-0.000 000 000 762 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 762 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
-0.000 000 000 762 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:

|-0.000 000 000 762 8| = 0.000 000 000 762 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: 0.000 000 000 762 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) 0.000 000 000 762 8 × 2 = 0 + 0.000 000 001 525 6;
  • 2) 0.000 000 001 525 6 × 2 = 0 + 0.000 000 003 051 2;
  • 3) 0.000 000 003 051 2 × 2 = 0 + 0.000 000 006 102 4;
  • 4) 0.000 000 006 102 4 × 2 = 0 + 0.000 000 012 204 8;
  • 5) 0.000 000 012 204 8 × 2 = 0 + 0.000 000 024 409 6;
  • 6) 0.000 000 024 409 6 × 2 = 0 + 0.000 000 048 819 2;
  • 7) 0.000 000 048 819 2 × 2 = 0 + 0.000 000 097 638 4;
  • 8) 0.000 000 097 638 4 × 2 = 0 + 0.000 000 195 276 8;
  • 9) 0.000 000 195 276 8 × 2 = 0 + 0.000 000 390 553 6;
  • 10) 0.000 000 390 553 6 × 2 = 0 + 0.000 000 781 107 2;
  • 11) 0.000 000 781 107 2 × 2 = 0 + 0.000 001 562 214 4;
  • 12) 0.000 001 562 214 4 × 2 = 0 + 0.000 003 124 428 8;
  • 13) 0.000 003 124 428 8 × 2 = 0 + 0.000 006 248 857 6;
  • 14) 0.000 006 248 857 6 × 2 = 0 + 0.000 012 497 715 2;
  • 15) 0.000 012 497 715 2 × 2 = 0 + 0.000 024 995 430 4;
  • 16) 0.000 024 995 430 4 × 2 = 0 + 0.000 049 990 860 8;
  • 17) 0.000 049 990 860 8 × 2 = 0 + 0.000 099 981 721 6;
  • 18) 0.000 099 981 721 6 × 2 = 0 + 0.000 199 963 443 2;
  • 19) 0.000 199 963 443 2 × 2 = 0 + 0.000 399 926 886 4;
  • 20) 0.000 399 926 886 4 × 2 = 0 + 0.000 799 853 772 8;
  • 21) 0.000 799 853 772 8 × 2 = 0 + 0.001 599 707 545 6;
  • 22) 0.001 599 707 545 6 × 2 = 0 + 0.003 199 415 091 2;
  • 23) 0.003 199 415 091 2 × 2 = 0 + 0.006 398 830 182 4;
  • 24) 0.006 398 830 182 4 × 2 = 0 + 0.012 797 660 364 8;
  • 25) 0.012 797 660 364 8 × 2 = 0 + 0.025 595 320 729 6;
  • 26) 0.025 595 320 729 6 × 2 = 0 + 0.051 190 641 459 2;
  • 27) 0.051 190 641 459 2 × 2 = 0 + 0.102 381 282 918 4;
  • 28) 0.102 381 282 918 4 × 2 = 0 + 0.204 762 565 836 8;
  • 29) 0.204 762 565 836 8 × 2 = 0 + 0.409 525 131 673 6;
  • 30) 0.409 525 131 673 6 × 2 = 0 + 0.819 050 263 347 2;
  • 31) 0.819 050 263 347 2 × 2 = 1 + 0.638 100 526 694 4;
  • 32) 0.638 100 526 694 4 × 2 = 1 + 0.276 201 053 388 8;
  • 33) 0.276 201 053 388 8 × 2 = 0 + 0.552 402 106 777 6;
  • 34) 0.552 402 106 777 6 × 2 = 1 + 0.104 804 213 555 2;
  • 35) 0.104 804 213 555 2 × 2 = 0 + 0.209 608 427 110 4;
  • 36) 0.209 608 427 110 4 × 2 = 0 + 0.419 216 854 220 8;
  • 37) 0.419 216 854 220 8 × 2 = 0 + 0.838 433 708 441 6;
  • 38) 0.838 433 708 441 6 × 2 = 1 + 0.676 867 416 883 2;
  • 39) 0.676 867 416 883 2 × 2 = 1 + 0.353 734 833 766 4;
  • 40) 0.353 734 833 766 4 × 2 = 0 + 0.707 469 667 532 8;
  • 41) 0.707 469 667 532 8 × 2 = 1 + 0.414 939 335 065 6;
  • 42) 0.414 939 335 065 6 × 2 = 0 + 0.829 878 670 131 2;
  • 43) 0.829 878 670 131 2 × 2 = 1 + 0.659 757 340 262 4;
  • 44) 0.659 757 340 262 4 × 2 = 1 + 0.319 514 680 524 8;
  • 45) 0.319 514 680 524 8 × 2 = 0 + 0.639 029 361 049 6;
  • 46) 0.639 029 361 049 6 × 2 = 1 + 0.278 058 722 099 2;
  • 47) 0.278 058 722 099 2 × 2 = 0 + 0.556 117 444 198 4;
  • 48) 0.556 117 444 198 4 × 2 = 1 + 0.112 234 888 396 8;
  • 49) 0.112 234 888 396 8 × 2 = 0 + 0.224 469 776 793 6;
  • 50) 0.224 469 776 793 6 × 2 = 0 + 0.448 939 553 587 2;
  • 51) 0.448 939 553 587 2 × 2 = 0 + 0.897 879 107 174 4;
  • 52) 0.897 879 107 174 4 × 2 = 1 + 0.795 758 214 348 8;
  • 53) 0.795 758 214 348 8 × 2 = 1 + 0.591 516 428 697 6;
  • 54) 0.591 516 428 697 6 × 2 = 1 + 0.183 032 857 395 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:


0.000 000 000 762 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0110 1011 0101 0001 11(2)

6. Positive number before normalization:

0.000 000 000 762 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0110 1011 0101 0001 11(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:


0.000 000 000 762 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0110 1011 0101 0001 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0110 1011 0101 0001 11(2) × 20 =

1.1010 0011 0101 1010 1000 111(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):
1.1010 0011 0101 1010 1000 111


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. 101 0001 1010 1101 0100 0111 =

101 0001 1010 1101 0100 0111


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) =
101 0001 1010 1101 0100 0111


Decimal number -0.000 000 000 762 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 0001 1010 1101 0100 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