-0.000 000 000 741 19 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 19(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
-0.000 000 000 741 19(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:

|-0.000 000 000 741 19| = 0.000 000 000 741 19


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: 0.000 000 000 741 19.

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) 0.000 000 000 741 19 × 2 = 0 + 0.000 000 001 482 38;
  • 2) 0.000 000 001 482 38 × 2 = 0 + 0.000 000 002 964 76;
  • 3) 0.000 000 002 964 76 × 2 = 0 + 0.000 000 005 929 52;
  • 4) 0.000 000 005 929 52 × 2 = 0 + 0.000 000 011 859 04;
  • 5) 0.000 000 011 859 04 × 2 = 0 + 0.000 000 023 718 08;
  • 6) 0.000 000 023 718 08 × 2 = 0 + 0.000 000 047 436 16;
  • 7) 0.000 000 047 436 16 × 2 = 0 + 0.000 000 094 872 32;
  • 8) 0.000 000 094 872 32 × 2 = 0 + 0.000 000 189 744 64;
  • 9) 0.000 000 189 744 64 × 2 = 0 + 0.000 000 379 489 28;
  • 10) 0.000 000 379 489 28 × 2 = 0 + 0.000 000 758 978 56;
  • 11) 0.000 000 758 978 56 × 2 = 0 + 0.000 001 517 957 12;
  • 12) 0.000 001 517 957 12 × 2 = 0 + 0.000 003 035 914 24;
  • 13) 0.000 003 035 914 24 × 2 = 0 + 0.000 006 071 828 48;
  • 14) 0.000 006 071 828 48 × 2 = 0 + 0.000 012 143 656 96;
  • 15) 0.000 012 143 656 96 × 2 = 0 + 0.000 024 287 313 92;
  • 16) 0.000 024 287 313 92 × 2 = 0 + 0.000 048 574 627 84;
  • 17) 0.000 048 574 627 84 × 2 = 0 + 0.000 097 149 255 68;
  • 18) 0.000 097 149 255 68 × 2 = 0 + 0.000 194 298 511 36;
  • 19) 0.000 194 298 511 36 × 2 = 0 + 0.000 388 597 022 72;
  • 20) 0.000 388 597 022 72 × 2 = 0 + 0.000 777 194 045 44;
  • 21) 0.000 777 194 045 44 × 2 = 0 + 0.001 554 388 090 88;
  • 22) 0.001 554 388 090 88 × 2 = 0 + 0.003 108 776 181 76;
  • 23) 0.003 108 776 181 76 × 2 = 0 + 0.006 217 552 363 52;
  • 24) 0.006 217 552 363 52 × 2 = 0 + 0.012 435 104 727 04;
  • 25) 0.012 435 104 727 04 × 2 = 0 + 0.024 870 209 454 08;
  • 26) 0.024 870 209 454 08 × 2 = 0 + 0.049 740 418 908 16;
  • 27) 0.049 740 418 908 16 × 2 = 0 + 0.099 480 837 816 32;
  • 28) 0.099 480 837 816 32 × 2 = 0 + 0.198 961 675 632 64;
  • 29) 0.198 961 675 632 64 × 2 = 0 + 0.397 923 351 265 28;
  • 30) 0.397 923 351 265 28 × 2 = 0 + 0.795 846 702 530 56;
  • 31) 0.795 846 702 530 56 × 2 = 1 + 0.591 693 405 061 12;
  • 32) 0.591 693 405 061 12 × 2 = 1 + 0.183 386 810 122 24;
  • 33) 0.183 386 810 122 24 × 2 = 0 + 0.366 773 620 244 48;
  • 34) 0.366 773 620 244 48 × 2 = 0 + 0.733 547 240 488 96;
  • 35) 0.733 547 240 488 96 × 2 = 1 + 0.467 094 480 977 92;
  • 36) 0.467 094 480 977 92 × 2 = 0 + 0.934 188 961 955 84;
  • 37) 0.934 188 961 955 84 × 2 = 1 + 0.868 377 923 911 68;
  • 38) 0.868 377 923 911 68 × 2 = 1 + 0.736 755 847 823 36;
  • 39) 0.736 755 847 823 36 × 2 = 1 + 0.473 511 695 646 72;
  • 40) 0.473 511 695 646 72 × 2 = 0 + 0.947 023 391 293 44;
  • 41) 0.947 023 391 293 44 × 2 = 1 + 0.894 046 782 586 88;
  • 42) 0.894 046 782 586 88 × 2 = 1 + 0.788 093 565 173 76;
  • 43) 0.788 093 565 173 76 × 2 = 1 + 0.576 187 130 347 52;
  • 44) 0.576 187 130 347 52 × 2 = 1 + 0.152 374 260 695 04;
  • 45) 0.152 374 260 695 04 × 2 = 0 + 0.304 748 521 390 08;
  • 46) 0.304 748 521 390 08 × 2 = 0 + 0.609 497 042 780 16;
  • 47) 0.609 497 042 780 16 × 2 = 1 + 0.218 994 085 560 32;
  • 48) 0.218 994 085 560 32 × 2 = 0 + 0.437 988 171 120 64;
  • 49) 0.437 988 171 120 64 × 2 = 0 + 0.875 976 342 241 28;
  • 50) 0.875 976 342 241 28 × 2 = 1 + 0.751 952 684 482 56;
  • 51) 0.751 952 684 482 56 × 2 = 1 + 0.503 905 368 965 12;
  • 52) 0.503 905 368 965 12 × 2 = 1 + 0.007 810 737 930 24;
  • 53) 0.007 810 737 930 24 × 2 = 0 + 0.015 621 475 860 48;
  • 54) 0.015 621 475 860 48 × 2 = 0 + 0.031 242 951 720 96;

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:


0.000 000 000 741 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1111 0010 0111 00(2)

6. Positive number before normalization:

0.000 000 000 741 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1111 0010 0111 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:


0.000 000 000 741 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1111 0010 0111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1111 0010 0111 00(2) × 20 =

1.1001 0111 0111 1001 0011 100(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 0111 1001 0011 100


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 1011 1100 1001 1100 =

100 1011 1011 1100 1001 1100


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 1011 1100 1001 1100


Decimal number -0.000 000 000 741 19 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1011 1100 1001 1100

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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