-0.000 000 000 741 01 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 01(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
-0.000 000 000 741 01(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:

|-0.000 000 000 741 01| = 0.000 000 000 741 01


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: 0.000 000 000 741 01.

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) 0.000 000 000 741 01 × 2 = 0 + 0.000 000 001 482 02;
  • 2) 0.000 000 001 482 02 × 2 = 0 + 0.000 000 002 964 04;
  • 3) 0.000 000 002 964 04 × 2 = 0 + 0.000 000 005 928 08;
  • 4) 0.000 000 005 928 08 × 2 = 0 + 0.000 000 011 856 16;
  • 5) 0.000 000 011 856 16 × 2 = 0 + 0.000 000 023 712 32;
  • 6) 0.000 000 023 712 32 × 2 = 0 + 0.000 000 047 424 64;
  • 7) 0.000 000 047 424 64 × 2 = 0 + 0.000 000 094 849 28;
  • 8) 0.000 000 094 849 28 × 2 = 0 + 0.000 000 189 698 56;
  • 9) 0.000 000 189 698 56 × 2 = 0 + 0.000 000 379 397 12;
  • 10) 0.000 000 379 397 12 × 2 = 0 + 0.000 000 758 794 24;
  • 11) 0.000 000 758 794 24 × 2 = 0 + 0.000 001 517 588 48;
  • 12) 0.000 001 517 588 48 × 2 = 0 + 0.000 003 035 176 96;
  • 13) 0.000 003 035 176 96 × 2 = 0 + 0.000 006 070 353 92;
  • 14) 0.000 006 070 353 92 × 2 = 0 + 0.000 012 140 707 84;
  • 15) 0.000 012 140 707 84 × 2 = 0 + 0.000 024 281 415 68;
  • 16) 0.000 024 281 415 68 × 2 = 0 + 0.000 048 562 831 36;
  • 17) 0.000 048 562 831 36 × 2 = 0 + 0.000 097 125 662 72;
  • 18) 0.000 097 125 662 72 × 2 = 0 + 0.000 194 251 325 44;
  • 19) 0.000 194 251 325 44 × 2 = 0 + 0.000 388 502 650 88;
  • 20) 0.000 388 502 650 88 × 2 = 0 + 0.000 777 005 301 76;
  • 21) 0.000 777 005 301 76 × 2 = 0 + 0.001 554 010 603 52;
  • 22) 0.001 554 010 603 52 × 2 = 0 + 0.003 108 021 207 04;
  • 23) 0.003 108 021 207 04 × 2 = 0 + 0.006 216 042 414 08;
  • 24) 0.006 216 042 414 08 × 2 = 0 + 0.012 432 084 828 16;
  • 25) 0.012 432 084 828 16 × 2 = 0 + 0.024 864 169 656 32;
  • 26) 0.024 864 169 656 32 × 2 = 0 + 0.049 728 339 312 64;
  • 27) 0.049 728 339 312 64 × 2 = 0 + 0.099 456 678 625 28;
  • 28) 0.099 456 678 625 28 × 2 = 0 + 0.198 913 357 250 56;
  • 29) 0.198 913 357 250 56 × 2 = 0 + 0.397 826 714 501 12;
  • 30) 0.397 826 714 501 12 × 2 = 0 + 0.795 653 429 002 24;
  • 31) 0.795 653 429 002 24 × 2 = 1 + 0.591 306 858 004 48;
  • 32) 0.591 306 858 004 48 × 2 = 1 + 0.182 613 716 008 96;
  • 33) 0.182 613 716 008 96 × 2 = 0 + 0.365 227 432 017 92;
  • 34) 0.365 227 432 017 92 × 2 = 0 + 0.730 454 864 035 84;
  • 35) 0.730 454 864 035 84 × 2 = 1 + 0.460 909 728 071 68;
  • 36) 0.460 909 728 071 68 × 2 = 0 + 0.921 819 456 143 36;
  • 37) 0.921 819 456 143 36 × 2 = 1 + 0.843 638 912 286 72;
  • 38) 0.843 638 912 286 72 × 2 = 1 + 0.687 277 824 573 44;
  • 39) 0.687 277 824 573 44 × 2 = 1 + 0.374 555 649 146 88;
  • 40) 0.374 555 649 146 88 × 2 = 0 + 0.749 111 298 293 76;
  • 41) 0.749 111 298 293 76 × 2 = 1 + 0.498 222 596 587 52;
  • 42) 0.498 222 596 587 52 × 2 = 0 + 0.996 445 193 175 04;
  • 43) 0.996 445 193 175 04 × 2 = 1 + 0.992 890 386 350 08;
  • 44) 0.992 890 386 350 08 × 2 = 1 + 0.985 780 772 700 16;
  • 45) 0.985 780 772 700 16 × 2 = 1 + 0.971 561 545 400 32;
  • 46) 0.971 561 545 400 32 × 2 = 1 + 0.943 123 090 800 64;
  • 47) 0.943 123 090 800 64 × 2 = 1 + 0.886 246 181 601 28;
  • 48) 0.886 246 181 601 28 × 2 = 1 + 0.772 492 363 202 56;
  • 49) 0.772 492 363 202 56 × 2 = 1 + 0.544 984 726 405 12;
  • 50) 0.544 984 726 405 12 × 2 = 1 + 0.089 969 452 810 24;
  • 51) 0.089 969 452 810 24 × 2 = 0 + 0.179 938 905 620 48;
  • 52) 0.179 938 905 620 48 × 2 = 0 + 0.359 877 811 240 96;
  • 53) 0.359 877 811 240 96 × 2 = 0 + 0.719 755 622 481 92;
  • 54) 0.719 755 622 481 92 × 2 = 1 + 0.439 511 244 963 84;

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:


0.000 000 000 741 01(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1011 1111 1100 01(2)

6. Positive number before normalization:

0.000 000 000 741 01(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1011 1111 1100 01(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:


0.000 000 000 741 01(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1011 1111 1100 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1011 1111 1100 01(2) × 20 =

1.1001 0111 0101 1111 1110 001(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 0111 0101 1111 1110 001


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 1010 1111 1111 0001 =

100 1011 1010 1111 1111 0001


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 1010 1111 1111 0001


Decimal number -0.000 000 000 741 01 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1010 1111 1111 0001

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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