-0.000 000 100 100 140 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 100 100 140 8(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
-0.000 000 100 100 140 8(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:

|-0.000 000 100 100 140 8| = 0.000 000 100 100 140 8


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: 0.000 000 100 100 140 8.

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) 0.000 000 100 100 140 8 × 2 = 0 + 0.000 000 200 200 281 6;
  • 2) 0.000 000 200 200 281 6 × 2 = 0 + 0.000 000 400 400 563 2;
  • 3) 0.000 000 400 400 563 2 × 2 = 0 + 0.000 000 800 801 126 4;
  • 4) 0.000 000 800 801 126 4 × 2 = 0 + 0.000 001 601 602 252 8;
  • 5) 0.000 001 601 602 252 8 × 2 = 0 + 0.000 003 203 204 505 6;
  • 6) 0.000 003 203 204 505 6 × 2 = 0 + 0.000 006 406 409 011 2;
  • 7) 0.000 006 406 409 011 2 × 2 = 0 + 0.000 012 812 818 022 4;
  • 8) 0.000 012 812 818 022 4 × 2 = 0 + 0.000 025 625 636 044 8;
  • 9) 0.000 025 625 636 044 8 × 2 = 0 + 0.000 051 251 272 089 6;
  • 10) 0.000 051 251 272 089 6 × 2 = 0 + 0.000 102 502 544 179 2;
  • 11) 0.000 102 502 544 179 2 × 2 = 0 + 0.000 205 005 088 358 4;
  • 12) 0.000 205 005 088 358 4 × 2 = 0 + 0.000 410 010 176 716 8;
  • 13) 0.000 410 010 176 716 8 × 2 = 0 + 0.000 820 020 353 433 6;
  • 14) 0.000 820 020 353 433 6 × 2 = 0 + 0.001 640 040 706 867 2;
  • 15) 0.001 640 040 706 867 2 × 2 = 0 + 0.003 280 081 413 734 4;
  • 16) 0.003 280 081 413 734 4 × 2 = 0 + 0.006 560 162 827 468 8;
  • 17) 0.006 560 162 827 468 8 × 2 = 0 + 0.013 120 325 654 937 6;
  • 18) 0.013 120 325 654 937 6 × 2 = 0 + 0.026 240 651 309 875 2;
  • 19) 0.026 240 651 309 875 2 × 2 = 0 + 0.052 481 302 619 750 4;
  • 20) 0.052 481 302 619 750 4 × 2 = 0 + 0.104 962 605 239 500 8;
  • 21) 0.104 962 605 239 500 8 × 2 = 0 + 0.209 925 210 479 001 6;
  • 22) 0.209 925 210 479 001 6 × 2 = 0 + 0.419 850 420 958 003 2;
  • 23) 0.419 850 420 958 003 2 × 2 = 0 + 0.839 700 841 916 006 4;
  • 24) 0.839 700 841 916 006 4 × 2 = 1 + 0.679 401 683 832 012 8;
  • 25) 0.679 401 683 832 012 8 × 2 = 1 + 0.358 803 367 664 025 6;
  • 26) 0.358 803 367 664 025 6 × 2 = 0 + 0.717 606 735 328 051 2;
  • 27) 0.717 606 735 328 051 2 × 2 = 1 + 0.435 213 470 656 102 4;
  • 28) 0.435 213 470 656 102 4 × 2 = 0 + 0.870 426 941 312 204 8;
  • 29) 0.870 426 941 312 204 8 × 2 = 1 + 0.740 853 882 624 409 6;
  • 30) 0.740 853 882 624 409 6 × 2 = 1 + 0.481 707 765 248 819 2;
  • 31) 0.481 707 765 248 819 2 × 2 = 0 + 0.963 415 530 497 638 4;
  • 32) 0.963 415 530 497 638 4 × 2 = 1 + 0.926 831 060 995 276 8;
  • 33) 0.926 831 060 995 276 8 × 2 = 1 + 0.853 662 121 990 553 6;
  • 34) 0.853 662 121 990 553 6 × 2 = 1 + 0.707 324 243 981 107 2;
  • 35) 0.707 324 243 981 107 2 × 2 = 1 + 0.414 648 487 962 214 4;
  • 36) 0.414 648 487 962 214 4 × 2 = 0 + 0.829 296 975 924 428 8;
  • 37) 0.829 296 975 924 428 8 × 2 = 1 + 0.658 593 951 848 857 6;
  • 38) 0.658 593 951 848 857 6 × 2 = 1 + 0.317 187 903 697 715 2;
  • 39) 0.317 187 903 697 715 2 × 2 = 0 + 0.634 375 807 395 430 4;
  • 40) 0.634 375 807 395 430 4 × 2 = 1 + 0.268 751 614 790 860 8;
  • 41) 0.268 751 614 790 860 8 × 2 = 0 + 0.537 503 229 581 721 6;
  • 42) 0.537 503 229 581 721 6 × 2 = 1 + 0.075 006 459 163 443 2;
  • 43) 0.075 006 459 163 443 2 × 2 = 0 + 0.150 012 918 326 886 4;
  • 44) 0.150 012 918 326 886 4 × 2 = 0 + 0.300 025 836 653 772 8;
  • 45) 0.300 025 836 653 772 8 × 2 = 0 + 0.600 051 673 307 545 6;
  • 46) 0.600 051 673 307 545 6 × 2 = 1 + 0.200 103 346 615 091 2;
  • 47) 0.200 103 346 615 091 2 × 2 = 0 + 0.400 206 693 230 182 4;

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:


0.000 000 100 100 140 8(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0100 010(2)

6. Positive number before normalization:

0.000 000 100 100 140 8(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0100 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 100 100 140 8(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0100 010(2) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0100 010(2) × 20 =

1.1010 1101 1110 1101 0100 010(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1010 1101 1110 1101 0100 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 101 0110 1111 0110 1010 0010 =

101 0110 1111 0110 1010 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 0111

Mantissa (23 bits) =
101 0110 1111 0110 1010 0010


Decimal number -0.000 000 100 100 140 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 101 0110 1111 0110 1010 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