-0.000 000 000 744 68 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 744 68(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
-0.000 000 000 744 68(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:

|-0.000 000 000 744 68| = 0.000 000 000 744 68


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: 0.000 000 000 744 68.

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) 0.000 000 000 744 68 × 2 = 0 + 0.000 000 001 489 36;
  • 2) 0.000 000 001 489 36 × 2 = 0 + 0.000 000 002 978 72;
  • 3) 0.000 000 002 978 72 × 2 = 0 + 0.000 000 005 957 44;
  • 4) 0.000 000 005 957 44 × 2 = 0 + 0.000 000 011 914 88;
  • 5) 0.000 000 011 914 88 × 2 = 0 + 0.000 000 023 829 76;
  • 6) 0.000 000 023 829 76 × 2 = 0 + 0.000 000 047 659 52;
  • 7) 0.000 000 047 659 52 × 2 = 0 + 0.000 000 095 319 04;
  • 8) 0.000 000 095 319 04 × 2 = 0 + 0.000 000 190 638 08;
  • 9) 0.000 000 190 638 08 × 2 = 0 + 0.000 000 381 276 16;
  • 10) 0.000 000 381 276 16 × 2 = 0 + 0.000 000 762 552 32;
  • 11) 0.000 000 762 552 32 × 2 = 0 + 0.000 001 525 104 64;
  • 12) 0.000 001 525 104 64 × 2 = 0 + 0.000 003 050 209 28;
  • 13) 0.000 003 050 209 28 × 2 = 0 + 0.000 006 100 418 56;
  • 14) 0.000 006 100 418 56 × 2 = 0 + 0.000 012 200 837 12;
  • 15) 0.000 012 200 837 12 × 2 = 0 + 0.000 024 401 674 24;
  • 16) 0.000 024 401 674 24 × 2 = 0 + 0.000 048 803 348 48;
  • 17) 0.000 048 803 348 48 × 2 = 0 + 0.000 097 606 696 96;
  • 18) 0.000 097 606 696 96 × 2 = 0 + 0.000 195 213 393 92;
  • 19) 0.000 195 213 393 92 × 2 = 0 + 0.000 390 426 787 84;
  • 20) 0.000 390 426 787 84 × 2 = 0 + 0.000 780 853 575 68;
  • 21) 0.000 780 853 575 68 × 2 = 0 + 0.001 561 707 151 36;
  • 22) 0.001 561 707 151 36 × 2 = 0 + 0.003 123 414 302 72;
  • 23) 0.003 123 414 302 72 × 2 = 0 + 0.006 246 828 605 44;
  • 24) 0.006 246 828 605 44 × 2 = 0 + 0.012 493 657 210 88;
  • 25) 0.012 493 657 210 88 × 2 = 0 + 0.024 987 314 421 76;
  • 26) 0.024 987 314 421 76 × 2 = 0 + 0.049 974 628 843 52;
  • 27) 0.049 974 628 843 52 × 2 = 0 + 0.099 949 257 687 04;
  • 28) 0.099 949 257 687 04 × 2 = 0 + 0.199 898 515 374 08;
  • 29) 0.199 898 515 374 08 × 2 = 0 + 0.399 797 030 748 16;
  • 30) 0.399 797 030 748 16 × 2 = 0 + 0.799 594 061 496 32;
  • 31) 0.799 594 061 496 32 × 2 = 1 + 0.599 188 122 992 64;
  • 32) 0.599 188 122 992 64 × 2 = 1 + 0.198 376 245 985 28;
  • 33) 0.198 376 245 985 28 × 2 = 0 + 0.396 752 491 970 56;
  • 34) 0.396 752 491 970 56 × 2 = 0 + 0.793 504 983 941 12;
  • 35) 0.793 504 983 941 12 × 2 = 1 + 0.587 009 967 882 24;
  • 36) 0.587 009 967 882 24 × 2 = 1 + 0.174 019 935 764 48;
  • 37) 0.174 019 935 764 48 × 2 = 0 + 0.348 039 871 528 96;
  • 38) 0.348 039 871 528 96 × 2 = 0 + 0.696 079 743 057 92;
  • 39) 0.696 079 743 057 92 × 2 = 1 + 0.392 159 486 115 84;
  • 40) 0.392 159 486 115 84 × 2 = 0 + 0.784 318 972 231 68;
  • 41) 0.784 318 972 231 68 × 2 = 1 + 0.568 637 944 463 36;
  • 42) 0.568 637 944 463 36 × 2 = 1 + 0.137 275 888 926 72;
  • 43) 0.137 275 888 926 72 × 2 = 0 + 0.274 551 777 853 44;
  • 44) 0.274 551 777 853 44 × 2 = 0 + 0.549 103 555 706 88;
  • 45) 0.549 103 555 706 88 × 2 = 1 + 0.098 207 111 413 76;
  • 46) 0.098 207 111 413 76 × 2 = 0 + 0.196 414 222 827 52;
  • 47) 0.196 414 222 827 52 × 2 = 0 + 0.392 828 445 655 04;
  • 48) 0.392 828 445 655 04 × 2 = 0 + 0.785 656 891 310 08;
  • 49) 0.785 656 891 310 08 × 2 = 1 + 0.571 313 782 620 16;
  • 50) 0.571 313 782 620 16 × 2 = 1 + 0.142 627 565 240 32;
  • 51) 0.142 627 565 240 32 × 2 = 0 + 0.285 255 130 480 64;
  • 52) 0.285 255 130 480 64 × 2 = 0 + 0.570 510 260 961 28;
  • 53) 0.570 510 260 961 28 × 2 = 1 + 0.141 020 521 922 56;
  • 54) 0.141 020 521 922 56 × 2 = 0 + 0.282 041 043 845 12;

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:


0.000 000 000 744 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1100 1000 1100 10(2)

6. Positive number before normalization:

0.000 000 000 744 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1100 1000 1100 10(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:


0.000 000 000 744 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1100 1000 1100 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1100 1000 1100 10(2) × 20 =

1.1001 1001 0110 0100 0110 010(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 1001 0110 0100 0110 010


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 1011 0010 0011 0010 =

100 1100 1011 0010 0011 0010


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 1011 0010 0011 0010


Decimal number -0.000 000 000 744 68 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 1011 0010 0011 0010

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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