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

Convert decimal -0.000 000 000 742 32(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 32(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 32| = 0.000 000 000 742 32


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 32.

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 32 × 2 = 0 + 0.000 000 001 484 64;
  • 2) 0.000 000 001 484 64 × 2 = 0 + 0.000 000 002 969 28;
  • 3) 0.000 000 002 969 28 × 2 = 0 + 0.000 000 005 938 56;
  • 4) 0.000 000 005 938 56 × 2 = 0 + 0.000 000 011 877 12;
  • 5) 0.000 000 011 877 12 × 2 = 0 + 0.000 000 023 754 24;
  • 6) 0.000 000 023 754 24 × 2 = 0 + 0.000 000 047 508 48;
  • 7) 0.000 000 047 508 48 × 2 = 0 + 0.000 000 095 016 96;
  • 8) 0.000 000 095 016 96 × 2 = 0 + 0.000 000 190 033 92;
  • 9) 0.000 000 190 033 92 × 2 = 0 + 0.000 000 380 067 84;
  • 10) 0.000 000 380 067 84 × 2 = 0 + 0.000 000 760 135 68;
  • 11) 0.000 000 760 135 68 × 2 = 0 + 0.000 001 520 271 36;
  • 12) 0.000 001 520 271 36 × 2 = 0 + 0.000 003 040 542 72;
  • 13) 0.000 003 040 542 72 × 2 = 0 + 0.000 006 081 085 44;
  • 14) 0.000 006 081 085 44 × 2 = 0 + 0.000 012 162 170 88;
  • 15) 0.000 012 162 170 88 × 2 = 0 + 0.000 024 324 341 76;
  • 16) 0.000 024 324 341 76 × 2 = 0 + 0.000 048 648 683 52;
  • 17) 0.000 048 648 683 52 × 2 = 0 + 0.000 097 297 367 04;
  • 18) 0.000 097 297 367 04 × 2 = 0 + 0.000 194 594 734 08;
  • 19) 0.000 194 594 734 08 × 2 = 0 + 0.000 389 189 468 16;
  • 20) 0.000 389 189 468 16 × 2 = 0 + 0.000 778 378 936 32;
  • 21) 0.000 778 378 936 32 × 2 = 0 + 0.001 556 757 872 64;
  • 22) 0.001 556 757 872 64 × 2 = 0 + 0.003 113 515 745 28;
  • 23) 0.003 113 515 745 28 × 2 = 0 + 0.006 227 031 490 56;
  • 24) 0.006 227 031 490 56 × 2 = 0 + 0.012 454 062 981 12;
  • 25) 0.012 454 062 981 12 × 2 = 0 + 0.024 908 125 962 24;
  • 26) 0.024 908 125 962 24 × 2 = 0 + 0.049 816 251 924 48;
  • 27) 0.049 816 251 924 48 × 2 = 0 + 0.099 632 503 848 96;
  • 28) 0.099 632 503 848 96 × 2 = 0 + 0.199 265 007 697 92;
  • 29) 0.199 265 007 697 92 × 2 = 0 + 0.398 530 015 395 84;
  • 30) 0.398 530 015 395 84 × 2 = 0 + 0.797 060 030 791 68;
  • 31) 0.797 060 030 791 68 × 2 = 1 + 0.594 120 061 583 36;
  • 32) 0.594 120 061 583 36 × 2 = 1 + 0.188 240 123 166 72;
  • 33) 0.188 240 123 166 72 × 2 = 0 + 0.376 480 246 333 44;
  • 34) 0.376 480 246 333 44 × 2 = 0 + 0.752 960 492 666 88;
  • 35) 0.752 960 492 666 88 × 2 = 1 + 0.505 920 985 333 76;
  • 36) 0.505 920 985 333 76 × 2 = 1 + 0.011 841 970 667 52;
  • 37) 0.011 841 970 667 52 × 2 = 0 + 0.023 683 941 335 04;
  • 38) 0.023 683 941 335 04 × 2 = 0 + 0.047 367 882 670 08;
  • 39) 0.047 367 882 670 08 × 2 = 0 + 0.094 735 765 340 16;
  • 40) 0.094 735 765 340 16 × 2 = 0 + 0.189 471 530 680 32;
  • 41) 0.189 471 530 680 32 × 2 = 0 + 0.378 943 061 360 64;
  • 42) 0.378 943 061 360 64 × 2 = 0 + 0.757 886 122 721 28;
  • 43) 0.757 886 122 721 28 × 2 = 1 + 0.515 772 245 442 56;
  • 44) 0.515 772 245 442 56 × 2 = 1 + 0.031 544 490 885 12;
  • 45) 0.031 544 490 885 12 × 2 = 0 + 0.063 088 981 770 24;
  • 46) 0.063 088 981 770 24 × 2 = 0 + 0.126 177 963 540 48;
  • 47) 0.126 177 963 540 48 × 2 = 0 + 0.252 355 927 080 96;
  • 48) 0.252 355 927 080 96 × 2 = 0 + 0.504 711 854 161 92;
  • 49) 0.504 711 854 161 92 × 2 = 1 + 0.009 423 708 323 84;
  • 50) 0.009 423 708 323 84 × 2 = 0 + 0.018 847 416 647 68;
  • 51) 0.018 847 416 647 68 × 2 = 0 + 0.037 694 833 295 36;
  • 52) 0.037 694 833 295 36 × 2 = 0 + 0.075 389 666 590 72;
  • 53) 0.075 389 666 590 72 × 2 = 0 + 0.150 779 333 181 44;
  • 54) 0.150 779 333 181 44 × 2 = 0 + 0.301 558 666 362 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 - 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 32(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0011 0000 1000 00(2)

6. Positive number before normalization:

0.000 000 000 742 32(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0011 0000 1000 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 32(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0011 0000 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0011 0000 1000 00(2) × 20 =

1.1001 1000 0001 1000 0100 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 0001 1000 0100 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 1100 0010 0000 =

100 1100 0000 1100 0010 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 1100 0000 1100 0010 0000


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

1 - 0110 0000 - 100 1100 0000 1100 0010 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