-0.000 000 000 744 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 744 3(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 3(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 3| = 0.000 000 000 744 3


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 3.

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 3 × 2 = 0 + 0.000 000 001 488 6;
  • 2) 0.000 000 001 488 6 × 2 = 0 + 0.000 000 002 977 2;
  • 3) 0.000 000 002 977 2 × 2 = 0 + 0.000 000 005 954 4;
  • 4) 0.000 000 005 954 4 × 2 = 0 + 0.000 000 011 908 8;
  • 5) 0.000 000 011 908 8 × 2 = 0 + 0.000 000 023 817 6;
  • 6) 0.000 000 023 817 6 × 2 = 0 + 0.000 000 047 635 2;
  • 7) 0.000 000 047 635 2 × 2 = 0 + 0.000 000 095 270 4;
  • 8) 0.000 000 095 270 4 × 2 = 0 + 0.000 000 190 540 8;
  • 9) 0.000 000 190 540 8 × 2 = 0 + 0.000 000 381 081 6;
  • 10) 0.000 000 381 081 6 × 2 = 0 + 0.000 000 762 163 2;
  • 11) 0.000 000 762 163 2 × 2 = 0 + 0.000 001 524 326 4;
  • 12) 0.000 001 524 326 4 × 2 = 0 + 0.000 003 048 652 8;
  • 13) 0.000 003 048 652 8 × 2 = 0 + 0.000 006 097 305 6;
  • 14) 0.000 006 097 305 6 × 2 = 0 + 0.000 012 194 611 2;
  • 15) 0.000 012 194 611 2 × 2 = 0 + 0.000 024 389 222 4;
  • 16) 0.000 024 389 222 4 × 2 = 0 + 0.000 048 778 444 8;
  • 17) 0.000 048 778 444 8 × 2 = 0 + 0.000 097 556 889 6;
  • 18) 0.000 097 556 889 6 × 2 = 0 + 0.000 195 113 779 2;
  • 19) 0.000 195 113 779 2 × 2 = 0 + 0.000 390 227 558 4;
  • 20) 0.000 390 227 558 4 × 2 = 0 + 0.000 780 455 116 8;
  • 21) 0.000 780 455 116 8 × 2 = 0 + 0.001 560 910 233 6;
  • 22) 0.001 560 910 233 6 × 2 = 0 + 0.003 121 820 467 2;
  • 23) 0.003 121 820 467 2 × 2 = 0 + 0.006 243 640 934 4;
  • 24) 0.006 243 640 934 4 × 2 = 0 + 0.012 487 281 868 8;
  • 25) 0.012 487 281 868 8 × 2 = 0 + 0.024 974 563 737 6;
  • 26) 0.024 974 563 737 6 × 2 = 0 + 0.049 949 127 475 2;
  • 27) 0.049 949 127 475 2 × 2 = 0 + 0.099 898 254 950 4;
  • 28) 0.099 898 254 950 4 × 2 = 0 + 0.199 796 509 900 8;
  • 29) 0.199 796 509 900 8 × 2 = 0 + 0.399 593 019 801 6;
  • 30) 0.399 593 019 801 6 × 2 = 0 + 0.799 186 039 603 2;
  • 31) 0.799 186 039 603 2 × 2 = 1 + 0.598 372 079 206 4;
  • 32) 0.598 372 079 206 4 × 2 = 1 + 0.196 744 158 412 8;
  • 33) 0.196 744 158 412 8 × 2 = 0 + 0.393 488 316 825 6;
  • 34) 0.393 488 316 825 6 × 2 = 0 + 0.786 976 633 651 2;
  • 35) 0.786 976 633 651 2 × 2 = 1 + 0.573 953 267 302 4;
  • 36) 0.573 953 267 302 4 × 2 = 1 + 0.147 906 534 604 8;
  • 37) 0.147 906 534 604 8 × 2 = 0 + 0.295 813 069 209 6;
  • 38) 0.295 813 069 209 6 × 2 = 0 + 0.591 626 138 419 2;
  • 39) 0.591 626 138 419 2 × 2 = 1 + 0.183 252 276 838 4;
  • 40) 0.183 252 276 838 4 × 2 = 0 + 0.366 504 553 676 8;
  • 41) 0.366 504 553 676 8 × 2 = 0 + 0.733 009 107 353 6;
  • 42) 0.733 009 107 353 6 × 2 = 1 + 0.466 018 214 707 2;
  • 43) 0.466 018 214 707 2 × 2 = 0 + 0.932 036 429 414 4;
  • 44) 0.932 036 429 414 4 × 2 = 1 + 0.864 072 858 828 8;
  • 45) 0.864 072 858 828 8 × 2 = 1 + 0.728 145 717 657 6;
  • 46) 0.728 145 717 657 6 × 2 = 1 + 0.456 291 435 315 2;
  • 47) 0.456 291 435 315 2 × 2 = 0 + 0.912 582 870 630 4;
  • 48) 0.912 582 870 630 4 × 2 = 1 + 0.825 165 741 260 8;
  • 49) 0.825 165 741 260 8 × 2 = 1 + 0.650 331 482 521 6;
  • 50) 0.650 331 482 521 6 × 2 = 1 + 0.300 662 965 043 2;
  • 51) 0.300 662 965 043 2 × 2 = 0 + 0.601 325 930 086 4;
  • 52) 0.601 325 930 086 4 × 2 = 1 + 0.202 651 860 172 8;
  • 53) 0.202 651 860 172 8 × 2 = 0 + 0.405 303 720 345 6;
  • 54) 0.405 303 720 345 6 × 2 = 0 + 0.810 607 440 691 2;

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 3(10) =

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

6. Positive number before normalization:

0.000 000 000 744 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 0101 1101 1101 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:


0.000 000 000 744 3(10) =

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

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

1.1001 1001 0010 1110 1110 100(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 0010 1110 1110 100


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 1001 0111 0111 0100 =

100 1100 1001 0111 0111 0100


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 1001 0111 0111 0100


Decimal number -0.000 000 000 744 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 1001 0111 0111 0100

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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