-0.000 000 000 917 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 917(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
-0.000 000 000 917(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:

|-0.000 000 000 917| = 0.000 000 000 917


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: 0.000 000 000 917.

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) 0.000 000 000 917 × 2 = 0 + 0.000 000 001 834;
  • 2) 0.000 000 001 834 × 2 = 0 + 0.000 000 003 668;
  • 3) 0.000 000 003 668 × 2 = 0 + 0.000 000 007 336;
  • 4) 0.000 000 007 336 × 2 = 0 + 0.000 000 014 672;
  • 5) 0.000 000 014 672 × 2 = 0 + 0.000 000 029 344;
  • 6) 0.000 000 029 344 × 2 = 0 + 0.000 000 058 688;
  • 7) 0.000 000 058 688 × 2 = 0 + 0.000 000 117 376;
  • 8) 0.000 000 117 376 × 2 = 0 + 0.000 000 234 752;
  • 9) 0.000 000 234 752 × 2 = 0 + 0.000 000 469 504;
  • 10) 0.000 000 469 504 × 2 = 0 + 0.000 000 939 008;
  • 11) 0.000 000 939 008 × 2 = 0 + 0.000 001 878 016;
  • 12) 0.000 001 878 016 × 2 = 0 + 0.000 003 756 032;
  • 13) 0.000 003 756 032 × 2 = 0 + 0.000 007 512 064;
  • 14) 0.000 007 512 064 × 2 = 0 + 0.000 015 024 128;
  • 15) 0.000 015 024 128 × 2 = 0 + 0.000 030 048 256;
  • 16) 0.000 030 048 256 × 2 = 0 + 0.000 060 096 512;
  • 17) 0.000 060 096 512 × 2 = 0 + 0.000 120 193 024;
  • 18) 0.000 120 193 024 × 2 = 0 + 0.000 240 386 048;
  • 19) 0.000 240 386 048 × 2 = 0 + 0.000 480 772 096;
  • 20) 0.000 480 772 096 × 2 = 0 + 0.000 961 544 192;
  • 21) 0.000 961 544 192 × 2 = 0 + 0.001 923 088 384;
  • 22) 0.001 923 088 384 × 2 = 0 + 0.003 846 176 768;
  • 23) 0.003 846 176 768 × 2 = 0 + 0.007 692 353 536;
  • 24) 0.007 692 353 536 × 2 = 0 + 0.015 384 707 072;
  • 25) 0.015 384 707 072 × 2 = 0 + 0.030 769 414 144;
  • 26) 0.030 769 414 144 × 2 = 0 + 0.061 538 828 288;
  • 27) 0.061 538 828 288 × 2 = 0 + 0.123 077 656 576;
  • 28) 0.123 077 656 576 × 2 = 0 + 0.246 155 313 152;
  • 29) 0.246 155 313 152 × 2 = 0 + 0.492 310 626 304;
  • 30) 0.492 310 626 304 × 2 = 0 + 0.984 621 252 608;
  • 31) 0.984 621 252 608 × 2 = 1 + 0.969 242 505 216;
  • 32) 0.969 242 505 216 × 2 = 1 + 0.938 485 010 432;
  • 33) 0.938 485 010 432 × 2 = 1 + 0.876 970 020 864;
  • 34) 0.876 970 020 864 × 2 = 1 + 0.753 940 041 728;
  • 35) 0.753 940 041 728 × 2 = 1 + 0.507 880 083 456;
  • 36) 0.507 880 083 456 × 2 = 1 + 0.015 760 166 912;
  • 37) 0.015 760 166 912 × 2 = 0 + 0.031 520 333 824;
  • 38) 0.031 520 333 824 × 2 = 0 + 0.063 040 667 648;
  • 39) 0.063 040 667 648 × 2 = 0 + 0.126 081 335 296;
  • 40) 0.126 081 335 296 × 2 = 0 + 0.252 162 670 592;
  • 41) 0.252 162 670 592 × 2 = 0 + 0.504 325 341 184;
  • 42) 0.504 325 341 184 × 2 = 1 + 0.008 650 682 368;
  • 43) 0.008 650 682 368 × 2 = 0 + 0.017 301 364 736;
  • 44) 0.017 301 364 736 × 2 = 0 + 0.034 602 729 472;
  • 45) 0.034 602 729 472 × 2 = 0 + 0.069 205 458 944;
  • 46) 0.069 205 458 944 × 2 = 0 + 0.138 410 917 888;
  • 47) 0.138 410 917 888 × 2 = 0 + 0.276 821 835 776;
  • 48) 0.276 821 835 776 × 2 = 0 + 0.553 643 671 552;
  • 49) 0.553 643 671 552 × 2 = 1 + 0.107 287 343 104;
  • 50) 0.107 287 343 104 × 2 = 0 + 0.214 574 686 208;
  • 51) 0.214 574 686 208 × 2 = 0 + 0.429 149 372 416;
  • 52) 0.429 149 372 416 × 2 = 0 + 0.858 298 744 832;
  • 53) 0.858 298 744 832 × 2 = 1 + 0.716 597 489 664;
  • 54) 0.716 597 489 664 × 2 = 1 + 0.433 194 979 328;

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:


0.000 000 000 917(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1111 0000 0100 0000 1000 11(2)

6. Positive number before normalization:

0.000 000 000 917(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1111 0000 0100 0000 1000 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:


0.000 000 000 917(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1111 0000 0100 0000 1000 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1111 0000 0100 0000 1000 11(2) × 20 =

1.1111 1000 0010 0000 0100 011(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):
1.1111 1000 0010 0000 0100 011


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. 111 1100 0001 0000 0010 0011 =

111 1100 0001 0000 0010 0011


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) =
111 1100 0001 0000 0010 0011


Decimal number -0.000 000 000 917 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 111 1100 0001 0000 0010 0011

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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