-0.000 000 000 743 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 743 7(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
-0.000 000 000 743 7(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:

|-0.000 000 000 743 7| = 0.000 000 000 743 7


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: 0.000 000 000 743 7.

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) 0.000 000 000 743 7 × 2 = 0 + 0.000 000 001 487 4;
  • 2) 0.000 000 001 487 4 × 2 = 0 + 0.000 000 002 974 8;
  • 3) 0.000 000 002 974 8 × 2 = 0 + 0.000 000 005 949 6;
  • 4) 0.000 000 005 949 6 × 2 = 0 + 0.000 000 011 899 2;
  • 5) 0.000 000 011 899 2 × 2 = 0 + 0.000 000 023 798 4;
  • 6) 0.000 000 023 798 4 × 2 = 0 + 0.000 000 047 596 8;
  • 7) 0.000 000 047 596 8 × 2 = 0 + 0.000 000 095 193 6;
  • 8) 0.000 000 095 193 6 × 2 = 0 + 0.000 000 190 387 2;
  • 9) 0.000 000 190 387 2 × 2 = 0 + 0.000 000 380 774 4;
  • 10) 0.000 000 380 774 4 × 2 = 0 + 0.000 000 761 548 8;
  • 11) 0.000 000 761 548 8 × 2 = 0 + 0.000 001 523 097 6;
  • 12) 0.000 001 523 097 6 × 2 = 0 + 0.000 003 046 195 2;
  • 13) 0.000 003 046 195 2 × 2 = 0 + 0.000 006 092 390 4;
  • 14) 0.000 006 092 390 4 × 2 = 0 + 0.000 012 184 780 8;
  • 15) 0.000 012 184 780 8 × 2 = 0 + 0.000 024 369 561 6;
  • 16) 0.000 024 369 561 6 × 2 = 0 + 0.000 048 739 123 2;
  • 17) 0.000 048 739 123 2 × 2 = 0 + 0.000 097 478 246 4;
  • 18) 0.000 097 478 246 4 × 2 = 0 + 0.000 194 956 492 8;
  • 19) 0.000 194 956 492 8 × 2 = 0 + 0.000 389 912 985 6;
  • 20) 0.000 389 912 985 6 × 2 = 0 + 0.000 779 825 971 2;
  • 21) 0.000 779 825 971 2 × 2 = 0 + 0.001 559 651 942 4;
  • 22) 0.001 559 651 942 4 × 2 = 0 + 0.003 119 303 884 8;
  • 23) 0.003 119 303 884 8 × 2 = 0 + 0.006 238 607 769 6;
  • 24) 0.006 238 607 769 6 × 2 = 0 + 0.012 477 215 539 2;
  • 25) 0.012 477 215 539 2 × 2 = 0 + 0.024 954 431 078 4;
  • 26) 0.024 954 431 078 4 × 2 = 0 + 0.049 908 862 156 8;
  • 27) 0.049 908 862 156 8 × 2 = 0 + 0.099 817 724 313 6;
  • 28) 0.099 817 724 313 6 × 2 = 0 + 0.199 635 448 627 2;
  • 29) 0.199 635 448 627 2 × 2 = 0 + 0.399 270 897 254 4;
  • 30) 0.399 270 897 254 4 × 2 = 0 + 0.798 541 794 508 8;
  • 31) 0.798 541 794 508 8 × 2 = 1 + 0.597 083 589 017 6;
  • 32) 0.597 083 589 017 6 × 2 = 1 + 0.194 167 178 035 2;
  • 33) 0.194 167 178 035 2 × 2 = 0 + 0.388 334 356 070 4;
  • 34) 0.388 334 356 070 4 × 2 = 0 + 0.776 668 712 140 8;
  • 35) 0.776 668 712 140 8 × 2 = 1 + 0.553 337 424 281 6;
  • 36) 0.553 337 424 281 6 × 2 = 1 + 0.106 674 848 563 2;
  • 37) 0.106 674 848 563 2 × 2 = 0 + 0.213 349 697 126 4;
  • 38) 0.213 349 697 126 4 × 2 = 0 + 0.426 699 394 252 8;
  • 39) 0.426 699 394 252 8 × 2 = 0 + 0.853 398 788 505 6;
  • 40) 0.853 398 788 505 6 × 2 = 1 + 0.706 797 577 011 2;
  • 41) 0.706 797 577 011 2 × 2 = 1 + 0.413 595 154 022 4;
  • 42) 0.413 595 154 022 4 × 2 = 0 + 0.827 190 308 044 8;
  • 43) 0.827 190 308 044 8 × 2 = 1 + 0.654 380 616 089 6;
  • 44) 0.654 380 616 089 6 × 2 = 1 + 0.308 761 232 179 2;
  • 45) 0.308 761 232 179 2 × 2 = 0 + 0.617 522 464 358 4;
  • 46) 0.617 522 464 358 4 × 2 = 1 + 0.235 044 928 716 8;
  • 47) 0.235 044 928 716 8 × 2 = 0 + 0.470 089 857 433 6;
  • 48) 0.470 089 857 433 6 × 2 = 0 + 0.940 179 714 867 2;
  • 49) 0.940 179 714 867 2 × 2 = 1 + 0.880 359 429 734 4;
  • 50) 0.880 359 429 734 4 × 2 = 1 + 0.760 718 859 468 8;
  • 51) 0.760 718 859 468 8 × 2 = 1 + 0.521 437 718 937 6;
  • 52) 0.521 437 718 937 6 × 2 = 1 + 0.042 875 437 875 2;
  • 53) 0.042 875 437 875 2 × 2 = 0 + 0.085 750 875 750 4;
  • 54) 0.085 750 875 750 4 × 2 = 0 + 0.171 501 751 500 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:


0.000 000 000 743 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0001 1011 0100 1111 00(2)

6. Positive number before normalization:

0.000 000 000 743 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0001 1011 0100 1111 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:


0.000 000 000 743 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0001 1011 0100 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0001 1011 0100 1111 00(2) × 20 =

1.1001 1000 1101 1010 0111 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 1000 1101 1010 0111 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 1100 0110 1101 0011 1100 =

100 1100 0110 1101 0011 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 1100 0110 1101 0011 1100


Decimal number -0.000 000 000 743 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0110 1101 0011 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