-0.000 000 000 740 12 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 740 12(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
-0.000 000 000 740 12(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:

|-0.000 000 000 740 12| = 0.000 000 000 740 12


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: 0.000 000 000 740 12.

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) 0.000 000 000 740 12 × 2 = 0 + 0.000 000 001 480 24;
  • 2) 0.000 000 001 480 24 × 2 = 0 + 0.000 000 002 960 48;
  • 3) 0.000 000 002 960 48 × 2 = 0 + 0.000 000 005 920 96;
  • 4) 0.000 000 005 920 96 × 2 = 0 + 0.000 000 011 841 92;
  • 5) 0.000 000 011 841 92 × 2 = 0 + 0.000 000 023 683 84;
  • 6) 0.000 000 023 683 84 × 2 = 0 + 0.000 000 047 367 68;
  • 7) 0.000 000 047 367 68 × 2 = 0 + 0.000 000 094 735 36;
  • 8) 0.000 000 094 735 36 × 2 = 0 + 0.000 000 189 470 72;
  • 9) 0.000 000 189 470 72 × 2 = 0 + 0.000 000 378 941 44;
  • 10) 0.000 000 378 941 44 × 2 = 0 + 0.000 000 757 882 88;
  • 11) 0.000 000 757 882 88 × 2 = 0 + 0.000 001 515 765 76;
  • 12) 0.000 001 515 765 76 × 2 = 0 + 0.000 003 031 531 52;
  • 13) 0.000 003 031 531 52 × 2 = 0 + 0.000 006 063 063 04;
  • 14) 0.000 006 063 063 04 × 2 = 0 + 0.000 012 126 126 08;
  • 15) 0.000 012 126 126 08 × 2 = 0 + 0.000 024 252 252 16;
  • 16) 0.000 024 252 252 16 × 2 = 0 + 0.000 048 504 504 32;
  • 17) 0.000 048 504 504 32 × 2 = 0 + 0.000 097 009 008 64;
  • 18) 0.000 097 009 008 64 × 2 = 0 + 0.000 194 018 017 28;
  • 19) 0.000 194 018 017 28 × 2 = 0 + 0.000 388 036 034 56;
  • 20) 0.000 388 036 034 56 × 2 = 0 + 0.000 776 072 069 12;
  • 21) 0.000 776 072 069 12 × 2 = 0 + 0.001 552 144 138 24;
  • 22) 0.001 552 144 138 24 × 2 = 0 + 0.003 104 288 276 48;
  • 23) 0.003 104 288 276 48 × 2 = 0 + 0.006 208 576 552 96;
  • 24) 0.006 208 576 552 96 × 2 = 0 + 0.012 417 153 105 92;
  • 25) 0.012 417 153 105 92 × 2 = 0 + 0.024 834 306 211 84;
  • 26) 0.024 834 306 211 84 × 2 = 0 + 0.049 668 612 423 68;
  • 27) 0.049 668 612 423 68 × 2 = 0 + 0.099 337 224 847 36;
  • 28) 0.099 337 224 847 36 × 2 = 0 + 0.198 674 449 694 72;
  • 29) 0.198 674 449 694 72 × 2 = 0 + 0.397 348 899 389 44;
  • 30) 0.397 348 899 389 44 × 2 = 0 + 0.794 697 798 778 88;
  • 31) 0.794 697 798 778 88 × 2 = 1 + 0.589 395 597 557 76;
  • 32) 0.589 395 597 557 76 × 2 = 1 + 0.178 791 195 115 52;
  • 33) 0.178 791 195 115 52 × 2 = 0 + 0.357 582 390 231 04;
  • 34) 0.357 582 390 231 04 × 2 = 0 + 0.715 164 780 462 08;
  • 35) 0.715 164 780 462 08 × 2 = 1 + 0.430 329 560 924 16;
  • 36) 0.430 329 560 924 16 × 2 = 0 + 0.860 659 121 848 32;
  • 37) 0.860 659 121 848 32 × 2 = 1 + 0.721 318 243 696 64;
  • 38) 0.721 318 243 696 64 × 2 = 1 + 0.442 636 487 393 28;
  • 39) 0.442 636 487 393 28 × 2 = 0 + 0.885 272 974 786 56;
  • 40) 0.885 272 974 786 56 × 2 = 1 + 0.770 545 949 573 12;
  • 41) 0.770 545 949 573 12 × 2 = 1 + 0.541 091 899 146 24;
  • 42) 0.541 091 899 146 24 × 2 = 1 + 0.082 183 798 292 48;
  • 43) 0.082 183 798 292 48 × 2 = 0 + 0.164 367 596 584 96;
  • 44) 0.164 367 596 584 96 × 2 = 0 + 0.328 735 193 169 92;
  • 45) 0.328 735 193 169 92 × 2 = 0 + 0.657 470 386 339 84;
  • 46) 0.657 470 386 339 84 × 2 = 1 + 0.314 940 772 679 68;
  • 47) 0.314 940 772 679 68 × 2 = 0 + 0.629 881 545 359 36;
  • 48) 0.629 881 545 359 36 × 2 = 1 + 0.259 763 090 718 72;
  • 49) 0.259 763 090 718 72 × 2 = 0 + 0.519 526 181 437 44;
  • 50) 0.519 526 181 437 44 × 2 = 1 + 0.039 052 362 874 88;
  • 51) 0.039 052 362 874 88 × 2 = 0 + 0.078 104 725 749 76;
  • 52) 0.078 104 725 749 76 × 2 = 0 + 0.156 209 451 499 52;
  • 53) 0.156 209 451 499 52 × 2 = 0 + 0.312 418 902 999 04;
  • 54) 0.312 418 902 999 04 × 2 = 0 + 0.624 837 805 998 08;

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:


0.000 000 000 740 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1100 0101 0100 00(2)

6. Positive number before normalization:

0.000 000 000 740 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1100 0101 0100 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:


0.000 000 000 740 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1100 0101 0100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1100 0101 0100 00(2) × 20 =

1.1001 0110 1110 0010 1010 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 0110 1110 0010 1010 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 1011 0111 0001 0101 0000 =

100 1011 0111 0001 0101 0000


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 1011 0111 0001 0101 0000


Decimal number -0.000 000 000 740 12 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 0111 0001 0101 0000

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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