-0.000 000 000 747 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 747 2(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
-0.000 000 000 747 2(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:

|-0.000 000 000 747 2| = 0.000 000 000 747 2


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: 0.000 000 000 747 2.

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) 0.000 000 000 747 2 × 2 = 0 + 0.000 000 001 494 4;
  • 2) 0.000 000 001 494 4 × 2 = 0 + 0.000 000 002 988 8;
  • 3) 0.000 000 002 988 8 × 2 = 0 + 0.000 000 005 977 6;
  • 4) 0.000 000 005 977 6 × 2 = 0 + 0.000 000 011 955 2;
  • 5) 0.000 000 011 955 2 × 2 = 0 + 0.000 000 023 910 4;
  • 6) 0.000 000 023 910 4 × 2 = 0 + 0.000 000 047 820 8;
  • 7) 0.000 000 047 820 8 × 2 = 0 + 0.000 000 095 641 6;
  • 8) 0.000 000 095 641 6 × 2 = 0 + 0.000 000 191 283 2;
  • 9) 0.000 000 191 283 2 × 2 = 0 + 0.000 000 382 566 4;
  • 10) 0.000 000 382 566 4 × 2 = 0 + 0.000 000 765 132 8;
  • 11) 0.000 000 765 132 8 × 2 = 0 + 0.000 001 530 265 6;
  • 12) 0.000 001 530 265 6 × 2 = 0 + 0.000 003 060 531 2;
  • 13) 0.000 003 060 531 2 × 2 = 0 + 0.000 006 121 062 4;
  • 14) 0.000 006 121 062 4 × 2 = 0 + 0.000 012 242 124 8;
  • 15) 0.000 012 242 124 8 × 2 = 0 + 0.000 024 484 249 6;
  • 16) 0.000 024 484 249 6 × 2 = 0 + 0.000 048 968 499 2;
  • 17) 0.000 048 968 499 2 × 2 = 0 + 0.000 097 936 998 4;
  • 18) 0.000 097 936 998 4 × 2 = 0 + 0.000 195 873 996 8;
  • 19) 0.000 195 873 996 8 × 2 = 0 + 0.000 391 747 993 6;
  • 20) 0.000 391 747 993 6 × 2 = 0 + 0.000 783 495 987 2;
  • 21) 0.000 783 495 987 2 × 2 = 0 + 0.001 566 991 974 4;
  • 22) 0.001 566 991 974 4 × 2 = 0 + 0.003 133 983 948 8;
  • 23) 0.003 133 983 948 8 × 2 = 0 + 0.006 267 967 897 6;
  • 24) 0.006 267 967 897 6 × 2 = 0 + 0.012 535 935 795 2;
  • 25) 0.012 535 935 795 2 × 2 = 0 + 0.025 071 871 590 4;
  • 26) 0.025 071 871 590 4 × 2 = 0 + 0.050 143 743 180 8;
  • 27) 0.050 143 743 180 8 × 2 = 0 + 0.100 287 486 361 6;
  • 28) 0.100 287 486 361 6 × 2 = 0 + 0.200 574 972 723 2;
  • 29) 0.200 574 972 723 2 × 2 = 0 + 0.401 149 945 446 4;
  • 30) 0.401 149 945 446 4 × 2 = 0 + 0.802 299 890 892 8;
  • 31) 0.802 299 890 892 8 × 2 = 1 + 0.604 599 781 785 6;
  • 32) 0.604 599 781 785 6 × 2 = 1 + 0.209 199 563 571 2;
  • 33) 0.209 199 563 571 2 × 2 = 0 + 0.418 399 127 142 4;
  • 34) 0.418 399 127 142 4 × 2 = 0 + 0.836 798 254 284 8;
  • 35) 0.836 798 254 284 8 × 2 = 1 + 0.673 596 508 569 6;
  • 36) 0.673 596 508 569 6 × 2 = 1 + 0.347 193 017 139 2;
  • 37) 0.347 193 017 139 2 × 2 = 0 + 0.694 386 034 278 4;
  • 38) 0.694 386 034 278 4 × 2 = 1 + 0.388 772 068 556 8;
  • 39) 0.388 772 068 556 8 × 2 = 0 + 0.777 544 137 113 6;
  • 40) 0.777 544 137 113 6 × 2 = 1 + 0.555 088 274 227 2;
  • 41) 0.555 088 274 227 2 × 2 = 1 + 0.110 176 548 454 4;
  • 42) 0.110 176 548 454 4 × 2 = 0 + 0.220 353 096 908 8;
  • 43) 0.220 353 096 908 8 × 2 = 0 + 0.440 706 193 817 6;
  • 44) 0.440 706 193 817 6 × 2 = 0 + 0.881 412 387 635 2;
  • 45) 0.881 412 387 635 2 × 2 = 1 + 0.762 824 775 270 4;
  • 46) 0.762 824 775 270 4 × 2 = 1 + 0.525 649 550 540 8;
  • 47) 0.525 649 550 540 8 × 2 = 1 + 0.051 299 101 081 6;
  • 48) 0.051 299 101 081 6 × 2 = 0 + 0.102 598 202 163 2;
  • 49) 0.102 598 202 163 2 × 2 = 0 + 0.205 196 404 326 4;
  • 50) 0.205 196 404 326 4 × 2 = 0 + 0.410 392 808 652 8;
  • 51) 0.410 392 808 652 8 × 2 = 0 + 0.820 785 617 305 6;
  • 52) 0.820 785 617 305 6 × 2 = 1 + 0.641 571 234 611 2;
  • 53) 0.641 571 234 611 2 × 2 = 1 + 0.283 142 469 222 4;
  • 54) 0.283 142 469 222 4 × 2 = 0 + 0.566 284 938 444 8;

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:


0.000 000 000 747 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0101 1000 1110 0001 10(2)

6. Positive number before normalization:

0.000 000 000 747 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0101 1000 1110 0001 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:


0.000 000 000 747 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0101 1000 1110 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0101 1000 1110 0001 10(2) × 20 =

1.1001 1010 1100 0111 0000 110(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 1010 1100 0111 0000 110


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 1101 0110 0011 1000 0110 =

100 1101 0110 0011 1000 0110


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 1101 0110 0011 1000 0110


Decimal number -0.000 000 000 747 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1101 0110 0011 1000 0110

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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