Base ten decimal number 0.000 000 000 002 345 converted to 32 bit single precision IEEE 754 binary floating point standard

How to convert the decimal number 0.000 000 000 002 345(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: 0. Divide the number repeatedly by 2, keeping track of each remainder, until 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, by taking 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: 0.000 000 000 002 345. 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) 0.000 000 000 002 345 × 2 = 0 + 0.000 000 000 004 69;
  • 2) 0.000 000 000 004 69 × 2 = 0 + 0.000 000 000 009 38;
  • 3) 0.000 000 000 009 38 × 2 = 0 + 0.000 000 000 018 76;
  • 4) 0.000 000 000 018 76 × 2 = 0 + 0.000 000 000 037 52;
  • 5) 0.000 000 000 037 52 × 2 = 0 + 0.000 000 000 075 04;
  • 6) 0.000 000 000 075 04 × 2 = 0 + 0.000 000 000 150 08;
  • 7) 0.000 000 000 150 08 × 2 = 0 + 0.000 000 000 300 16;
  • 8) 0.000 000 000 300 16 × 2 = 0 + 0.000 000 000 600 32;
  • 9) 0.000 000 000 600 32 × 2 = 0 + 0.000 000 001 200 64;
  • 10) 0.000 000 001 200 64 × 2 = 0 + 0.000 000 002 401 28;
  • 11) 0.000 000 002 401 28 × 2 = 0 + 0.000 000 004 802 56;
  • 12) 0.000 000 004 802 56 × 2 = 0 + 0.000 000 009 605 12;
  • 13) 0.000 000 009 605 12 × 2 = 0 + 0.000 000 019 210 24;
  • 14) 0.000 000 019 210 24 × 2 = 0 + 0.000 000 038 420 48;
  • 15) 0.000 000 038 420 48 × 2 = 0 + 0.000 000 076 840 96;
  • 16) 0.000 000 076 840 96 × 2 = 0 + 0.000 000 153 681 92;
  • 17) 0.000 000 153 681 92 × 2 = 0 + 0.000 000 307 363 84;
  • 18) 0.000 000 307 363 84 × 2 = 0 + 0.000 000 614 727 68;
  • 19) 0.000 000 614 727 68 × 2 = 0 + 0.000 001 229 455 36;
  • 20) 0.000 001 229 455 36 × 2 = 0 + 0.000 002 458 910 72;
  • 21) 0.000 002 458 910 72 × 2 = 0 + 0.000 004 917 821 44;
  • 22) 0.000 004 917 821 44 × 2 = 0 + 0.000 009 835 642 88;
  • 23) 0.000 009 835 642 88 × 2 = 0 + 0.000 019 671 285 76;
  • 24) 0.000 019 671 285 76 × 2 = 0 + 0.000 039 342 571 52;
  • 25) 0.000 039 342 571 52 × 2 = 0 + 0.000 078 685 143 04;
  • 26) 0.000 078 685 143 04 × 2 = 0 + 0.000 157 370 286 08;
  • 27) 0.000 157 370 286 08 × 2 = 0 + 0.000 314 740 572 16;
  • 28) 0.000 314 740 572 16 × 2 = 0 + 0.000 629 481 144 32;
  • 29) 0.000 629 481 144 32 × 2 = 0 + 0.001 258 962 288 64;
  • 30) 0.001 258 962 288 64 × 2 = 0 + 0.002 517 924 577 28;
  • 31) 0.002 517 924 577 28 × 2 = 0 + 0.005 035 849 154 56;
  • 32) 0.005 035 849 154 56 × 2 = 0 + 0.010 071 698 309 12;
  • 33) 0.010 071 698 309 12 × 2 = 0 + 0.020 143 396 618 24;
  • 34) 0.020 143 396 618 24 × 2 = 0 + 0.040 286 793 236 48;
  • 35) 0.040 286 793 236 48 × 2 = 0 + 0.080 573 586 472 96;
  • 36) 0.080 573 586 472 96 × 2 = 0 + 0.161 147 172 945 92;
  • 37) 0.161 147 172 945 92 × 2 = 0 + 0.322 294 345 891 84;
  • 38) 0.322 294 345 891 84 × 2 = 0 + 0.644 588 691 783 68;
  • 39) 0.644 588 691 783 68 × 2 = 1 + 0.289 177 383 567 36;
  • 40) 0.289 177 383 567 36 × 2 = 0 + 0.578 354 767 134 72;
  • 41) 0.578 354 767 134 72 × 2 = 1 + 0.156 709 534 269 44;
  • 42) 0.156 709 534 269 44 × 2 = 0 + 0.313 419 068 538 88;
  • 43) 0.313 419 068 538 88 × 2 = 0 + 0.626 838 137 077 76;
  • 44) 0.626 838 137 077 76 × 2 = 1 + 0.253 676 274 155 52;
  • 45) 0.253 676 274 155 52 × 2 = 0 + 0.507 352 548 311 04;
  • 46) 0.507 352 548 311 04 × 2 = 1 + 0.014 705 096 622 08;
  • 47) 0.014 705 096 622 08 × 2 = 0 + 0.029 410 193 244 16;
  • 48) 0.029 410 193 244 16 × 2 = 0 + 0.058 820 386 488 32;
  • 49) 0.058 820 386 488 32 × 2 = 0 + 0.117 640 772 976 64;
  • 50) 0.117 640 772 976 64 × 2 = 0 + 0.235 281 545 953 28;
  • 51) 0.235 281 545 953 28 × 2 = 0 + 0.470 563 091 906 56;
  • 52) 0.470 563 091 906 56 × 2 = 0 + 0.941 126 183 813 12;
  • 53) 0.941 126 183 813 12 × 2 = 1 + 0.882 252 367 626 24;
  • 54) 0.882 252 367 626 24 × 2 = 1 + 0.764 504 735 252 48;
  • 55) 0.764 504 735 252 48 × 2 = 1 + 0.529 009 470 504 96;
  • 56) 0.529 009 470 504 96 × 2 = 1 + 0.058 018 941 009 92;
  • 57) 0.058 018 941 009 92 × 2 = 0 + 0.116 037 882 019 84;
  • 58) 0.116 037 882 019 84 × 2 = 0 + 0.232 075 764 039 68;
  • 59) 0.232 075 764 039 68 × 2 = 0 + 0.464 151 528 079 36;
  • 60) 0.464 151 528 079 36 × 2 = 0 + 0.928 303 056 158 72;
  • 61) 0.928 303 056 158 72 × 2 = 1 + 0.856 606 112 317 44;
  • 62) 0.856 606 112 317 44 × 2 = 1 + 0.713 212 224 634 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...)

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:

0.000 000 000 002 345(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1001 0100 0000 1111 0000 11(2)

Positive number before normalization:

0.000 000 000 002 345(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1001 0100 0000 1111 0000 11(2)

5. Normalize the binary representation of the number, shifting the decimal mark 39 positions to the right so that only one non zero digit remains to the left of it:

0.000 000 000 002 345(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1001 0100 0000 1111 0000 11(2) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1001 0100 0000 1111 0000 11(2) × 20 =


1.0100 1010 0000 0111 1000 011(2) × 2-39

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): -39


Mantissa (not normalized): 1.0100 1010 0000 0111 1000 011

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 =


-39 + 2(8-1) - 1 =


(-39 + 127)(10) =


88(10)


  • division = quotient + remainder;
  • 88 ÷ 2 = 44 + 0;
  • 44 ÷ 2 = 22 + 0;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


88(10) =


0101 1000(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, only if necessary (not the case here):

Mantissa (normalized) =


1. 010 0101 0000 0011 1100 0011 =


010 0101 0000 0011 1100 0011

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) =
0101 1000


Mantissa (23 bits) =
010 0101 0000 0011 1100 0011

Number 0.000 000 000 002 345, a decimal, converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:


0 - 0101 1000 - 010 0101 0000 0011 1100 0011

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 1

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 1

      1
    • 1

      0

Convert decimal numbers from base ten 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), 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