-0.000 000 000 744 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 744 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
-0.000 000 000 744 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:

|-0.000 000 000 744 2| = 0.000 000 000 744 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: 0.000 000 000 744 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) 0.000 000 000 744 2 × 2 = 0 + 0.000 000 001 488 4;
  • 2) 0.000 000 001 488 4 × 2 = 0 + 0.000 000 002 976 8;
  • 3) 0.000 000 002 976 8 × 2 = 0 + 0.000 000 005 953 6;
  • 4) 0.000 000 005 953 6 × 2 = 0 + 0.000 000 011 907 2;
  • 5) 0.000 000 011 907 2 × 2 = 0 + 0.000 000 023 814 4;
  • 6) 0.000 000 023 814 4 × 2 = 0 + 0.000 000 047 628 8;
  • 7) 0.000 000 047 628 8 × 2 = 0 + 0.000 000 095 257 6;
  • 8) 0.000 000 095 257 6 × 2 = 0 + 0.000 000 190 515 2;
  • 9) 0.000 000 190 515 2 × 2 = 0 + 0.000 000 381 030 4;
  • 10) 0.000 000 381 030 4 × 2 = 0 + 0.000 000 762 060 8;
  • 11) 0.000 000 762 060 8 × 2 = 0 + 0.000 001 524 121 6;
  • 12) 0.000 001 524 121 6 × 2 = 0 + 0.000 003 048 243 2;
  • 13) 0.000 003 048 243 2 × 2 = 0 + 0.000 006 096 486 4;
  • 14) 0.000 006 096 486 4 × 2 = 0 + 0.000 012 192 972 8;
  • 15) 0.000 012 192 972 8 × 2 = 0 + 0.000 024 385 945 6;
  • 16) 0.000 024 385 945 6 × 2 = 0 + 0.000 048 771 891 2;
  • 17) 0.000 048 771 891 2 × 2 = 0 + 0.000 097 543 782 4;
  • 18) 0.000 097 543 782 4 × 2 = 0 + 0.000 195 087 564 8;
  • 19) 0.000 195 087 564 8 × 2 = 0 + 0.000 390 175 129 6;
  • 20) 0.000 390 175 129 6 × 2 = 0 + 0.000 780 350 259 2;
  • 21) 0.000 780 350 259 2 × 2 = 0 + 0.001 560 700 518 4;
  • 22) 0.001 560 700 518 4 × 2 = 0 + 0.003 121 401 036 8;
  • 23) 0.003 121 401 036 8 × 2 = 0 + 0.006 242 802 073 6;
  • 24) 0.006 242 802 073 6 × 2 = 0 + 0.012 485 604 147 2;
  • 25) 0.012 485 604 147 2 × 2 = 0 + 0.024 971 208 294 4;
  • 26) 0.024 971 208 294 4 × 2 = 0 + 0.049 942 416 588 8;
  • 27) 0.049 942 416 588 8 × 2 = 0 + 0.099 884 833 177 6;
  • 28) 0.099 884 833 177 6 × 2 = 0 + 0.199 769 666 355 2;
  • 29) 0.199 769 666 355 2 × 2 = 0 + 0.399 539 332 710 4;
  • 30) 0.399 539 332 710 4 × 2 = 0 + 0.799 078 665 420 8;
  • 31) 0.799 078 665 420 8 × 2 = 1 + 0.598 157 330 841 6;
  • 32) 0.598 157 330 841 6 × 2 = 1 + 0.196 314 661 683 2;
  • 33) 0.196 314 661 683 2 × 2 = 0 + 0.392 629 323 366 4;
  • 34) 0.392 629 323 366 4 × 2 = 0 + 0.785 258 646 732 8;
  • 35) 0.785 258 646 732 8 × 2 = 1 + 0.570 517 293 465 6;
  • 36) 0.570 517 293 465 6 × 2 = 1 + 0.141 034 586 931 2;
  • 37) 0.141 034 586 931 2 × 2 = 0 + 0.282 069 173 862 4;
  • 38) 0.282 069 173 862 4 × 2 = 0 + 0.564 138 347 724 8;
  • 39) 0.564 138 347 724 8 × 2 = 1 + 0.128 276 695 449 6;
  • 40) 0.128 276 695 449 6 × 2 = 0 + 0.256 553 390 899 2;
  • 41) 0.256 553 390 899 2 × 2 = 0 + 0.513 106 781 798 4;
  • 42) 0.513 106 781 798 4 × 2 = 1 + 0.026 213 563 596 8;
  • 43) 0.026 213 563 596 8 × 2 = 0 + 0.052 427 127 193 6;
  • 44) 0.052 427 127 193 6 × 2 = 0 + 0.104 854 254 387 2;
  • 45) 0.104 854 254 387 2 × 2 = 0 + 0.209 708 508 774 4;
  • 46) 0.209 708 508 774 4 × 2 = 0 + 0.419 417 017 548 8;
  • 47) 0.419 417 017 548 8 × 2 = 0 + 0.838 834 035 097 6;
  • 48) 0.838 834 035 097 6 × 2 = 1 + 0.677 668 070 195 2;
  • 49) 0.677 668 070 195 2 × 2 = 1 + 0.355 336 140 390 4;
  • 50) 0.355 336 140 390 4 × 2 = 0 + 0.710 672 280 780 8;
  • 51) 0.710 672 280 780 8 × 2 = 1 + 0.421 344 561 561 6;
  • 52) 0.421 344 561 561 6 × 2 = 0 + 0.842 689 123 123 2;
  • 53) 0.842 689 123 123 2 × 2 = 1 + 0.685 378 246 246 4;
  • 54) 0.685 378 246 246 4 × 2 = 1 + 0.370 756 492 492 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:


0.000 000 000 744 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 0100 0001 1010 11(2)

6. Positive number before normalization:

0.000 000 000 744 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 0100 0001 1010 11(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:


0.000 000 000 744 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 0100 0001 1010 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 0100 0001 1010 11(2) × 20 =

1.1001 1001 0010 0000 1101 011(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 1001 0010 0000 1101 011


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 1001 0000 0110 1011 =

100 1100 1001 0000 0110 1011


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 1001 0000 0110 1011


Decimal number -0.000 000 000 744 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 1001 0000 0110 1011

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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