-0.000 000 000 687 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 687(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
-0.000 000 000 687(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:

|-0.000 000 000 687| = 0.000 000 000 687


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: 0.000 000 000 687.

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) 0.000 000 000 687 × 2 = 0 + 0.000 000 001 374;
  • 2) 0.000 000 001 374 × 2 = 0 + 0.000 000 002 748;
  • 3) 0.000 000 002 748 × 2 = 0 + 0.000 000 005 496;
  • 4) 0.000 000 005 496 × 2 = 0 + 0.000 000 010 992;
  • 5) 0.000 000 010 992 × 2 = 0 + 0.000 000 021 984;
  • 6) 0.000 000 021 984 × 2 = 0 + 0.000 000 043 968;
  • 7) 0.000 000 043 968 × 2 = 0 + 0.000 000 087 936;
  • 8) 0.000 000 087 936 × 2 = 0 + 0.000 000 175 872;
  • 9) 0.000 000 175 872 × 2 = 0 + 0.000 000 351 744;
  • 10) 0.000 000 351 744 × 2 = 0 + 0.000 000 703 488;
  • 11) 0.000 000 703 488 × 2 = 0 + 0.000 001 406 976;
  • 12) 0.000 001 406 976 × 2 = 0 + 0.000 002 813 952;
  • 13) 0.000 002 813 952 × 2 = 0 + 0.000 005 627 904;
  • 14) 0.000 005 627 904 × 2 = 0 + 0.000 011 255 808;
  • 15) 0.000 011 255 808 × 2 = 0 + 0.000 022 511 616;
  • 16) 0.000 022 511 616 × 2 = 0 + 0.000 045 023 232;
  • 17) 0.000 045 023 232 × 2 = 0 + 0.000 090 046 464;
  • 18) 0.000 090 046 464 × 2 = 0 + 0.000 180 092 928;
  • 19) 0.000 180 092 928 × 2 = 0 + 0.000 360 185 856;
  • 20) 0.000 360 185 856 × 2 = 0 + 0.000 720 371 712;
  • 21) 0.000 720 371 712 × 2 = 0 + 0.001 440 743 424;
  • 22) 0.001 440 743 424 × 2 = 0 + 0.002 881 486 848;
  • 23) 0.002 881 486 848 × 2 = 0 + 0.005 762 973 696;
  • 24) 0.005 762 973 696 × 2 = 0 + 0.011 525 947 392;
  • 25) 0.011 525 947 392 × 2 = 0 + 0.023 051 894 784;
  • 26) 0.023 051 894 784 × 2 = 0 + 0.046 103 789 568;
  • 27) 0.046 103 789 568 × 2 = 0 + 0.092 207 579 136;
  • 28) 0.092 207 579 136 × 2 = 0 + 0.184 415 158 272;
  • 29) 0.184 415 158 272 × 2 = 0 + 0.368 830 316 544;
  • 30) 0.368 830 316 544 × 2 = 0 + 0.737 660 633 088;
  • 31) 0.737 660 633 088 × 2 = 1 + 0.475 321 266 176;
  • 32) 0.475 321 266 176 × 2 = 0 + 0.950 642 532 352;
  • 33) 0.950 642 532 352 × 2 = 1 + 0.901 285 064 704;
  • 34) 0.901 285 064 704 × 2 = 1 + 0.802 570 129 408;
  • 35) 0.802 570 129 408 × 2 = 1 + 0.605 140 258 816;
  • 36) 0.605 140 258 816 × 2 = 1 + 0.210 280 517 632;
  • 37) 0.210 280 517 632 × 2 = 0 + 0.420 561 035 264;
  • 38) 0.420 561 035 264 × 2 = 0 + 0.841 122 070 528;
  • 39) 0.841 122 070 528 × 2 = 1 + 0.682 244 141 056;
  • 40) 0.682 244 141 056 × 2 = 1 + 0.364 488 282 112;
  • 41) 0.364 488 282 112 × 2 = 0 + 0.728 976 564 224;
  • 42) 0.728 976 564 224 × 2 = 1 + 0.457 953 128 448;
  • 43) 0.457 953 128 448 × 2 = 0 + 0.915 906 256 896;
  • 44) 0.915 906 256 896 × 2 = 1 + 0.831 812 513 792;
  • 45) 0.831 812 513 792 × 2 = 1 + 0.663 625 027 584;
  • 46) 0.663 625 027 584 × 2 = 1 + 0.327 250 055 168;
  • 47) 0.327 250 055 168 × 2 = 0 + 0.654 500 110 336;
  • 48) 0.654 500 110 336 × 2 = 1 + 0.309 000 220 672;
  • 49) 0.309 000 220 672 × 2 = 0 + 0.618 000 441 344;
  • 50) 0.618 000 441 344 × 2 = 1 + 0.236 000 882 688;
  • 51) 0.236 000 882 688 × 2 = 0 + 0.472 001 765 376;
  • 52) 0.472 001 765 376 × 2 = 0 + 0.944 003 530 752;
  • 53) 0.944 003 530 752 × 2 = 1 + 0.888 007 061 504;
  • 54) 0.888 007 061 504 × 2 = 1 + 0.776 014 123 008;

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:


0.000 000 000 687(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 0011 0101 1101 0100 11(2)

6. Positive number before normalization:

0.000 000 000 687(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 0011 0101 1101 0100 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:


0.000 000 000 687(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 0011 0101 1101 0100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 0011 0101 1101 0100 11(2) × 20 =

1.0111 1001 1010 1110 1010 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):
1.0111 1001 1010 1110 1010 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. 011 1100 1101 0111 0101 0011 =

011 1100 1101 0111 0101 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) =
011 1100 1101 0111 0101 0011


Decimal number -0.000 000 000 687 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 011 1100 1101 0111 0101 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