-0.000 000 000 742 43 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 43(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 43(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 43| = 0.000 000 000 742 43


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 43.

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 43 × 2 = 0 + 0.000 000 001 484 86;
  • 2) 0.000 000 001 484 86 × 2 = 0 + 0.000 000 002 969 72;
  • 3) 0.000 000 002 969 72 × 2 = 0 + 0.000 000 005 939 44;
  • 4) 0.000 000 005 939 44 × 2 = 0 + 0.000 000 011 878 88;
  • 5) 0.000 000 011 878 88 × 2 = 0 + 0.000 000 023 757 76;
  • 6) 0.000 000 023 757 76 × 2 = 0 + 0.000 000 047 515 52;
  • 7) 0.000 000 047 515 52 × 2 = 0 + 0.000 000 095 031 04;
  • 8) 0.000 000 095 031 04 × 2 = 0 + 0.000 000 190 062 08;
  • 9) 0.000 000 190 062 08 × 2 = 0 + 0.000 000 380 124 16;
  • 10) 0.000 000 380 124 16 × 2 = 0 + 0.000 000 760 248 32;
  • 11) 0.000 000 760 248 32 × 2 = 0 + 0.000 001 520 496 64;
  • 12) 0.000 001 520 496 64 × 2 = 0 + 0.000 003 040 993 28;
  • 13) 0.000 003 040 993 28 × 2 = 0 + 0.000 006 081 986 56;
  • 14) 0.000 006 081 986 56 × 2 = 0 + 0.000 012 163 973 12;
  • 15) 0.000 012 163 973 12 × 2 = 0 + 0.000 024 327 946 24;
  • 16) 0.000 024 327 946 24 × 2 = 0 + 0.000 048 655 892 48;
  • 17) 0.000 048 655 892 48 × 2 = 0 + 0.000 097 311 784 96;
  • 18) 0.000 097 311 784 96 × 2 = 0 + 0.000 194 623 569 92;
  • 19) 0.000 194 623 569 92 × 2 = 0 + 0.000 389 247 139 84;
  • 20) 0.000 389 247 139 84 × 2 = 0 + 0.000 778 494 279 68;
  • 21) 0.000 778 494 279 68 × 2 = 0 + 0.001 556 988 559 36;
  • 22) 0.001 556 988 559 36 × 2 = 0 + 0.003 113 977 118 72;
  • 23) 0.003 113 977 118 72 × 2 = 0 + 0.006 227 954 237 44;
  • 24) 0.006 227 954 237 44 × 2 = 0 + 0.012 455 908 474 88;
  • 25) 0.012 455 908 474 88 × 2 = 0 + 0.024 911 816 949 76;
  • 26) 0.024 911 816 949 76 × 2 = 0 + 0.049 823 633 899 52;
  • 27) 0.049 823 633 899 52 × 2 = 0 + 0.099 647 267 799 04;
  • 28) 0.099 647 267 799 04 × 2 = 0 + 0.199 294 535 598 08;
  • 29) 0.199 294 535 598 08 × 2 = 0 + 0.398 589 071 196 16;
  • 30) 0.398 589 071 196 16 × 2 = 0 + 0.797 178 142 392 32;
  • 31) 0.797 178 142 392 32 × 2 = 1 + 0.594 356 284 784 64;
  • 32) 0.594 356 284 784 64 × 2 = 1 + 0.188 712 569 569 28;
  • 33) 0.188 712 569 569 28 × 2 = 0 + 0.377 425 139 138 56;
  • 34) 0.377 425 139 138 56 × 2 = 0 + 0.754 850 278 277 12;
  • 35) 0.754 850 278 277 12 × 2 = 1 + 0.509 700 556 554 24;
  • 36) 0.509 700 556 554 24 × 2 = 1 + 0.019 401 113 108 48;
  • 37) 0.019 401 113 108 48 × 2 = 0 + 0.038 802 226 216 96;
  • 38) 0.038 802 226 216 96 × 2 = 0 + 0.077 604 452 433 92;
  • 39) 0.077 604 452 433 92 × 2 = 0 + 0.155 208 904 867 84;
  • 40) 0.155 208 904 867 84 × 2 = 0 + 0.310 417 809 735 68;
  • 41) 0.310 417 809 735 68 × 2 = 0 + 0.620 835 619 471 36;
  • 42) 0.620 835 619 471 36 × 2 = 1 + 0.241 671 238 942 72;
  • 43) 0.241 671 238 942 72 × 2 = 0 + 0.483 342 477 885 44;
  • 44) 0.483 342 477 885 44 × 2 = 0 + 0.966 684 955 770 88;
  • 45) 0.966 684 955 770 88 × 2 = 1 + 0.933 369 911 541 76;
  • 46) 0.933 369 911 541 76 × 2 = 1 + 0.866 739 823 083 52;
  • 47) 0.866 739 823 083 52 × 2 = 1 + 0.733 479 646 167 04;
  • 48) 0.733 479 646 167 04 × 2 = 1 + 0.466 959 292 334 08;
  • 49) 0.466 959 292 334 08 × 2 = 0 + 0.933 918 584 668 16;
  • 50) 0.933 918 584 668 16 × 2 = 1 + 0.867 837 169 336 32;
  • 51) 0.867 837 169 336 32 × 2 = 1 + 0.735 674 338 672 64;
  • 52) 0.735 674 338 672 64 × 2 = 1 + 0.471 348 677 345 28;
  • 53) 0.471 348 677 345 28 × 2 = 0 + 0.942 697 354 690 56;
  • 54) 0.942 697 354 690 56 × 2 = 1 + 0.885 394 709 381 12;

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 43(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0100 1111 0111 01(2)

6. Positive number before normalization:

0.000 000 000 742 43(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0100 1111 0111 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 43(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0100 1111 0111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0100 1111 0111 01(2) × 20 =

1.1001 1000 0010 0111 1011 101(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 0010 0111 1011 101


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 0001 0011 1101 1101 =

100 1100 0001 0011 1101 1101


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 0001 0011 1101 1101


Decimal number -0.000 000 000 742 43 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0001 0011 1101 1101

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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