0.000 000 110 009 21 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 009 21(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
0.000 000 110 009 21(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: 0.000 000 110 009 21.

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) 0.000 000 110 009 21 × 2 = 0 + 0.000 000 220 018 42;
  • 2) 0.000 000 220 018 42 × 2 = 0 + 0.000 000 440 036 84;
  • 3) 0.000 000 440 036 84 × 2 = 0 + 0.000 000 880 073 68;
  • 4) 0.000 000 880 073 68 × 2 = 0 + 0.000 001 760 147 36;
  • 5) 0.000 001 760 147 36 × 2 = 0 + 0.000 003 520 294 72;
  • 6) 0.000 003 520 294 72 × 2 = 0 + 0.000 007 040 589 44;
  • 7) 0.000 007 040 589 44 × 2 = 0 + 0.000 014 081 178 88;
  • 8) 0.000 014 081 178 88 × 2 = 0 + 0.000 028 162 357 76;
  • 9) 0.000 028 162 357 76 × 2 = 0 + 0.000 056 324 715 52;
  • 10) 0.000 056 324 715 52 × 2 = 0 + 0.000 112 649 431 04;
  • 11) 0.000 112 649 431 04 × 2 = 0 + 0.000 225 298 862 08;
  • 12) 0.000 225 298 862 08 × 2 = 0 + 0.000 450 597 724 16;
  • 13) 0.000 450 597 724 16 × 2 = 0 + 0.000 901 195 448 32;
  • 14) 0.000 901 195 448 32 × 2 = 0 + 0.001 802 390 896 64;
  • 15) 0.001 802 390 896 64 × 2 = 0 + 0.003 604 781 793 28;
  • 16) 0.003 604 781 793 28 × 2 = 0 + 0.007 209 563 586 56;
  • 17) 0.007 209 563 586 56 × 2 = 0 + 0.014 419 127 173 12;
  • 18) 0.014 419 127 173 12 × 2 = 0 + 0.028 838 254 346 24;
  • 19) 0.028 838 254 346 24 × 2 = 0 + 0.057 676 508 692 48;
  • 20) 0.057 676 508 692 48 × 2 = 0 + 0.115 353 017 384 96;
  • 21) 0.115 353 017 384 96 × 2 = 0 + 0.230 706 034 769 92;
  • 22) 0.230 706 034 769 92 × 2 = 0 + 0.461 412 069 539 84;
  • 23) 0.461 412 069 539 84 × 2 = 0 + 0.922 824 139 079 68;
  • 24) 0.922 824 139 079 68 × 2 = 1 + 0.845 648 278 159 36;
  • 25) 0.845 648 278 159 36 × 2 = 1 + 0.691 296 556 318 72;
  • 26) 0.691 296 556 318 72 × 2 = 1 + 0.382 593 112 637 44;
  • 27) 0.382 593 112 637 44 × 2 = 0 + 0.765 186 225 274 88;
  • 28) 0.765 186 225 274 88 × 2 = 1 + 0.530 372 450 549 76;
  • 29) 0.530 372 450 549 76 × 2 = 1 + 0.060 744 901 099 52;
  • 30) 0.060 744 901 099 52 × 2 = 0 + 0.121 489 802 199 04;
  • 31) 0.121 489 802 199 04 × 2 = 0 + 0.242 979 604 398 08;
  • 32) 0.242 979 604 398 08 × 2 = 0 + 0.485 959 208 796 16;
  • 33) 0.485 959 208 796 16 × 2 = 0 + 0.971 918 417 592 32;
  • 34) 0.971 918 417 592 32 × 2 = 1 + 0.943 836 835 184 64;
  • 35) 0.943 836 835 184 64 × 2 = 1 + 0.887 673 670 369 28;
  • 36) 0.887 673 670 369 28 × 2 = 1 + 0.775 347 340 738 56;
  • 37) 0.775 347 340 738 56 × 2 = 1 + 0.550 694 681 477 12;
  • 38) 0.550 694 681 477 12 × 2 = 1 + 0.101 389 362 954 24;
  • 39) 0.101 389 362 954 24 × 2 = 0 + 0.202 778 725 908 48;
  • 40) 0.202 778 725 908 48 × 2 = 0 + 0.405 557 451 816 96;
  • 41) 0.405 557 451 816 96 × 2 = 0 + 0.811 114 903 633 92;
  • 42) 0.811 114 903 633 92 × 2 = 1 + 0.622 229 807 267 84;
  • 43) 0.622 229 807 267 84 × 2 = 1 + 0.244 459 614 535 68;
  • 44) 0.244 459 614 535 68 × 2 = 0 + 0.488 919 229 071 36;
  • 45) 0.488 919 229 071 36 × 2 = 0 + 0.977 838 458 142 72;
  • 46) 0.977 838 458 142 72 × 2 = 1 + 0.955 676 916 285 44;
  • 47) 0.955 676 916 285 44 × 2 = 1 + 0.911 353 832 570 88;

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:


0.000 000 110 009 21(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0110 011(2)

5. Positive number before normalization:

0.000 000 110 009 21(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0110 011(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:


0.000 000 110 009 21(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0110 011(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0110 011(2) × 20 =

1.1101 1000 0111 1100 0110 011(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):
1.1101 1000 0111 1100 0110 011


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. 110 1100 0011 1110 0011 0011 =

110 1100 0011 1110 0011 0011


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) =
110 1100 0011 1110 0011 0011


Decimal number 0.000 000 110 009 21 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 0011 0011

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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