Convert 1.000 000 119 209 289 550 781 26 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

How to convert the decimal number 1.000 000 119 209 289 550 781 26(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (base 2) the integer part: 1. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

1(10) =


1(2)

3. Convert to binary (base 2) the fractional part: 0.000 000 119 209 289 550 781 26. Multiply it 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) 0.000 000 119 209 289 550 781 26 × 2 = 0 + 0.000 000 238 418 579 101 562 52;
  • 2) 0.000 000 238 418 579 101 562 52 × 2 = 0 + 0.000 000 476 837 158 203 125 04;
  • 3) 0.000 000 476 837 158 203 125 04 × 2 = 0 + 0.000 000 953 674 316 406 250 08;
  • 4) 0.000 000 953 674 316 406 250 08 × 2 = 0 + 0.000 001 907 348 632 812 500 16;
  • 5) 0.000 001 907 348 632 812 500 16 × 2 = 0 + 0.000 003 814 697 265 625 000 32;
  • 6) 0.000 003 814 697 265 625 000 32 × 2 = 0 + 0.000 007 629 394 531 250 000 64;
  • 7) 0.000 007 629 394 531 250 000 64 × 2 = 0 + 0.000 015 258 789 062 500 001 28;
  • 8) 0.000 015 258 789 062 500 001 28 × 2 = 0 + 0.000 030 517 578 125 000 002 56;
  • 9) 0.000 030 517 578 125 000 002 56 × 2 = 0 + 0.000 061 035 156 250 000 005 12;
  • 10) 0.000 061 035 156 250 000 005 12 × 2 = 0 + 0.000 122 070 312 500 000 010 24;
  • 11) 0.000 122 070 312 500 000 010 24 × 2 = 0 + 0.000 244 140 625 000 000 020 48;
  • 12) 0.000 244 140 625 000 000 020 48 × 2 = 0 + 0.000 488 281 250 000 000 040 96;
  • 13) 0.000 488 281 250 000 000 040 96 × 2 = 0 + 0.000 976 562 500 000 000 081 92;
  • 14) 0.000 976 562 500 000 000 081 92 × 2 = 0 + 0.001 953 125 000 000 000 163 84;
  • 15) 0.001 953 125 000 000 000 163 84 × 2 = 0 + 0.003 906 250 000 000 000 327 68;
  • 16) 0.003 906 250 000 000 000 327 68 × 2 = 0 + 0.007 812 500 000 000 000 655 36;
  • 17) 0.007 812 500 000 000 000 655 36 × 2 = 0 + 0.015 625 000 000 000 001 310 72;
  • 18) 0.015 625 000 000 000 001 310 72 × 2 = 0 + 0.031 250 000 000 000 002 621 44;
  • 19) 0.031 250 000 000 000 002 621 44 × 2 = 0 + 0.062 500 000 000 000 005 242 88;
  • 20) 0.062 500 000 000 000 005 242 88 × 2 = 0 + 0.125 000 000 000 000 010 485 76;
  • 21) 0.125 000 000 000 000 010 485 76 × 2 = 0 + 0.250 000 000 000 000 020 971 52;
  • 22) 0.250 000 000 000 000 020 971 52 × 2 = 0 + 0.500 000 000 000 000 041 943 04;
  • 23) 0.500 000 000 000 000 041 943 04 × 2 = 1 + 0.000 000 000 000 000 083 886 08;
  • 24) 0.000 000 000 000 000 083 886 08 × 2 = 0 + 0.000 000 000 000 000 167 772 16;

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...)

4. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.000 000 119 209 289 550 781 26(10) =


0.0000 0000 0000 0000 0000 0010(2)

Positive number before normalization:

1.000 000 119 209 289 550 781 26(10) =


1.0000 0000 0000 0000 0000 0010(2)

5. Normalize the binary representation of the number, shifting the decimal mark 0 positions to the left so that only one non zero digit remains to the left of it:

1.000 000 119 209 289 550 781 26(10) =


1.0000 0000 0000 0000 0000 0010(2) =


1.0000 0000 0000 0000 0000 0010(2) × 20

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): 0


Mantissa (not normalized): 1.0000 0000 0000 0000 0000 0010

6. 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:

Exponent (adjusted) =


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


0 + 2(8-1) - 1 =


(0 + 127)(10) =


127(10)


  • division = quotient + remainder;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


127(10) =


0111 1111(2)

7. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...):

Mantissa (normalized) =


1. 000 0000 0000 0000 0000 0001 0 =


000 0000 0000 0000 0000 0001

Conclusion:

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) =
0111 1111


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

Number 1.000 000 119 209 289 550 781 26 converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:
0 - 0111 1111 - 000 0000 0000 0000 0000 0001

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 1

      23
  • Mantissa (23 bits):

    • 0

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 1

      0

1.000 000 119 209 289 550 781 25 = ? ... 1.000 000 119 209 289 550 781 27 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

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