-0.000 000 000 742 152 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 152 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
-0.000 000 000 742 152 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:

|-0.000 000 000 742 152 2| = 0.000 000 000 742 152 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: 0.000 000 000 742 152 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) 0.000 000 000 742 152 2 × 2 = 0 + 0.000 000 001 484 304 4;
  • 2) 0.000 000 001 484 304 4 × 2 = 0 + 0.000 000 002 968 608 8;
  • 3) 0.000 000 002 968 608 8 × 2 = 0 + 0.000 000 005 937 217 6;
  • 4) 0.000 000 005 937 217 6 × 2 = 0 + 0.000 000 011 874 435 2;
  • 5) 0.000 000 011 874 435 2 × 2 = 0 + 0.000 000 023 748 870 4;
  • 6) 0.000 000 023 748 870 4 × 2 = 0 + 0.000 000 047 497 740 8;
  • 7) 0.000 000 047 497 740 8 × 2 = 0 + 0.000 000 094 995 481 6;
  • 8) 0.000 000 094 995 481 6 × 2 = 0 + 0.000 000 189 990 963 2;
  • 9) 0.000 000 189 990 963 2 × 2 = 0 + 0.000 000 379 981 926 4;
  • 10) 0.000 000 379 981 926 4 × 2 = 0 + 0.000 000 759 963 852 8;
  • 11) 0.000 000 759 963 852 8 × 2 = 0 + 0.000 001 519 927 705 6;
  • 12) 0.000 001 519 927 705 6 × 2 = 0 + 0.000 003 039 855 411 2;
  • 13) 0.000 003 039 855 411 2 × 2 = 0 + 0.000 006 079 710 822 4;
  • 14) 0.000 006 079 710 822 4 × 2 = 0 + 0.000 012 159 421 644 8;
  • 15) 0.000 012 159 421 644 8 × 2 = 0 + 0.000 024 318 843 289 6;
  • 16) 0.000 024 318 843 289 6 × 2 = 0 + 0.000 048 637 686 579 2;
  • 17) 0.000 048 637 686 579 2 × 2 = 0 + 0.000 097 275 373 158 4;
  • 18) 0.000 097 275 373 158 4 × 2 = 0 + 0.000 194 550 746 316 8;
  • 19) 0.000 194 550 746 316 8 × 2 = 0 + 0.000 389 101 492 633 6;
  • 20) 0.000 389 101 492 633 6 × 2 = 0 + 0.000 778 202 985 267 2;
  • 21) 0.000 778 202 985 267 2 × 2 = 0 + 0.001 556 405 970 534 4;
  • 22) 0.001 556 405 970 534 4 × 2 = 0 + 0.003 112 811 941 068 8;
  • 23) 0.003 112 811 941 068 8 × 2 = 0 + 0.006 225 623 882 137 6;
  • 24) 0.006 225 623 882 137 6 × 2 = 0 + 0.012 451 247 764 275 2;
  • 25) 0.012 451 247 764 275 2 × 2 = 0 + 0.024 902 495 528 550 4;
  • 26) 0.024 902 495 528 550 4 × 2 = 0 + 0.049 804 991 057 100 8;
  • 27) 0.049 804 991 057 100 8 × 2 = 0 + 0.099 609 982 114 201 6;
  • 28) 0.099 609 982 114 201 6 × 2 = 0 + 0.199 219 964 228 403 2;
  • 29) 0.199 219 964 228 403 2 × 2 = 0 + 0.398 439 928 456 806 4;
  • 30) 0.398 439 928 456 806 4 × 2 = 0 + 0.796 879 856 913 612 8;
  • 31) 0.796 879 856 913 612 8 × 2 = 1 + 0.593 759 713 827 225 6;
  • 32) 0.593 759 713 827 225 6 × 2 = 1 + 0.187 519 427 654 451 2;
  • 33) 0.187 519 427 654 451 2 × 2 = 0 + 0.375 038 855 308 902 4;
  • 34) 0.375 038 855 308 902 4 × 2 = 0 + 0.750 077 710 617 804 8;
  • 35) 0.750 077 710 617 804 8 × 2 = 1 + 0.500 155 421 235 609 6;
  • 36) 0.500 155 421 235 609 6 × 2 = 1 + 0.000 310 842 471 219 2;
  • 37) 0.000 310 842 471 219 2 × 2 = 0 + 0.000 621 684 942 438 4;
  • 38) 0.000 621 684 942 438 4 × 2 = 0 + 0.001 243 369 884 876 8;
  • 39) 0.001 243 369 884 876 8 × 2 = 0 + 0.002 486 739 769 753 6;
  • 40) 0.002 486 739 769 753 6 × 2 = 0 + 0.004 973 479 539 507 2;
  • 41) 0.004 973 479 539 507 2 × 2 = 0 + 0.009 946 959 079 014 4;
  • 42) 0.009 946 959 079 014 4 × 2 = 0 + 0.019 893 918 158 028 8;
  • 43) 0.019 893 918 158 028 8 × 2 = 0 + 0.039 787 836 316 057 6;
  • 44) 0.039 787 836 316 057 6 × 2 = 0 + 0.079 575 672 632 115 2;
  • 45) 0.079 575 672 632 115 2 × 2 = 0 + 0.159 151 345 264 230 4;
  • 46) 0.159 151 345 264 230 4 × 2 = 0 + 0.318 302 690 528 460 8;
  • 47) 0.318 302 690 528 460 8 × 2 = 0 + 0.636 605 381 056 921 6;
  • 48) 0.636 605 381 056 921 6 × 2 = 1 + 0.273 210 762 113 843 2;
  • 49) 0.273 210 762 113 843 2 × 2 = 0 + 0.546 421 524 227 686 4;
  • 50) 0.546 421 524 227 686 4 × 2 = 1 + 0.092 843 048 455 372 8;
  • 51) 0.092 843 048 455 372 8 × 2 = 0 + 0.185 686 096 910 745 6;
  • 52) 0.185 686 096 910 745 6 × 2 = 0 + 0.371 372 193 821 491 2;
  • 53) 0.371 372 193 821 491 2 × 2 = 0 + 0.742 744 387 642 982 4;
  • 54) 0.742 744 387 642 982 4 × 2 = 1 + 0.485 488 775 285 964 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:


0.000 000 000 742 152 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0001 0100 01(2)

6. Positive number before normalization:

0.000 000 000 742 152 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0001 0100 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:


0.000 000 000 742 152 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0001 0100 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0001 0100 01(2) × 20 =

1.1001 1000 0000 0000 1010 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 1000 0000 0000 1010 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 1100 0000 0000 0101 0001 =

100 1100 0000 0000 0101 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 1100 0000 0000 0101 0001


Decimal number -0.000 000 000 742 152 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0101 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