0.000 000 110 009 79 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 009 79(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 79(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 79.

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 79 × 2 = 0 + 0.000 000 220 019 58;
  • 2) 0.000 000 220 019 58 × 2 = 0 + 0.000 000 440 039 16;
  • 3) 0.000 000 440 039 16 × 2 = 0 + 0.000 000 880 078 32;
  • 4) 0.000 000 880 078 32 × 2 = 0 + 0.000 001 760 156 64;
  • 5) 0.000 001 760 156 64 × 2 = 0 + 0.000 003 520 313 28;
  • 6) 0.000 003 520 313 28 × 2 = 0 + 0.000 007 040 626 56;
  • 7) 0.000 007 040 626 56 × 2 = 0 + 0.000 014 081 253 12;
  • 8) 0.000 014 081 253 12 × 2 = 0 + 0.000 028 162 506 24;
  • 9) 0.000 028 162 506 24 × 2 = 0 + 0.000 056 325 012 48;
  • 10) 0.000 056 325 012 48 × 2 = 0 + 0.000 112 650 024 96;
  • 11) 0.000 112 650 024 96 × 2 = 0 + 0.000 225 300 049 92;
  • 12) 0.000 225 300 049 92 × 2 = 0 + 0.000 450 600 099 84;
  • 13) 0.000 450 600 099 84 × 2 = 0 + 0.000 901 200 199 68;
  • 14) 0.000 901 200 199 68 × 2 = 0 + 0.001 802 400 399 36;
  • 15) 0.001 802 400 399 36 × 2 = 0 + 0.003 604 800 798 72;
  • 16) 0.003 604 800 798 72 × 2 = 0 + 0.007 209 601 597 44;
  • 17) 0.007 209 601 597 44 × 2 = 0 + 0.014 419 203 194 88;
  • 18) 0.014 419 203 194 88 × 2 = 0 + 0.028 838 406 389 76;
  • 19) 0.028 838 406 389 76 × 2 = 0 + 0.057 676 812 779 52;
  • 20) 0.057 676 812 779 52 × 2 = 0 + 0.115 353 625 559 04;
  • 21) 0.115 353 625 559 04 × 2 = 0 + 0.230 707 251 118 08;
  • 22) 0.230 707 251 118 08 × 2 = 0 + 0.461 414 502 236 16;
  • 23) 0.461 414 502 236 16 × 2 = 0 + 0.922 829 004 472 32;
  • 24) 0.922 829 004 472 32 × 2 = 1 + 0.845 658 008 944 64;
  • 25) 0.845 658 008 944 64 × 2 = 1 + 0.691 316 017 889 28;
  • 26) 0.691 316 017 889 28 × 2 = 1 + 0.382 632 035 778 56;
  • 27) 0.382 632 035 778 56 × 2 = 0 + 0.765 264 071 557 12;
  • 28) 0.765 264 071 557 12 × 2 = 1 + 0.530 528 143 114 24;
  • 29) 0.530 528 143 114 24 × 2 = 1 + 0.061 056 286 228 48;
  • 30) 0.061 056 286 228 48 × 2 = 0 + 0.122 112 572 456 96;
  • 31) 0.122 112 572 456 96 × 2 = 0 + 0.244 225 144 913 92;
  • 32) 0.244 225 144 913 92 × 2 = 0 + 0.488 450 289 827 84;
  • 33) 0.488 450 289 827 84 × 2 = 0 + 0.976 900 579 655 68;
  • 34) 0.976 900 579 655 68 × 2 = 1 + 0.953 801 159 311 36;
  • 35) 0.953 801 159 311 36 × 2 = 1 + 0.907 602 318 622 72;
  • 36) 0.907 602 318 622 72 × 2 = 1 + 0.815 204 637 245 44;
  • 37) 0.815 204 637 245 44 × 2 = 1 + 0.630 409 274 490 88;
  • 38) 0.630 409 274 490 88 × 2 = 1 + 0.260 818 548 981 76;
  • 39) 0.260 818 548 981 76 × 2 = 0 + 0.521 637 097 963 52;
  • 40) 0.521 637 097 963 52 × 2 = 1 + 0.043 274 195 927 04;
  • 41) 0.043 274 195 927 04 × 2 = 0 + 0.086 548 391 854 08;
  • 42) 0.086 548 391 854 08 × 2 = 0 + 0.173 096 783 708 16;
  • 43) 0.173 096 783 708 16 × 2 = 0 + 0.346 193 567 416 32;
  • 44) 0.346 193 567 416 32 × 2 = 0 + 0.692 387 134 832 64;
  • 45) 0.692 387 134 832 64 × 2 = 1 + 0.384 774 269 665 28;
  • 46) 0.384 774 269 665 28 × 2 = 0 + 0.769 548 539 330 56;
  • 47) 0.769 548 539 330 56 × 2 = 1 + 0.539 097 078 661 12;

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 79(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 0000 101(2)

5. Positive number before normalization:

0.000 000 110 009 79(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 0000 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:


0.000 000 110 009 79(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 0000 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 0000 101(2) × 20 =

1.1101 1000 0111 1101 0000 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):
1.1101 1000 0111 1101 0000 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. 110 1100 0011 1110 1000 0101 =

110 1100 0011 1110 1000 0101


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 1000 0101


Decimal number 0.000 000 110 009 79 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 1000 0101

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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