-0.000 000 000 735 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 735 1(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
-0.000 000 000 735 1(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:

|-0.000 000 000 735 1| = 0.000 000 000 735 1


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: 0.000 000 000 735 1.

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) 0.000 000 000 735 1 × 2 = 0 + 0.000 000 001 470 2;
  • 2) 0.000 000 001 470 2 × 2 = 0 + 0.000 000 002 940 4;
  • 3) 0.000 000 002 940 4 × 2 = 0 + 0.000 000 005 880 8;
  • 4) 0.000 000 005 880 8 × 2 = 0 + 0.000 000 011 761 6;
  • 5) 0.000 000 011 761 6 × 2 = 0 + 0.000 000 023 523 2;
  • 6) 0.000 000 023 523 2 × 2 = 0 + 0.000 000 047 046 4;
  • 7) 0.000 000 047 046 4 × 2 = 0 + 0.000 000 094 092 8;
  • 8) 0.000 000 094 092 8 × 2 = 0 + 0.000 000 188 185 6;
  • 9) 0.000 000 188 185 6 × 2 = 0 + 0.000 000 376 371 2;
  • 10) 0.000 000 376 371 2 × 2 = 0 + 0.000 000 752 742 4;
  • 11) 0.000 000 752 742 4 × 2 = 0 + 0.000 001 505 484 8;
  • 12) 0.000 001 505 484 8 × 2 = 0 + 0.000 003 010 969 6;
  • 13) 0.000 003 010 969 6 × 2 = 0 + 0.000 006 021 939 2;
  • 14) 0.000 006 021 939 2 × 2 = 0 + 0.000 012 043 878 4;
  • 15) 0.000 012 043 878 4 × 2 = 0 + 0.000 024 087 756 8;
  • 16) 0.000 024 087 756 8 × 2 = 0 + 0.000 048 175 513 6;
  • 17) 0.000 048 175 513 6 × 2 = 0 + 0.000 096 351 027 2;
  • 18) 0.000 096 351 027 2 × 2 = 0 + 0.000 192 702 054 4;
  • 19) 0.000 192 702 054 4 × 2 = 0 + 0.000 385 404 108 8;
  • 20) 0.000 385 404 108 8 × 2 = 0 + 0.000 770 808 217 6;
  • 21) 0.000 770 808 217 6 × 2 = 0 + 0.001 541 616 435 2;
  • 22) 0.001 541 616 435 2 × 2 = 0 + 0.003 083 232 870 4;
  • 23) 0.003 083 232 870 4 × 2 = 0 + 0.006 166 465 740 8;
  • 24) 0.006 166 465 740 8 × 2 = 0 + 0.012 332 931 481 6;
  • 25) 0.012 332 931 481 6 × 2 = 0 + 0.024 665 862 963 2;
  • 26) 0.024 665 862 963 2 × 2 = 0 + 0.049 331 725 926 4;
  • 27) 0.049 331 725 926 4 × 2 = 0 + 0.098 663 451 852 8;
  • 28) 0.098 663 451 852 8 × 2 = 0 + 0.197 326 903 705 6;
  • 29) 0.197 326 903 705 6 × 2 = 0 + 0.394 653 807 411 2;
  • 30) 0.394 653 807 411 2 × 2 = 0 + 0.789 307 614 822 4;
  • 31) 0.789 307 614 822 4 × 2 = 1 + 0.578 615 229 644 8;
  • 32) 0.578 615 229 644 8 × 2 = 1 + 0.157 230 459 289 6;
  • 33) 0.157 230 459 289 6 × 2 = 0 + 0.314 460 918 579 2;
  • 34) 0.314 460 918 579 2 × 2 = 0 + 0.628 921 837 158 4;
  • 35) 0.628 921 837 158 4 × 2 = 1 + 0.257 843 674 316 8;
  • 36) 0.257 843 674 316 8 × 2 = 0 + 0.515 687 348 633 6;
  • 37) 0.515 687 348 633 6 × 2 = 1 + 0.031 374 697 267 2;
  • 38) 0.031 374 697 267 2 × 2 = 0 + 0.062 749 394 534 4;
  • 39) 0.062 749 394 534 4 × 2 = 0 + 0.125 498 789 068 8;
  • 40) 0.125 498 789 068 8 × 2 = 0 + 0.250 997 578 137 6;
  • 41) 0.250 997 578 137 6 × 2 = 0 + 0.501 995 156 275 2;
  • 42) 0.501 995 156 275 2 × 2 = 1 + 0.003 990 312 550 4;
  • 43) 0.003 990 312 550 4 × 2 = 0 + 0.007 980 625 100 8;
  • 44) 0.007 980 625 100 8 × 2 = 0 + 0.015 961 250 201 6;
  • 45) 0.015 961 250 201 6 × 2 = 0 + 0.031 922 500 403 2;
  • 46) 0.031 922 500 403 2 × 2 = 0 + 0.063 845 000 806 4;
  • 47) 0.063 845 000 806 4 × 2 = 0 + 0.127 690 001 612 8;
  • 48) 0.127 690 001 612 8 × 2 = 0 + 0.255 380 003 225 6;
  • 49) 0.255 380 003 225 6 × 2 = 0 + 0.510 760 006 451 2;
  • 50) 0.510 760 006 451 2 × 2 = 1 + 0.021 520 012 902 4;
  • 51) 0.021 520 012 902 4 × 2 = 0 + 0.043 040 025 804 8;
  • 52) 0.043 040 025 804 8 × 2 = 0 + 0.086 080 051 609 6;
  • 53) 0.086 080 051 609 6 × 2 = 0 + 0.172 160 103 219 2;
  • 54) 0.172 160 103 219 2 × 2 = 0 + 0.344 320 206 438 4;

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:


0.000 000 000 735 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1000 0100 0000 0100 00(2)

6. Positive number before normalization:

0.000 000 000 735 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1000 0100 0000 0100 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:


0.000 000 000 735 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1000 0100 0000 0100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1000 0100 0000 0100 00(2) × 20 =

1.1001 0100 0010 0000 0010 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):
1.1001 0100 0010 0000 0010 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 1010 0001 0000 0001 0000 =

100 1010 0001 0000 0001 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 1010 0001 0000 0001 0000


Decimal number -0.000 000 000 735 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1010 0001 0000 0001 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