-0.000 000 000 706 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 706(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
-0.000 000 000 706(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:

|-0.000 000 000 706| = 0.000 000 000 706


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: 0.000 000 000 706.

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) 0.000 000 000 706 × 2 = 0 + 0.000 000 001 412;
  • 2) 0.000 000 001 412 × 2 = 0 + 0.000 000 002 824;
  • 3) 0.000 000 002 824 × 2 = 0 + 0.000 000 005 648;
  • 4) 0.000 000 005 648 × 2 = 0 + 0.000 000 011 296;
  • 5) 0.000 000 011 296 × 2 = 0 + 0.000 000 022 592;
  • 6) 0.000 000 022 592 × 2 = 0 + 0.000 000 045 184;
  • 7) 0.000 000 045 184 × 2 = 0 + 0.000 000 090 368;
  • 8) 0.000 000 090 368 × 2 = 0 + 0.000 000 180 736;
  • 9) 0.000 000 180 736 × 2 = 0 + 0.000 000 361 472;
  • 10) 0.000 000 361 472 × 2 = 0 + 0.000 000 722 944;
  • 11) 0.000 000 722 944 × 2 = 0 + 0.000 001 445 888;
  • 12) 0.000 001 445 888 × 2 = 0 + 0.000 002 891 776;
  • 13) 0.000 002 891 776 × 2 = 0 + 0.000 005 783 552;
  • 14) 0.000 005 783 552 × 2 = 0 + 0.000 011 567 104;
  • 15) 0.000 011 567 104 × 2 = 0 + 0.000 023 134 208;
  • 16) 0.000 023 134 208 × 2 = 0 + 0.000 046 268 416;
  • 17) 0.000 046 268 416 × 2 = 0 + 0.000 092 536 832;
  • 18) 0.000 092 536 832 × 2 = 0 + 0.000 185 073 664;
  • 19) 0.000 185 073 664 × 2 = 0 + 0.000 370 147 328;
  • 20) 0.000 370 147 328 × 2 = 0 + 0.000 740 294 656;
  • 21) 0.000 740 294 656 × 2 = 0 + 0.001 480 589 312;
  • 22) 0.001 480 589 312 × 2 = 0 + 0.002 961 178 624;
  • 23) 0.002 961 178 624 × 2 = 0 + 0.005 922 357 248;
  • 24) 0.005 922 357 248 × 2 = 0 + 0.011 844 714 496;
  • 25) 0.011 844 714 496 × 2 = 0 + 0.023 689 428 992;
  • 26) 0.023 689 428 992 × 2 = 0 + 0.047 378 857 984;
  • 27) 0.047 378 857 984 × 2 = 0 + 0.094 757 715 968;
  • 28) 0.094 757 715 968 × 2 = 0 + 0.189 515 431 936;
  • 29) 0.189 515 431 936 × 2 = 0 + 0.379 030 863 872;
  • 30) 0.379 030 863 872 × 2 = 0 + 0.758 061 727 744;
  • 31) 0.758 061 727 744 × 2 = 1 + 0.516 123 455 488;
  • 32) 0.516 123 455 488 × 2 = 1 + 0.032 246 910 976;
  • 33) 0.032 246 910 976 × 2 = 0 + 0.064 493 821 952;
  • 34) 0.064 493 821 952 × 2 = 0 + 0.128 987 643 904;
  • 35) 0.128 987 643 904 × 2 = 0 + 0.257 975 287 808;
  • 36) 0.257 975 287 808 × 2 = 0 + 0.515 950 575 616;
  • 37) 0.515 950 575 616 × 2 = 1 + 0.031 901 151 232;
  • 38) 0.031 901 151 232 × 2 = 0 + 0.063 802 302 464;
  • 39) 0.063 802 302 464 × 2 = 0 + 0.127 604 604 928;
  • 40) 0.127 604 604 928 × 2 = 0 + 0.255 209 209 856;
  • 41) 0.255 209 209 856 × 2 = 0 + 0.510 418 419 712;
  • 42) 0.510 418 419 712 × 2 = 1 + 0.020 836 839 424;
  • 43) 0.020 836 839 424 × 2 = 0 + 0.041 673 678 848;
  • 44) 0.041 673 678 848 × 2 = 0 + 0.083 347 357 696;
  • 45) 0.083 347 357 696 × 2 = 0 + 0.166 694 715 392;
  • 46) 0.166 694 715 392 × 2 = 0 + 0.333 389 430 784;
  • 47) 0.333 389 430 784 × 2 = 0 + 0.666 778 861 568;
  • 48) 0.666 778 861 568 × 2 = 1 + 0.333 557 723 136;
  • 49) 0.333 557 723 136 × 2 = 0 + 0.667 115 446 272;
  • 50) 0.667 115 446 272 × 2 = 1 + 0.334 230 892 544;
  • 51) 0.334 230 892 544 × 2 = 0 + 0.668 461 785 088;
  • 52) 0.668 461 785 088 × 2 = 1 + 0.336 923 570 176;
  • 53) 0.336 923 570 176 × 2 = 0 + 0.673 847 140 352;
  • 54) 0.673 847 140 352 × 2 = 1 + 0.347 694 280 704;

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:


0.000 000 000 706(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 1000 0100 0001 0101 01(2)

6. Positive number before normalization:

0.000 000 000 706(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 1000 0100 0001 0101 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:


0.000 000 000 706(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 1000 0100 0001 0101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 1000 0100 0001 0101 01(2) × 20 =

1.1000 0100 0010 0000 1010 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):
1.1000 0100 0010 0000 1010 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 0010 0001 0000 0101 0101 =

100 0010 0001 0000 0101 0101


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 0010 0001 0000 0101 0101


Decimal number -0.000 000 000 706 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 0010 0001 0000 0101 0101

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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