8.108 695 652 173 912 193 916 294 199 66 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 8.108 695 652 173 912 193 916 294 199 66(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
8.108 695 652 173 912 193 916 294 199 66(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. First, convert to binary (in base 2) the integer part: 8.
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;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

8(10) =

1000(2)


3. Convert to binary (base 2) the fractional part: 0.108 695 652 173 912 193 916 294 199 66.

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) 0.108 695 652 173 912 193 916 294 199 66 × 2 = 0 + 0.217 391 304 347 824 387 832 588 399 32;
  • 2) 0.217 391 304 347 824 387 832 588 399 32 × 2 = 0 + 0.434 782 608 695 648 775 665 176 798 64;
  • 3) 0.434 782 608 695 648 775 665 176 798 64 × 2 = 0 + 0.869 565 217 391 297 551 330 353 597 28;
  • 4) 0.869 565 217 391 297 551 330 353 597 28 × 2 = 1 + 0.739 130 434 782 595 102 660 707 194 56;
  • 5) 0.739 130 434 782 595 102 660 707 194 56 × 2 = 1 + 0.478 260 869 565 190 205 321 414 389 12;
  • 6) 0.478 260 869 565 190 205 321 414 389 12 × 2 = 0 + 0.956 521 739 130 380 410 642 828 778 24;
  • 7) 0.956 521 739 130 380 410 642 828 778 24 × 2 = 1 + 0.913 043 478 260 760 821 285 657 556 48;
  • 8) 0.913 043 478 260 760 821 285 657 556 48 × 2 = 1 + 0.826 086 956 521 521 642 571 315 112 96;
  • 9) 0.826 086 956 521 521 642 571 315 112 96 × 2 = 1 + 0.652 173 913 043 043 285 142 630 225 92;
  • 10) 0.652 173 913 043 043 285 142 630 225 92 × 2 = 1 + 0.304 347 826 086 086 570 285 260 451 84;
  • 11) 0.304 347 826 086 086 570 285 260 451 84 × 2 = 0 + 0.608 695 652 172 173 140 570 520 903 68;
  • 12) 0.608 695 652 172 173 140 570 520 903 68 × 2 = 1 + 0.217 391 304 344 346 281 141 041 807 36;
  • 13) 0.217 391 304 344 346 281 141 041 807 36 × 2 = 0 + 0.434 782 608 688 692 562 282 083 614 72;
  • 14) 0.434 782 608 688 692 562 282 083 614 72 × 2 = 0 + 0.869 565 217 377 385 124 564 167 229 44;
  • 15) 0.869 565 217 377 385 124 564 167 229 44 × 2 = 1 + 0.739 130 434 754 770 249 128 334 458 88;
  • 16) 0.739 130 434 754 770 249 128 334 458 88 × 2 = 1 + 0.478 260 869 509 540 498 256 668 917 76;
  • 17) 0.478 260 869 509 540 498 256 668 917 76 × 2 = 0 + 0.956 521 739 019 080 996 513 337 835 52;
  • 18) 0.956 521 739 019 080 996 513 337 835 52 × 2 = 1 + 0.913 043 478 038 161 993 026 675 671 04;
  • 19) 0.913 043 478 038 161 993 026 675 671 04 × 2 = 1 + 0.826 086 956 076 323 986 053 351 342 08;
  • 20) 0.826 086 956 076 323 986 053 351 342 08 × 2 = 1 + 0.652 173 912 152 647 972 106 702 684 16;
  • 21) 0.652 173 912 152 647 972 106 702 684 16 × 2 = 1 + 0.304 347 824 305 295 944 213 405 368 32;
  • 22) 0.304 347 824 305 295 944 213 405 368 32 × 2 = 0 + 0.608 695 648 610 591 888 426 810 736 64;
  • 23) 0.608 695 648 610 591 888 426 810 736 64 × 2 = 1 + 0.217 391 297 221 183 776 853 621 473 28;
  • 24) 0.217 391 297 221 183 776 853 621 473 28 × 2 = 0 + 0.434 782 594 442 367 553 707 242 946 56;

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).


4. 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:


0.108 695 652 173 912 193 916 294 199 66(10) =

0.0001 1011 1101 0011 0111 1010(2)

5. Positive number before normalization:

8.108 695 652 173 912 193 916 294 199 66(10) =

1000.0001 1011 1101 0011 0111 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


8.108 695 652 173 912 193 916 294 199 66(10) =

1000.0001 1011 1101 0011 0111 1010(2) =

1000.0001 1011 1101 0011 0111 1010(2) × 20 =

1.0000 0011 0111 1010 0110 1111 010(2) × 23


7. 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 0 (a positive number)

Exponent (unadjusted): 3

Mantissa (not normalized):
1.0000 0011 0111 1010 0110 1111 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

3 + 2(8-1) - 1 =

(3 + 127)(10) =

130(10)


9. 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;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

130(10) =

1000 0010(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 000 0001 1011 1101 0011 0111 1010 =

000 0001 1011 1101 0011 0111


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0010

Mantissa (23 bits) =
000 0001 1011 1101 0011 0111


Decimal number 8.108 695 652 173 912 193 916 294 199 66 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0010 - 000 0001 1011 1101 0011 0111

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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