-0.000 000 000 767 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 767 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
-0.000 000 000 767 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:

|-0.000 000 000 767 2| = 0.000 000 000 767 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: 0.000 000 000 767 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) 0.000 000 000 767 2 × 2 = 0 + 0.000 000 001 534 4;
  • 2) 0.000 000 001 534 4 × 2 = 0 + 0.000 000 003 068 8;
  • 3) 0.000 000 003 068 8 × 2 = 0 + 0.000 000 006 137 6;
  • 4) 0.000 000 006 137 6 × 2 = 0 + 0.000 000 012 275 2;
  • 5) 0.000 000 012 275 2 × 2 = 0 + 0.000 000 024 550 4;
  • 6) 0.000 000 024 550 4 × 2 = 0 + 0.000 000 049 100 8;
  • 7) 0.000 000 049 100 8 × 2 = 0 + 0.000 000 098 201 6;
  • 8) 0.000 000 098 201 6 × 2 = 0 + 0.000 000 196 403 2;
  • 9) 0.000 000 196 403 2 × 2 = 0 + 0.000 000 392 806 4;
  • 10) 0.000 000 392 806 4 × 2 = 0 + 0.000 000 785 612 8;
  • 11) 0.000 000 785 612 8 × 2 = 0 + 0.000 001 571 225 6;
  • 12) 0.000 001 571 225 6 × 2 = 0 + 0.000 003 142 451 2;
  • 13) 0.000 003 142 451 2 × 2 = 0 + 0.000 006 284 902 4;
  • 14) 0.000 006 284 902 4 × 2 = 0 + 0.000 012 569 804 8;
  • 15) 0.000 012 569 804 8 × 2 = 0 + 0.000 025 139 609 6;
  • 16) 0.000 025 139 609 6 × 2 = 0 + 0.000 050 279 219 2;
  • 17) 0.000 050 279 219 2 × 2 = 0 + 0.000 100 558 438 4;
  • 18) 0.000 100 558 438 4 × 2 = 0 + 0.000 201 116 876 8;
  • 19) 0.000 201 116 876 8 × 2 = 0 + 0.000 402 233 753 6;
  • 20) 0.000 402 233 753 6 × 2 = 0 + 0.000 804 467 507 2;
  • 21) 0.000 804 467 507 2 × 2 = 0 + 0.001 608 935 014 4;
  • 22) 0.001 608 935 014 4 × 2 = 0 + 0.003 217 870 028 8;
  • 23) 0.003 217 870 028 8 × 2 = 0 + 0.006 435 740 057 6;
  • 24) 0.006 435 740 057 6 × 2 = 0 + 0.012 871 480 115 2;
  • 25) 0.012 871 480 115 2 × 2 = 0 + 0.025 742 960 230 4;
  • 26) 0.025 742 960 230 4 × 2 = 0 + 0.051 485 920 460 8;
  • 27) 0.051 485 920 460 8 × 2 = 0 + 0.102 971 840 921 6;
  • 28) 0.102 971 840 921 6 × 2 = 0 + 0.205 943 681 843 2;
  • 29) 0.205 943 681 843 2 × 2 = 0 + 0.411 887 363 686 4;
  • 30) 0.411 887 363 686 4 × 2 = 0 + 0.823 774 727 372 8;
  • 31) 0.823 774 727 372 8 × 2 = 1 + 0.647 549 454 745 6;
  • 32) 0.647 549 454 745 6 × 2 = 1 + 0.295 098 909 491 2;
  • 33) 0.295 098 909 491 2 × 2 = 0 + 0.590 197 818 982 4;
  • 34) 0.590 197 818 982 4 × 2 = 1 + 0.180 395 637 964 8;
  • 35) 0.180 395 637 964 8 × 2 = 0 + 0.360 791 275 929 6;
  • 36) 0.360 791 275 929 6 × 2 = 0 + 0.721 582 551 859 2;
  • 37) 0.721 582 551 859 2 × 2 = 1 + 0.443 165 103 718 4;
  • 38) 0.443 165 103 718 4 × 2 = 0 + 0.886 330 207 436 8;
  • 39) 0.886 330 207 436 8 × 2 = 1 + 0.772 660 414 873 6;
  • 40) 0.772 660 414 873 6 × 2 = 1 + 0.545 320 829 747 2;
  • 41) 0.545 320 829 747 2 × 2 = 1 + 0.090 641 659 494 4;
  • 42) 0.090 641 659 494 4 × 2 = 0 + 0.181 283 318 988 8;
  • 43) 0.181 283 318 988 8 × 2 = 0 + 0.362 566 637 977 6;
  • 44) 0.362 566 637 977 6 × 2 = 0 + 0.725 133 275 955 2;
  • 45) 0.725 133 275 955 2 × 2 = 1 + 0.450 266 551 910 4;
  • 46) 0.450 266 551 910 4 × 2 = 0 + 0.900 533 103 820 8;
  • 47) 0.900 533 103 820 8 × 2 = 1 + 0.801 066 207 641 6;
  • 48) 0.801 066 207 641 6 × 2 = 1 + 0.602 132 415 283 2;
  • 49) 0.602 132 415 283 2 × 2 = 1 + 0.204 264 830 566 4;
  • 50) 0.204 264 830 566 4 × 2 = 0 + 0.408 529 661 132 8;
  • 51) 0.408 529 661 132 8 × 2 = 0 + 0.817 059 322 265 6;
  • 52) 0.817 059 322 265 6 × 2 = 1 + 0.634 118 644 531 2;
  • 53) 0.634 118 644 531 2 × 2 = 1 + 0.268 237 289 062 4;
  • 54) 0.268 237 289 062 4 × 2 = 0 + 0.536 474 578 124 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:


0.000 000 000 767 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 1011 1000 1011 1001 10(2)

6. Positive number before normalization:

0.000 000 000 767 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 1011 1000 1011 1001 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:


0.000 000 000 767 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 1011 1000 1011 1001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 1011 1000 1011 1001 10(2) × 20 =

1.1010 0101 1100 0101 1100 110(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):
1.1010 0101 1100 0101 1100 110


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. 101 0010 1110 0010 1110 0110 =

101 0010 1110 0010 1110 0110


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) =
101 0010 1110 0010 1110 0110


Decimal number -0.000 000 000 767 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 0010 1110 0010 1110 0110

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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