-0.000 000 000 742 128 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 128 2(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
-0.000 000 000 742 128 2(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:

|-0.000 000 000 742 128 2| = 0.000 000 000 742 128 2


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: 0.000 000 000 742 128 2.

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) 0.000 000 000 742 128 2 × 2 = 0 + 0.000 000 001 484 256 4;
  • 2) 0.000 000 001 484 256 4 × 2 = 0 + 0.000 000 002 968 512 8;
  • 3) 0.000 000 002 968 512 8 × 2 = 0 + 0.000 000 005 937 025 6;
  • 4) 0.000 000 005 937 025 6 × 2 = 0 + 0.000 000 011 874 051 2;
  • 5) 0.000 000 011 874 051 2 × 2 = 0 + 0.000 000 023 748 102 4;
  • 6) 0.000 000 023 748 102 4 × 2 = 0 + 0.000 000 047 496 204 8;
  • 7) 0.000 000 047 496 204 8 × 2 = 0 + 0.000 000 094 992 409 6;
  • 8) 0.000 000 094 992 409 6 × 2 = 0 + 0.000 000 189 984 819 2;
  • 9) 0.000 000 189 984 819 2 × 2 = 0 + 0.000 000 379 969 638 4;
  • 10) 0.000 000 379 969 638 4 × 2 = 0 + 0.000 000 759 939 276 8;
  • 11) 0.000 000 759 939 276 8 × 2 = 0 + 0.000 001 519 878 553 6;
  • 12) 0.000 001 519 878 553 6 × 2 = 0 + 0.000 003 039 757 107 2;
  • 13) 0.000 003 039 757 107 2 × 2 = 0 + 0.000 006 079 514 214 4;
  • 14) 0.000 006 079 514 214 4 × 2 = 0 + 0.000 012 159 028 428 8;
  • 15) 0.000 012 159 028 428 8 × 2 = 0 + 0.000 024 318 056 857 6;
  • 16) 0.000 024 318 056 857 6 × 2 = 0 + 0.000 048 636 113 715 2;
  • 17) 0.000 048 636 113 715 2 × 2 = 0 + 0.000 097 272 227 430 4;
  • 18) 0.000 097 272 227 430 4 × 2 = 0 + 0.000 194 544 454 860 8;
  • 19) 0.000 194 544 454 860 8 × 2 = 0 + 0.000 389 088 909 721 6;
  • 20) 0.000 389 088 909 721 6 × 2 = 0 + 0.000 778 177 819 443 2;
  • 21) 0.000 778 177 819 443 2 × 2 = 0 + 0.001 556 355 638 886 4;
  • 22) 0.001 556 355 638 886 4 × 2 = 0 + 0.003 112 711 277 772 8;
  • 23) 0.003 112 711 277 772 8 × 2 = 0 + 0.006 225 422 555 545 6;
  • 24) 0.006 225 422 555 545 6 × 2 = 0 + 0.012 450 845 111 091 2;
  • 25) 0.012 450 845 111 091 2 × 2 = 0 + 0.024 901 690 222 182 4;
  • 26) 0.024 901 690 222 182 4 × 2 = 0 + 0.049 803 380 444 364 8;
  • 27) 0.049 803 380 444 364 8 × 2 = 0 + 0.099 606 760 888 729 6;
  • 28) 0.099 606 760 888 729 6 × 2 = 0 + 0.199 213 521 777 459 2;
  • 29) 0.199 213 521 777 459 2 × 2 = 0 + 0.398 427 043 554 918 4;
  • 30) 0.398 427 043 554 918 4 × 2 = 0 + 0.796 854 087 109 836 8;
  • 31) 0.796 854 087 109 836 8 × 2 = 1 + 0.593 708 174 219 673 6;
  • 32) 0.593 708 174 219 673 6 × 2 = 1 + 0.187 416 348 439 347 2;
  • 33) 0.187 416 348 439 347 2 × 2 = 0 + 0.374 832 696 878 694 4;
  • 34) 0.374 832 696 878 694 4 × 2 = 0 + 0.749 665 393 757 388 8;
  • 35) 0.749 665 393 757 388 8 × 2 = 1 + 0.499 330 787 514 777 6;
  • 36) 0.499 330 787 514 777 6 × 2 = 0 + 0.998 661 575 029 555 2;
  • 37) 0.998 661 575 029 555 2 × 2 = 1 + 0.997 323 150 059 110 4;
  • 38) 0.997 323 150 059 110 4 × 2 = 1 + 0.994 646 300 118 220 8;
  • 39) 0.994 646 300 118 220 8 × 2 = 1 + 0.989 292 600 236 441 6;
  • 40) 0.989 292 600 236 441 6 × 2 = 1 + 0.978 585 200 472 883 2;
  • 41) 0.978 585 200 472 883 2 × 2 = 1 + 0.957 170 400 945 766 4;
  • 42) 0.957 170 400 945 766 4 × 2 = 1 + 0.914 340 801 891 532 8;
  • 43) 0.914 340 801 891 532 8 × 2 = 1 + 0.828 681 603 783 065 6;
  • 44) 0.828 681 603 783 065 6 × 2 = 1 + 0.657 363 207 566 131 2;
  • 45) 0.657 363 207 566 131 2 × 2 = 1 + 0.314 726 415 132 262 4;
  • 46) 0.314 726 415 132 262 4 × 2 = 0 + 0.629 452 830 264 524 8;
  • 47) 0.629 452 830 264 524 8 × 2 = 1 + 0.258 905 660 529 049 6;
  • 48) 0.258 905 660 529 049 6 × 2 = 0 + 0.517 811 321 058 099 2;
  • 49) 0.517 811 321 058 099 2 × 2 = 1 + 0.035 622 642 116 198 4;
  • 50) 0.035 622 642 116 198 4 × 2 = 0 + 0.071 245 284 232 396 8;
  • 51) 0.071 245 284 232 396 8 × 2 = 0 + 0.142 490 568 464 793 6;
  • 52) 0.142 490 568 464 793 6 × 2 = 0 + 0.284 981 136 929 587 2;
  • 53) 0.284 981 136 929 587 2 × 2 = 0 + 0.569 962 273 859 174 4;
  • 54) 0.569 962 273 859 174 4 × 2 = 1 + 0.139 924 547 718 348 8;

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:


0.000 000 000 742 128 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1000 01(2)

6. Positive number before normalization:

0.000 000 000 742 128 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1000 01(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:


0.000 000 000 742 128 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1000 01(2) × 20 =

1.1001 0111 1111 1101 0100 001(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):
1.1001 0111 1111 1101 0100 001


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 1011 1111 1110 1010 0001 =

100 1011 1111 1110 1010 0001


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 1011 1111 1110 1010 0001


Decimal number -0.000 000 000 742 128 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1010 0001

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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