-0.000 000 000 742 12 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 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
-0.000 000 000 742 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:

|-0.000 000 000 742 12| = 0.000 000 000 742 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: 0.000 000 000 742 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) 0.000 000 000 742 12 × 2 = 0 + 0.000 000 001 484 24;
  • 2) 0.000 000 001 484 24 × 2 = 0 + 0.000 000 002 968 48;
  • 3) 0.000 000 002 968 48 × 2 = 0 + 0.000 000 005 936 96;
  • 4) 0.000 000 005 936 96 × 2 = 0 + 0.000 000 011 873 92;
  • 5) 0.000 000 011 873 92 × 2 = 0 + 0.000 000 023 747 84;
  • 6) 0.000 000 023 747 84 × 2 = 0 + 0.000 000 047 495 68;
  • 7) 0.000 000 047 495 68 × 2 = 0 + 0.000 000 094 991 36;
  • 8) 0.000 000 094 991 36 × 2 = 0 + 0.000 000 189 982 72;
  • 9) 0.000 000 189 982 72 × 2 = 0 + 0.000 000 379 965 44;
  • 10) 0.000 000 379 965 44 × 2 = 0 + 0.000 000 759 930 88;
  • 11) 0.000 000 759 930 88 × 2 = 0 + 0.000 001 519 861 76;
  • 12) 0.000 001 519 861 76 × 2 = 0 + 0.000 003 039 723 52;
  • 13) 0.000 003 039 723 52 × 2 = 0 + 0.000 006 079 447 04;
  • 14) 0.000 006 079 447 04 × 2 = 0 + 0.000 012 158 894 08;
  • 15) 0.000 012 158 894 08 × 2 = 0 + 0.000 024 317 788 16;
  • 16) 0.000 024 317 788 16 × 2 = 0 + 0.000 048 635 576 32;
  • 17) 0.000 048 635 576 32 × 2 = 0 + 0.000 097 271 152 64;
  • 18) 0.000 097 271 152 64 × 2 = 0 + 0.000 194 542 305 28;
  • 19) 0.000 194 542 305 28 × 2 = 0 + 0.000 389 084 610 56;
  • 20) 0.000 389 084 610 56 × 2 = 0 + 0.000 778 169 221 12;
  • 21) 0.000 778 169 221 12 × 2 = 0 + 0.001 556 338 442 24;
  • 22) 0.001 556 338 442 24 × 2 = 0 + 0.003 112 676 884 48;
  • 23) 0.003 112 676 884 48 × 2 = 0 + 0.006 225 353 768 96;
  • 24) 0.006 225 353 768 96 × 2 = 0 + 0.012 450 707 537 92;
  • 25) 0.012 450 707 537 92 × 2 = 0 + 0.024 901 415 075 84;
  • 26) 0.024 901 415 075 84 × 2 = 0 + 0.049 802 830 151 68;
  • 27) 0.049 802 830 151 68 × 2 = 0 + 0.099 605 660 303 36;
  • 28) 0.099 605 660 303 36 × 2 = 0 + 0.199 211 320 606 72;
  • 29) 0.199 211 320 606 72 × 2 = 0 + 0.398 422 641 213 44;
  • 30) 0.398 422 641 213 44 × 2 = 0 + 0.796 845 282 426 88;
  • 31) 0.796 845 282 426 88 × 2 = 1 + 0.593 690 564 853 76;
  • 32) 0.593 690 564 853 76 × 2 = 1 + 0.187 381 129 707 52;
  • 33) 0.187 381 129 707 52 × 2 = 0 + 0.374 762 259 415 04;
  • 34) 0.374 762 259 415 04 × 2 = 0 + 0.749 524 518 830 08;
  • 35) 0.749 524 518 830 08 × 2 = 1 + 0.499 049 037 660 16;
  • 36) 0.499 049 037 660 16 × 2 = 0 + 0.998 098 075 320 32;
  • 37) 0.998 098 075 320 32 × 2 = 1 + 0.996 196 150 640 64;
  • 38) 0.996 196 150 640 64 × 2 = 1 + 0.992 392 301 281 28;
  • 39) 0.992 392 301 281 28 × 2 = 1 + 0.984 784 602 562 56;
  • 40) 0.984 784 602 562 56 × 2 = 1 + 0.969 569 205 125 12;
  • 41) 0.969 569 205 125 12 × 2 = 1 + 0.939 138 410 250 24;
  • 42) 0.939 138 410 250 24 × 2 = 1 + 0.878 276 820 500 48;
  • 43) 0.878 276 820 500 48 × 2 = 1 + 0.756 553 641 000 96;
  • 44) 0.756 553 641 000 96 × 2 = 1 + 0.513 107 282 001 92;
  • 45) 0.513 107 282 001 92 × 2 = 1 + 0.026 214 564 003 84;
  • 46) 0.026 214 564 003 84 × 2 = 0 + 0.052 429 128 007 68;
  • 47) 0.052 429 128 007 68 × 2 = 0 + 0.104 858 256 015 36;
  • 48) 0.104 858 256 015 36 × 2 = 0 + 0.209 716 512 030 72;
  • 49) 0.209 716 512 030 72 × 2 = 0 + 0.419 433 024 061 44;
  • 50) 0.419 433 024 061 44 × 2 = 0 + 0.838 866 048 122 88;
  • 51) 0.838 866 048 122 88 × 2 = 1 + 0.677 732 096 245 76;
  • 52) 0.677 732 096 245 76 × 2 = 1 + 0.355 464 192 491 52;
  • 53) 0.355 464 192 491 52 × 2 = 0 + 0.710 928 384 983 04;
  • 54) 0.710 928 384 983 04 × 2 = 1 + 0.421 856 769 966 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:


0.000 000 000 742 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0011 01(2)

6. Positive number before normalization:

0.000 000 000 742 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0011 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:


0.000 000 000 742 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0011 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0011 01(2) × 20 =

1.1001 0111 1111 1100 0001 101(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 1111 1100 0001 101


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 1111 1110 0000 1101 =

100 1011 1111 1110 0000 1101


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 1111 1110 0000 1101


Decimal number -0.000 000 000 742 12 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 0000 1101

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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