-0.000 000 000 742 205 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 205(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
-0.000 000 000 742 205(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. Start with the positive version of the number:

|-0.000 000 000 742 205| = 0.000 000 000 742 205


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 205.

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) 0.000 000 000 742 205 × 2 = 0 + 0.000 000 001 484 41;
  • 2) 0.000 000 001 484 41 × 2 = 0 + 0.000 000 002 968 82;
  • 3) 0.000 000 002 968 82 × 2 = 0 + 0.000 000 005 937 64;
  • 4) 0.000 000 005 937 64 × 2 = 0 + 0.000 000 011 875 28;
  • 5) 0.000 000 011 875 28 × 2 = 0 + 0.000 000 023 750 56;
  • 6) 0.000 000 023 750 56 × 2 = 0 + 0.000 000 047 501 12;
  • 7) 0.000 000 047 501 12 × 2 = 0 + 0.000 000 095 002 24;
  • 8) 0.000 000 095 002 24 × 2 = 0 + 0.000 000 190 004 48;
  • 9) 0.000 000 190 004 48 × 2 = 0 + 0.000 000 380 008 96;
  • 10) 0.000 000 380 008 96 × 2 = 0 + 0.000 000 760 017 92;
  • 11) 0.000 000 760 017 92 × 2 = 0 + 0.000 001 520 035 84;
  • 12) 0.000 001 520 035 84 × 2 = 0 + 0.000 003 040 071 68;
  • 13) 0.000 003 040 071 68 × 2 = 0 + 0.000 006 080 143 36;
  • 14) 0.000 006 080 143 36 × 2 = 0 + 0.000 012 160 286 72;
  • 15) 0.000 012 160 286 72 × 2 = 0 + 0.000 024 320 573 44;
  • 16) 0.000 024 320 573 44 × 2 = 0 + 0.000 048 641 146 88;
  • 17) 0.000 048 641 146 88 × 2 = 0 + 0.000 097 282 293 76;
  • 18) 0.000 097 282 293 76 × 2 = 0 + 0.000 194 564 587 52;
  • 19) 0.000 194 564 587 52 × 2 = 0 + 0.000 389 129 175 04;
  • 20) 0.000 389 129 175 04 × 2 = 0 + 0.000 778 258 350 08;
  • 21) 0.000 778 258 350 08 × 2 = 0 + 0.001 556 516 700 16;
  • 22) 0.001 556 516 700 16 × 2 = 0 + 0.003 113 033 400 32;
  • 23) 0.003 113 033 400 32 × 2 = 0 + 0.006 226 066 800 64;
  • 24) 0.006 226 066 800 64 × 2 = 0 + 0.012 452 133 601 28;
  • 25) 0.012 452 133 601 28 × 2 = 0 + 0.024 904 267 202 56;
  • 26) 0.024 904 267 202 56 × 2 = 0 + 0.049 808 534 405 12;
  • 27) 0.049 808 534 405 12 × 2 = 0 + 0.099 617 068 810 24;
  • 28) 0.099 617 068 810 24 × 2 = 0 + 0.199 234 137 620 48;
  • 29) 0.199 234 137 620 48 × 2 = 0 + 0.398 468 275 240 96;
  • 30) 0.398 468 275 240 96 × 2 = 0 + 0.796 936 550 481 92;
  • 31) 0.796 936 550 481 92 × 2 = 1 + 0.593 873 100 963 84;
  • 32) 0.593 873 100 963 84 × 2 = 1 + 0.187 746 201 927 68;
  • 33) 0.187 746 201 927 68 × 2 = 0 + 0.375 492 403 855 36;
  • 34) 0.375 492 403 855 36 × 2 = 0 + 0.750 984 807 710 72;
  • 35) 0.750 984 807 710 72 × 2 = 1 + 0.501 969 615 421 44;
  • 36) 0.501 969 615 421 44 × 2 = 1 + 0.003 939 230 842 88;
  • 37) 0.003 939 230 842 88 × 2 = 0 + 0.007 878 461 685 76;
  • 38) 0.007 878 461 685 76 × 2 = 0 + 0.015 756 923 371 52;
  • 39) 0.015 756 923 371 52 × 2 = 0 + 0.031 513 846 743 04;
  • 40) 0.031 513 846 743 04 × 2 = 0 + 0.063 027 693 486 08;
  • 41) 0.063 027 693 486 08 × 2 = 0 + 0.126 055 386 972 16;
  • 42) 0.126 055 386 972 16 × 2 = 0 + 0.252 110 773 944 32;
  • 43) 0.252 110 773 944 32 × 2 = 0 + 0.504 221 547 888 64;
  • 44) 0.504 221 547 888 64 × 2 = 1 + 0.008 443 095 777 28;
  • 45) 0.008 443 095 777 28 × 2 = 0 + 0.016 886 191 554 56;
  • 46) 0.016 886 191 554 56 × 2 = 0 + 0.033 772 383 109 12;
  • 47) 0.033 772 383 109 12 × 2 = 0 + 0.067 544 766 218 24;
  • 48) 0.067 544 766 218 24 × 2 = 0 + 0.135 089 532 436 48;
  • 49) 0.135 089 532 436 48 × 2 = 0 + 0.270 179 064 872 96;
  • 50) 0.270 179 064 872 96 × 2 = 0 + 0.540 358 129 745 92;
  • 51) 0.540 358 129 745 92 × 2 = 1 + 0.080 716 259 491 84;
  • 52) 0.080 716 259 491 84 × 2 = 0 + 0.161 432 518 983 68;
  • 53) 0.161 432 518 983 68 × 2 = 0 + 0.322 865 037 967 36;
  • 54) 0.322 865 037 967 36 × 2 = 0 + 0.645 730 075 934 72;

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).


5. 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:


0.000 000 000 742 205(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0000 0010 00(2)

6. Positive number before normalization:

0.000 000 000 742 205(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0000 0010 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 205(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0000 0010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0000 0010 00(2) × 20 =

1.1001 1000 0000 1000 0001 000(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 1000 0000 1000 0001 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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. 100 1100 0000 0100 0000 1000 =

100 1100 0000 0100 0000 1000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1100 0000 0100 0000 1000


Decimal number -0.000 000 000 742 205 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0100 0000 1000

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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