 + 0.063 307 497 472;
  • 27) 0.063 307 497 472 × 2 = 0 + 0.126 614 994 944;
  • 28) 0.126 614 994 944 × 2 = 0 + 0.253 229 989 888;
  • 29) 0.253 229 989 888 × 2 = 0 + 0.506 459 979 776;
  • 30) 0.506 459 979 776 × 2 = 1 + 0.012 919 959 552;
  • 31) 0.012 919 959 552 × 2 = 0 + 0.025 839 919 104;
  • 32) 0.025 839 919 104 × 2 = 0 + 0.051 679 838 208;
  • 33) 0.051 679 838 208 × 2 = 0 + 0.103 359 676 416;
  • 34) 0.103 359 676 416 × 2 = 0 + 0.206 719 352 832;
  • 35) 0.206 719 352 832 × 2 = 0 + 0.413 438 705 664;
  • 36) 0.413 438 705 664 × 2 = 0 + 0.826 877 411 328;
  • 37) 0.826 877 411 328 × 2 = 1 + 0.653 754 822 656;
  • 38) 0.653 754 822 656 × 2 = 1 + 0.307 509 645 312;
  • 39) 0.307 509 645 312 × 2 = 0 + 0.615 019 290 624;
  • 40) 0.615 019 290 624 × 2 = 1 + 0.230 038 581 248;
  • 41) 0.230 038 581 248 × 2 = 0 + 0.460 077 162 496;
  • 42) 0.460 077 162 496 × 2 = 0 + 0.920 154 324 992;
  • 43) 0.920 154 324 992 × 2 = 1 + 0.840 308 649 984;
  • 44) 0.840 308 649 984 × 2 = 1 + 0.680 617 299 968;
  • 45) 0.680 617 299 968 × 2 = 1 + 0.361 234 599 936;
  • 46) 0.361 234 599 936 × 2 = 0 + 0.722 469 199 872;
  • 47) 0.722 469 199 872 × 2 = 1 + 0.444 938 399 744;

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:


0.000 000 060 548(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 1101 0011 101(2)

5. Positive number before normalization:

0.000 000 060 548(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 1101 0011 101(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:


0.000 000 060 548(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 1101 0011 101(2) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 1101 0011 101(2) × 20 =

1.0000 0100 0000 1101 0011 101(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 0100 0000 1101 0011 101


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 0010 0000 0110 1001 1101 =

000 0010 0000 0110 1001 1101


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 0010 0000 0110 1001 1101


Decimal number 0.000 000 060 548 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 000 0010 0000 0110 1001 1101

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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