0.000 000 110 010 07 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 010 07(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
0.000 000 110 010 07(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: 0.000 000 110 010 07.

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) 0.000 000 110 010 07 × 2 = 0 + 0.000 000 220 020 14;
  • 2) 0.000 000 220 020 14 × 2 = 0 + 0.000 000 440 040 28;
  • 3) 0.000 000 440 040 28 × 2 = 0 + 0.000 000 880 080 56;
  • 4) 0.000 000 880 080 56 × 2 = 0 + 0.000 001 760 161 12;
  • 5) 0.000 001 760 161 12 × 2 = 0 + 0.000 003 520 322 24;
  • 6) 0.000 003 520 322 24 × 2 = 0 + 0.000 007 040 644 48;
  • 7) 0.000 007 040 644 48 × 2 = 0 + 0.000 014 081 288 96;
  • 8) 0.000 014 081 288 96 × 2 = 0 + 0.000 028 162 577 92;
  • 9) 0.000 028 162 577 92 × 2 = 0 + 0.000 056 325 155 84;
  • 10) 0.000 056 325 155 84 × 2 = 0 + 0.000 112 650 311 68;
  • 11) 0.000 112 650 311 68 × 2 = 0 + 0.000 225 300 623 36;
  • 12) 0.000 225 300 623 36 × 2 = 0 + 0.000 450 601 246 72;
  • 13) 0.000 450 601 246 72 × 2 = 0 + 0.000 901 202 493 44;
  • 14) 0.000 901 202 493 44 × 2 = 0 + 0.001 802 404 986 88;
  • 15) 0.001 802 404 986 88 × 2 = 0 + 0.003 604 809 973 76;
  • 16) 0.003 604 809 973 76 × 2 = 0 + 0.007 209 619 947 52;
  • 17) 0.007 209 619 947 52 × 2 = 0 + 0.014 419 239 895 04;
  • 18) 0.014 419 239 895 04 × 2 = 0 + 0.028 838 479 790 08;
  • 19) 0.028 838 479 790 08 × 2 = 0 + 0.057 676 959 580 16;
  • 20) 0.057 676 959 580 16 × 2 = 0 + 0.115 353 919 160 32;
  • 21) 0.115 353 919 160 32 × 2 = 0 + 0.230 707 838 320 64;
  • 22) 0.230 707 838 320 64 × 2 = 0 + 0.461 415 676 641 28;
  • 23) 0.461 415 676 641 28 × 2 = 0 + 0.922 831 353 282 56;
  • 24) 0.922 831 353 282 56 × 2 = 1 + 0.845 662 706 565 12;
  • 25) 0.845 662 706 565 12 × 2 = 1 + 0.691 325 413 130 24;
  • 26) 0.691 325 413 130 24 × 2 = 1 + 0.382 650 826 260 48;
  • 27) 0.382 650 826 260 48 × 2 = 0 + 0.765 301 652 520 96;
  • 28) 0.765 301 652 520 96 × 2 = 1 + 0.530 603 305 041 92;
  • 29) 0.530 603 305 041 92 × 2 = 1 + 0.061 206 610 083 84;
  • 30) 0.061 206 610 083 84 × 2 = 0 + 0.122 413 220 167 68;
  • 31) 0.122 413 220 167 68 × 2 = 0 + 0.244 826 440 335 36;
  • 32) 0.244 826 440 335 36 × 2 = 0 + 0.489 652 880 670 72;
  • 33) 0.489 652 880 670 72 × 2 = 0 + 0.979 305 761 341 44;
  • 34) 0.979 305 761 341 44 × 2 = 1 + 0.958 611 522 682 88;
  • 35) 0.958 611 522 682 88 × 2 = 1 + 0.917 223 045 365 76;
  • 36) 0.917 223 045 365 76 × 2 = 1 + 0.834 446 090 731 52;
  • 37) 0.834 446 090 731 52 × 2 = 1 + 0.668 892 181 463 04;
  • 38) 0.668 892 181 463 04 × 2 = 1 + 0.337 784 362 926 08;
  • 39) 0.337 784 362 926 08 × 2 = 0 + 0.675 568 725 852 16;
  • 40) 0.675 568 725 852 16 × 2 = 1 + 0.351 137 451 704 32;
  • 41) 0.351 137 451 704 32 × 2 = 0 + 0.702 274 903 408 64;
  • 42) 0.702 274 903 408 64 × 2 = 1 + 0.404 549 806 817 28;
  • 43) 0.404 549 806 817 28 × 2 = 0 + 0.809 099 613 634 56;
  • 44) 0.809 099 613 634 56 × 2 = 1 + 0.618 199 227 269 12;
  • 45) 0.618 199 227 269 12 × 2 = 1 + 0.236 398 454 538 24;
  • 46) 0.236 398 454 538 24 × 2 = 0 + 0.472 796 909 076 48;
  • 47) 0.472 796 909 076 48 × 2 = 0 + 0.945 593 818 152 96;

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:


0.000 000 110 010 07(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 0101 100(2)

5. Positive number before normalization:

0.000 000 110 010 07(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 0101 100(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:


0.000 000 110 010 07(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 0101 100(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1101 0101 100(2) × 20 =

1.1101 1000 0111 1101 0101 100(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 0101 100


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 1010 1100 =

110 1100 0011 1110 1010 1100


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 1010 1100


Decimal number 0.000 000 110 010 07 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 1010 1100

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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