-0.000 000 000 259 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 259(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
-0.000 000 000 259(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:

|-0.000 000 000 259| = 0.000 000 000 259


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: 0.000 000 000 259.

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) 0.000 000 000 259 × 2 = 0 + 0.000 000 000 518;
  • 2) 0.000 000 000 518 × 2 = 0 + 0.000 000 001 036;
  • 3) 0.000 000 001 036 × 2 = 0 + 0.000 000 002 072;
  • 4) 0.000 000 002 072 × 2 = 0 + 0.000 000 004 144;
  • 5) 0.000 000 004 144 × 2 = 0 + 0.000 000 008 288;
  • 6) 0.000 000 008 288 × 2 = 0 + 0.000 000 016 576;
  • 7) 0.000 000 016 576 × 2 = 0 + 0.000 000 033 152;
  • 8) 0.000 000 033 152 × 2 = 0 + 0.000 000 066 304;
  • 9) 0.000 000 066 304 × 2 = 0 + 0.000 000 132 608;
  • 10) 0.000 000 132 608 × 2 = 0 + 0.000 000 265 216;
  • 11) 0.000 000 265 216 × 2 = 0 + 0.000 000 530 432;
  • 12) 0.000 000 530 432 × 2 = 0 + 0.000 001 060 864;
  • 13) 0.000 001 060 864 × 2 = 0 + 0.000 002 121 728;
  • 14) 0.000 002 121 728 × 2 = 0 + 0.000 004 243 456;
  • 15) 0.000 004 243 456 × 2 = 0 + 0.000 008 486 912;
  • 16) 0.000 008 486 912 × 2 = 0 + 0.000 016 973 824;
  • 17) 0.000 016 973 824 × 2 = 0 + 0.000 033 947 648;
  • 18) 0.000 033 947 648 × 2 = 0 + 0.000 067 895 296;
  • 19) 0.000 067 895 296 × 2 = 0 + 0.000 135 790 592;
  • 20) 0.000 135 790 592 × 2 = 0 + 0.000 271 581 184;
  • 21) 0.000 271 581 184 × 2 = 0 + 0.000 543 162 368;
  • 22) 0.000 543 162 368 × 2 = 0 + 0.001 086 324 736;
  • 23) 0.001 086 324 736 × 2 = 0 + 0.002 172 649 472;
  • 24) 0.002 172 649 472 × 2 = 0 + 0.004 345 298 944;
  • 25) 0.004 345 298 944 × 2 = 0 + 0.008 690 597 888;
  • 26) 0.008 690 597 888 × 2 = 0 + 0.017 381 195 776;
  • 27) 0.017 381 195 776 × 2 = 0 + 0.034 762 391 552;
  • 28) 0.034 762 391 552 × 2 = 0 + 0.069 524 783 104;
  • 29) 0.069 524 783 104 × 2 = 0 + 0.139 049 566 208;
  • 30) 0.139 049 566 208 × 2 = 0 + 0.278 099 132 416;
  • 31) 0.278 099 132 416 × 2 = 0 + 0.556 198 264 832;
  • 32) 0.556 198 264 832 × 2 = 1 + 0.112 396 529 664;
  • 33) 0.112 396 529 664 × 2 = 0 + 0.224 793 059 328;
  • 34) 0.224 793 059 328 × 2 = 0 + 0.449 586 118 656;
  • 35) 0.449 586 118 656 × 2 = 0 + 0.899 172 237 312;
  • 36) 0.899 172 237 312 × 2 = 1 + 0.798 344 474 624;
  • 37) 0.798 344 474 624 × 2 = 1 + 0.596 688 949 248;
  • 38) 0.596 688 949 248 × 2 = 1 + 0.193 377 898 496;
  • 39) 0.193 377 898 496 × 2 = 0 + 0.386 755 796 992;
  • 40) 0.386 755 796 992 × 2 = 0 + 0.773 511 593 984;
  • 41) 0.773 511 593 984 × 2 = 1 + 0.547 023 187 968;
  • 42) 0.547 023 187 968 × 2 = 1 + 0.094 046 375 936;
  • 43) 0.094 046 375 936 × 2 = 0 + 0.188 092 751 872;
  • 44) 0.188 092 751 872 × 2 = 0 + 0.376 185 503 744;
  • 45) 0.376 185 503 744 × 2 = 0 + 0.752 371 007 488;
  • 46) 0.752 371 007 488 × 2 = 1 + 0.504 742 014 976;
  • 47) 0.504 742 014 976 × 2 = 1 + 0.009 484 029 952;
  • 48) 0.009 484 029 952 × 2 = 0 + 0.018 968 059 904;
  • 49) 0.018 968 059 904 × 2 = 0 + 0.037 936 119 808;
  • 50) 0.037 936 119 808 × 2 = 0 + 0.075 872 239 616;
  • 51) 0.075 872 239 616 × 2 = 0 + 0.151 744 479 232;
  • 52) 0.151 744 479 232 × 2 = 0 + 0.303 488 958 464;
  • 53) 0.303 488 958 464 × 2 = 0 + 0.606 977 916 928;
  • 54) 0.606 977 916 928 × 2 = 1 + 0.213 955 833 856;
  • 55) 0.213 955 833 856 × 2 = 0 + 0.427 911 667 712;

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:


0.000 000 000 259(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0001 1100 1100 0110 0000 010(2)

6. Positive number before normalization:

0.000 000 000 259(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0001 1100 1100 0110 0000 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 259(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0001 1100 1100 0110 0000 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0001 1100 1100 0110 0000 010(2) × 20 =

1.0001 1100 1100 0110 0000 010(2) × 2-32


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): -32

Mantissa (not normalized):
1.0001 1100 1100 0110 0000 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(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;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

95(10) =

0101 1111(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. 000 1110 0110 0011 0000 0010 =

000 1110 0110 0011 0000 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) =
0101 1111

Mantissa (23 bits) =
000 1110 0110 0011 0000 0010


Decimal number -0.000 000 000 259 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1111 - 000 1110 0110 0011 0000 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