-0.000 000 000 699 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 699(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
-0.000 000 000 699(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:

|-0.000 000 000 699| = 0.000 000 000 699


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: 0.000 000 000 699.

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) 0.000 000 000 699 × 2 = 0 + 0.000 000 001 398;
  • 2) 0.000 000 001 398 × 2 = 0 + 0.000 000 002 796;
  • 3) 0.000 000 002 796 × 2 = 0 + 0.000 000 005 592;
  • 4) 0.000 000 005 592 × 2 = 0 + 0.000 000 011 184;
  • 5) 0.000 000 011 184 × 2 = 0 + 0.000 000 022 368;
  • 6) 0.000 000 022 368 × 2 = 0 + 0.000 000 044 736;
  • 7) 0.000 000 044 736 × 2 = 0 + 0.000 000 089 472;
  • 8) 0.000 000 089 472 × 2 = 0 + 0.000 000 178 944;
  • 9) 0.000 000 178 944 × 2 = 0 + 0.000 000 357 888;
  • 10) 0.000 000 357 888 × 2 = 0 + 0.000 000 715 776;
  • 11) 0.000 000 715 776 × 2 = 0 + 0.000 001 431 552;
  • 12) 0.000 001 431 552 × 2 = 0 + 0.000 002 863 104;
  • 13) 0.000 002 863 104 × 2 = 0 + 0.000 005 726 208;
  • 14) 0.000 005 726 208 × 2 = 0 + 0.000 011 452 416;
  • 15) 0.000 011 452 416 × 2 = 0 + 0.000 022 904 832;
  • 16) 0.000 022 904 832 × 2 = 0 + 0.000 045 809 664;
  • 17) 0.000 045 809 664 × 2 = 0 + 0.000 091 619 328;
  • 18) 0.000 091 619 328 × 2 = 0 + 0.000 183 238 656;
  • 19) 0.000 183 238 656 × 2 = 0 + 0.000 366 477 312;
  • 20) 0.000 366 477 312 × 2 = 0 + 0.000 732 954 624;
  • 21) 0.000 732 954 624 × 2 = 0 + 0.001 465 909 248;
  • 22) 0.001 465 909 248 × 2 = 0 + 0.002 931 818 496;
  • 23) 0.002 931 818 496 × 2 = 0 + 0.005 863 636 992;
  • 24) 0.005 863 636 992 × 2 = 0 + 0.011 727 273 984;
  • 25) 0.011 727 273 984 × 2 = 0 + 0.023 454 547 968;
  • 26) 0.023 454 547 968 × 2 = 0 + 0.046 909 095 936;
  • 27) 0.046 909 095 936 × 2 = 0 + 0.093 818 191 872;
  • 28) 0.093 818 191 872 × 2 = 0 + 0.187 636 383 744;
  • 29) 0.187 636 383 744 × 2 = 0 + 0.375 272 767 488;
  • 30) 0.375 272 767 488 × 2 = 0 + 0.750 545 534 976;
  • 31) 0.750 545 534 976 × 2 = 1 + 0.501 091 069 952;
  • 32) 0.501 091 069 952 × 2 = 1 + 0.002 182 139 904;
  • 33) 0.002 182 139 904 × 2 = 0 + 0.004 364 279 808;
  • 34) 0.004 364 279 808 × 2 = 0 + 0.008 728 559 616;
  • 35) 0.008 728 559 616 × 2 = 0 + 0.017 457 119 232;
  • 36) 0.017 457 119 232 × 2 = 0 + 0.034 914 238 464;
  • 37) 0.034 914 238 464 × 2 = 0 + 0.069 828 476 928;
  • 38) 0.069 828 476 928 × 2 = 0 + 0.139 656 953 856;
  • 39) 0.139 656 953 856 × 2 = 0 + 0.279 313 907 712;
  • 40) 0.279 313 907 712 × 2 = 0 + 0.558 627 815 424;
  • 41) 0.558 627 815 424 × 2 = 1 + 0.117 255 630 848;
  • 42) 0.117 255 630 848 × 2 = 0 + 0.234 511 261 696;
  • 43) 0.234 511 261 696 × 2 = 0 + 0.469 022 523 392;
  • 44) 0.469 022 523 392 × 2 = 0 + 0.938 045 046 784;
  • 45) 0.938 045 046 784 × 2 = 1 + 0.876 090 093 568;
  • 46) 0.876 090 093 568 × 2 = 1 + 0.752 180 187 136;
  • 47) 0.752 180 187 136 × 2 = 1 + 0.504 360 374 272;
  • 48) 0.504 360 374 272 × 2 = 1 + 0.008 720 748 544;
  • 49) 0.008 720 748 544 × 2 = 0 + 0.017 441 497 088;
  • 50) 0.017 441 497 088 × 2 = 0 + 0.034 882 994 176;
  • 51) 0.034 882 994 176 × 2 = 0 + 0.069 765 988 352;
  • 52) 0.069 765 988 352 × 2 = 0 + 0.139 531 976 704;
  • 53) 0.139 531 976 704 × 2 = 0 + 0.279 063 953 408;
  • 54) 0.279 063 953 408 × 2 = 0 + 0.558 127 906 816;

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:


0.000 000 000 699(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 1000 1111 0000 00(2)

6. Positive number before normalization:

0.000 000 000 699(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 1000 1111 0000 00(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:


0.000 000 000 699(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 1000 1111 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 1000 1111 0000 00(2) × 20 =

1.1000 0000 0100 0111 1000 000(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):
1.1000 0000 0100 0111 1000 000


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. 100 0000 0010 0011 1100 0000 =

100 0000 0010 0011 1100 0000


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) =
100 0000 0010 0011 1100 0000


Decimal number -0.000 000 000 699 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 0000 0010 0011 1100 0000

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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