-0.000 000 000 879 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 879(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
-0.000 000 000 879(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:

|-0.000 000 000 879| = 0.000 000 000 879


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: 0.000 000 000 879.

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) 0.000 000 000 879 × 2 = 0 + 0.000 000 001 758;
  • 2) 0.000 000 001 758 × 2 = 0 + 0.000 000 003 516;
  • 3) 0.000 000 003 516 × 2 = 0 + 0.000 000 007 032;
  • 4) 0.000 000 007 032 × 2 = 0 + 0.000 000 014 064;
  • 5) 0.000 000 014 064 × 2 = 0 + 0.000 000 028 128;
  • 6) 0.000 000 028 128 × 2 = 0 + 0.000 000 056 256;
  • 7) 0.000 000 056 256 × 2 = 0 + 0.000 000 112 512;
  • 8) 0.000 000 112 512 × 2 = 0 + 0.000 000 225 024;
  • 9) 0.000 000 225 024 × 2 = 0 + 0.000 000 450 048;
  • 10) 0.000 000 450 048 × 2 = 0 + 0.000 000 900 096;
  • 11) 0.000 000 900 096 × 2 = 0 + 0.000 001 800 192;
  • 12) 0.000 001 800 192 × 2 = 0 + 0.000 003 600 384;
  • 13) 0.000 003 600 384 × 2 = 0 + 0.000 007 200 768;
  • 14) 0.000 007 200 768 × 2 = 0 + 0.000 014 401 536;
  • 15) 0.000 014 401 536 × 2 = 0 + 0.000 028 803 072;
  • 16) 0.000 028 803 072 × 2 = 0 + 0.000 057 606 144;
  • 17) 0.000 057 606 144 × 2 = 0 + 0.000 115 212 288;
  • 18) 0.000 115 212 288 × 2 = 0 + 0.000 230 424 576;
  • 19) 0.000 230 424 576 × 2 = 0 + 0.000 460 849 152;
  • 20) 0.000 460 849 152 × 2 = 0 + 0.000 921 698 304;
  • 21) 0.000 921 698 304 × 2 = 0 + 0.001 843 396 608;
  • 22) 0.001 843 396 608 × 2 = 0 + 0.003 686 793 216;
  • 23) 0.003 686 793 216 × 2 = 0 + 0.007 373 586 432;
  • 24) 0.007 373 586 432 × 2 = 0 + 0.014 747 172 864;
  • 25) 0.014 747 172 864 × 2 = 0 + 0.029 494 345 728;
  • 26) 0.029 494 345 728 × 2 = 0 + 0.058 988 691 456;
  • 27) 0.058 988 691 456 × 2 = 0 + 0.117 977 382 912;
  • 28) 0.117 977 382 912 × 2 = 0 + 0.235 954 765 824;
  • 29) 0.235 954 765 824 × 2 = 0 + 0.471 909 531 648;
  • 30) 0.471 909 531 648 × 2 = 0 + 0.943 819 063 296;
  • 31) 0.943 819 063 296 × 2 = 1 + 0.887 638 126 592;
  • 32) 0.887 638 126 592 × 2 = 1 + 0.775 276 253 184;
  • 33) 0.775 276 253 184 × 2 = 1 + 0.550 552 506 368;
  • 34) 0.550 552 506 368 × 2 = 1 + 0.101 105 012 736;
  • 35) 0.101 105 012 736 × 2 = 0 + 0.202 210 025 472;
  • 36) 0.202 210 025 472 × 2 = 0 + 0.404 420 050 944;
  • 37) 0.404 420 050 944 × 2 = 0 + 0.808 840 101 888;
  • 38) 0.808 840 101 888 × 2 = 1 + 0.617 680 203 776;
  • 39) 0.617 680 203 776 × 2 = 1 + 0.235 360 407 552;
  • 40) 0.235 360 407 552 × 2 = 0 + 0.470 720 815 104;
  • 41) 0.470 720 815 104 × 2 = 0 + 0.941 441 630 208;
  • 42) 0.941 441 630 208 × 2 = 1 + 0.882 883 260 416;
  • 43) 0.882 883 260 416 × 2 = 1 + 0.765 766 520 832;
  • 44) 0.765 766 520 832 × 2 = 1 + 0.531 533 041 664;
  • 45) 0.531 533 041 664 × 2 = 1 + 0.063 066 083 328;
  • 46) 0.063 066 083 328 × 2 = 0 + 0.126 132 166 656;
  • 47) 0.126 132 166 656 × 2 = 0 + 0.252 264 333 312;
  • 48) 0.252 264 333 312 × 2 = 0 + 0.504 528 666 624;
  • 49) 0.504 528 666 624 × 2 = 1 + 0.009 057 333 248;
  • 50) 0.009 057 333 248 × 2 = 0 + 0.018 114 666 496;
  • 51) 0.018 114 666 496 × 2 = 0 + 0.036 229 332 992;
  • 52) 0.036 229 332 992 × 2 = 0 + 0.072 458 665 984;
  • 53) 0.072 458 665 984 × 2 = 0 + 0.144 917 331 968;
  • 54) 0.144 917 331 968 × 2 = 0 + 0.289 834 663 936;

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:


0.000 000 000 879(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 0110 0111 1000 1000 00(2)

6. Positive number before normalization:

0.000 000 000 879(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 0110 0111 1000 1000 00(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:


0.000 000 000 879(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 0110 0111 1000 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 0110 0111 1000 1000 00(2) × 20 =

1.1110 0011 0011 1100 0100 000(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):
1.1110 0011 0011 1100 0100 000


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. 111 0001 1001 1110 0010 0000 =

111 0001 1001 1110 0010 0000


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) =
111 0001 1001 1110 0010 0000


Decimal number -0.000 000 000 879 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 111 0001 1001 1110 0010 0000

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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