-0.000 000 000 742 21 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 21(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 21(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 21| = 0.000 000 000 742 21


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 21.

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 21 × 2 = 0 + 0.000 000 001 484 42;
  • 2) 0.000 000 001 484 42 × 2 = 0 + 0.000 000 002 968 84;
  • 3) 0.000 000 002 968 84 × 2 = 0 + 0.000 000 005 937 68;
  • 4) 0.000 000 005 937 68 × 2 = 0 + 0.000 000 011 875 36;
  • 5) 0.000 000 011 875 36 × 2 = 0 + 0.000 000 023 750 72;
  • 6) 0.000 000 023 750 72 × 2 = 0 + 0.000 000 047 501 44;
  • 7) 0.000 000 047 501 44 × 2 = 0 + 0.000 000 095 002 88;
  • 8) 0.000 000 095 002 88 × 2 = 0 + 0.000 000 190 005 76;
  • 9) 0.000 000 190 005 76 × 2 = 0 + 0.000 000 380 011 52;
  • 10) 0.000 000 380 011 52 × 2 = 0 + 0.000 000 760 023 04;
  • 11) 0.000 000 760 023 04 × 2 = 0 + 0.000 001 520 046 08;
  • 12) 0.000 001 520 046 08 × 2 = 0 + 0.000 003 040 092 16;
  • 13) 0.000 003 040 092 16 × 2 = 0 + 0.000 006 080 184 32;
  • 14) 0.000 006 080 184 32 × 2 = 0 + 0.000 012 160 368 64;
  • 15) 0.000 012 160 368 64 × 2 = 0 + 0.000 024 320 737 28;
  • 16) 0.000 024 320 737 28 × 2 = 0 + 0.000 048 641 474 56;
  • 17) 0.000 048 641 474 56 × 2 = 0 + 0.000 097 282 949 12;
  • 18) 0.000 097 282 949 12 × 2 = 0 + 0.000 194 565 898 24;
  • 19) 0.000 194 565 898 24 × 2 = 0 + 0.000 389 131 796 48;
  • 20) 0.000 389 131 796 48 × 2 = 0 + 0.000 778 263 592 96;
  • 21) 0.000 778 263 592 96 × 2 = 0 + 0.001 556 527 185 92;
  • 22) 0.001 556 527 185 92 × 2 = 0 + 0.003 113 054 371 84;
  • 23) 0.003 113 054 371 84 × 2 = 0 + 0.006 226 108 743 68;
  • 24) 0.006 226 108 743 68 × 2 = 0 + 0.012 452 217 487 36;
  • 25) 0.012 452 217 487 36 × 2 = 0 + 0.024 904 434 974 72;
  • 26) 0.024 904 434 974 72 × 2 = 0 + 0.049 808 869 949 44;
  • 27) 0.049 808 869 949 44 × 2 = 0 + 0.099 617 739 898 88;
  • 28) 0.099 617 739 898 88 × 2 = 0 + 0.199 235 479 797 76;
  • 29) 0.199 235 479 797 76 × 2 = 0 + 0.398 470 959 595 52;
  • 30) 0.398 470 959 595 52 × 2 = 0 + 0.796 941 919 191 04;
  • 31) 0.796 941 919 191 04 × 2 = 1 + 0.593 883 838 382 08;
  • 32) 0.593 883 838 382 08 × 2 = 1 + 0.187 767 676 764 16;
  • 33) 0.187 767 676 764 16 × 2 = 0 + 0.375 535 353 528 32;
  • 34) 0.375 535 353 528 32 × 2 = 0 + 0.751 070 707 056 64;
  • 35) 0.751 070 707 056 64 × 2 = 1 + 0.502 141 414 113 28;
  • 36) 0.502 141 414 113 28 × 2 = 1 + 0.004 282 828 226 56;
  • 37) 0.004 282 828 226 56 × 2 = 0 + 0.008 565 656 453 12;
  • 38) 0.008 565 656 453 12 × 2 = 0 + 0.017 131 312 906 24;
  • 39) 0.017 131 312 906 24 × 2 = 0 + 0.034 262 625 812 48;
  • 40) 0.034 262 625 812 48 × 2 = 0 + 0.068 525 251 624 96;
  • 41) 0.068 525 251 624 96 × 2 = 0 + 0.137 050 503 249 92;
  • 42) 0.137 050 503 249 92 × 2 = 0 + 0.274 101 006 499 84;
  • 43) 0.274 101 006 499 84 × 2 = 0 + 0.548 202 012 999 68;
  • 44) 0.548 202 012 999 68 × 2 = 1 + 0.096 404 025 999 36;
  • 45) 0.096 404 025 999 36 × 2 = 0 + 0.192 808 051 998 72;
  • 46) 0.192 808 051 998 72 × 2 = 0 + 0.385 616 103 997 44;
  • 47) 0.385 616 103 997 44 × 2 = 0 + 0.771 232 207 994 88;
  • 48) 0.771 232 207 994 88 × 2 = 1 + 0.542 464 415 989 76;
  • 49) 0.542 464 415 989 76 × 2 = 1 + 0.084 928 831 979 52;
  • 50) 0.084 928 831 979 52 × 2 = 0 + 0.169 857 663 959 04;
  • 51) 0.169 857 663 959 04 × 2 = 0 + 0.339 715 327 918 08;
  • 52) 0.339 715 327 918 08 × 2 = 0 + 0.679 430 655 836 16;
  • 53) 0.679 430 655 836 16 × 2 = 1 + 0.358 861 311 672 32;
  • 54) 0.358 861 311 672 32 × 2 = 0 + 0.717 722 623 344 64;

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 21(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0001 1000 10(2)

6. Positive number before normalization:

0.000 000 000 742 21(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0001 1000 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:


0.000 000 000 742 21(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0001 1000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0001 1000 10(2) × 20 =

1.1001 1000 0000 1000 1100 010(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 1000 1100 010


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 0100 0110 0010 =

100 1100 0000 0100 0110 0010


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 0100 0110 0010


Decimal number -0.000 000 000 742 21 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0100 0110 0010

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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