-0.000 000 000 741 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 9(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 9(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 9| = 0.000 000 000 741 9


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 9.

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 9 × 2 = 0 + 0.000 000 001 483 8;
  • 2) 0.000 000 001 483 8 × 2 = 0 + 0.000 000 002 967 6;
  • 3) 0.000 000 002 967 6 × 2 = 0 + 0.000 000 005 935 2;
  • 4) 0.000 000 005 935 2 × 2 = 0 + 0.000 000 011 870 4;
  • 5) 0.000 000 011 870 4 × 2 = 0 + 0.000 000 023 740 8;
  • 6) 0.000 000 023 740 8 × 2 = 0 + 0.000 000 047 481 6;
  • 7) 0.000 000 047 481 6 × 2 = 0 + 0.000 000 094 963 2;
  • 8) 0.000 000 094 963 2 × 2 = 0 + 0.000 000 189 926 4;
  • 9) 0.000 000 189 926 4 × 2 = 0 + 0.000 000 379 852 8;
  • 10) 0.000 000 379 852 8 × 2 = 0 + 0.000 000 759 705 6;
  • 11) 0.000 000 759 705 6 × 2 = 0 + 0.000 001 519 411 2;
  • 12) 0.000 001 519 411 2 × 2 = 0 + 0.000 003 038 822 4;
  • 13) 0.000 003 038 822 4 × 2 = 0 + 0.000 006 077 644 8;
  • 14) 0.000 006 077 644 8 × 2 = 0 + 0.000 012 155 289 6;
  • 15) 0.000 012 155 289 6 × 2 = 0 + 0.000 024 310 579 2;
  • 16) 0.000 024 310 579 2 × 2 = 0 + 0.000 048 621 158 4;
  • 17) 0.000 048 621 158 4 × 2 = 0 + 0.000 097 242 316 8;
  • 18) 0.000 097 242 316 8 × 2 = 0 + 0.000 194 484 633 6;
  • 19) 0.000 194 484 633 6 × 2 = 0 + 0.000 388 969 267 2;
  • 20) 0.000 388 969 267 2 × 2 = 0 + 0.000 777 938 534 4;
  • 21) 0.000 777 938 534 4 × 2 = 0 + 0.001 555 877 068 8;
  • 22) 0.001 555 877 068 8 × 2 = 0 + 0.003 111 754 137 6;
  • 23) 0.003 111 754 137 6 × 2 = 0 + 0.006 223 508 275 2;
  • 24) 0.006 223 508 275 2 × 2 = 0 + 0.012 447 016 550 4;
  • 25) 0.012 447 016 550 4 × 2 = 0 + 0.024 894 033 100 8;
  • 26) 0.024 894 033 100 8 × 2 = 0 + 0.049 788 066 201 6;
  • 27) 0.049 788 066 201 6 × 2 = 0 + 0.099 576 132 403 2;
  • 28) 0.099 576 132 403 2 × 2 = 0 + 0.199 152 264 806 4;
  • 29) 0.199 152 264 806 4 × 2 = 0 + 0.398 304 529 612 8;
  • 30) 0.398 304 529 612 8 × 2 = 0 + 0.796 609 059 225 6;
  • 31) 0.796 609 059 225 6 × 2 = 1 + 0.593 218 118 451 2;
  • 32) 0.593 218 118 451 2 × 2 = 1 + 0.186 436 236 902 4;
  • 33) 0.186 436 236 902 4 × 2 = 0 + 0.372 872 473 804 8;
  • 34) 0.372 872 473 804 8 × 2 = 0 + 0.745 744 947 609 6;
  • 35) 0.745 744 947 609 6 × 2 = 1 + 0.491 489 895 219 2;
  • 36) 0.491 489 895 219 2 × 2 = 0 + 0.982 979 790 438 4;
  • 37) 0.982 979 790 438 4 × 2 = 1 + 0.965 959 580 876 8;
  • 38) 0.965 959 580 876 8 × 2 = 1 + 0.931 919 161 753 6;
  • 39) 0.931 919 161 753 6 × 2 = 1 + 0.863 838 323 507 2;
  • 40) 0.863 838 323 507 2 × 2 = 1 + 0.727 676 647 014 4;
  • 41) 0.727 676 647 014 4 × 2 = 1 + 0.455 353 294 028 8;
  • 42) 0.455 353 294 028 8 × 2 = 0 + 0.910 706 588 057 6;
  • 43) 0.910 706 588 057 6 × 2 = 1 + 0.821 413 176 115 2;
  • 44) 0.821 413 176 115 2 × 2 = 1 + 0.642 826 352 230 4;
  • 45) 0.642 826 352 230 4 × 2 = 1 + 0.285 652 704 460 8;
  • 46) 0.285 652 704 460 8 × 2 = 0 + 0.571 305 408 921 6;
  • 47) 0.571 305 408 921 6 × 2 = 1 + 0.142 610 817 843 2;
  • 48) 0.142 610 817 843 2 × 2 = 0 + 0.285 221 635 686 4;
  • 49) 0.285 221 635 686 4 × 2 = 0 + 0.570 443 271 372 8;
  • 50) 0.570 443 271 372 8 × 2 = 1 + 0.140 886 542 745 6;
  • 51) 0.140 886 542 745 6 × 2 = 0 + 0.281 773 085 491 2;
  • 52) 0.281 773 085 491 2 × 2 = 0 + 0.563 546 170 982 4;
  • 53) 0.563 546 170 982 4 × 2 = 1 + 0.127 092 341 964 8;
  • 54) 0.127 092 341 964 8 × 2 = 0 + 0.254 184 683 929 6;

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 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1011 1010 0100 10(2)

6. Positive number before normalization:

0.000 000 000 741 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1011 1010 0100 10(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 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1011 1010 0100 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1011 1010 0100 10(2) × 20 =

1.1001 0111 1101 1101 0010 010(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 1101 1101 0010 010


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 1110 1110 1001 0010 =

100 1011 1110 1110 1001 0010


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 1110 1110 1001 0010


Decimal number -0.000 000 000 741 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1110 1110 1001 0010

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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