-0.000 000 000 59 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 59(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 59(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 59| = 0.000 000 000 59


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 59.

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 59 × 2 = 0 + 0.000 000 001 18;
  • 2) 0.000 000 001 18 × 2 = 0 + 0.000 000 002 36;
  • 3) 0.000 000 002 36 × 2 = 0 + 0.000 000 004 72;
  • 4) 0.000 000 004 72 × 2 = 0 + 0.000 000 009 44;
  • 5) 0.000 000 009 44 × 2 = 0 + 0.000 000 018 88;
  • 6) 0.000 000 018 88 × 2 = 0 + 0.000 000 037 76;
  • 7) 0.000 000 037 76 × 2 = 0 + 0.000 000 075 52;
  • 8) 0.000 000 075 52 × 2 = 0 + 0.000 000 151 04;
  • 9) 0.000 000 151 04 × 2 = 0 + 0.000 000 302 08;
  • 10) 0.000 000 302 08 × 2 = 0 + 0.000 000 604 16;
  • 11) 0.000 000 604 16 × 2 = 0 + 0.000 001 208 32;
  • 12) 0.000 001 208 32 × 2 = 0 + 0.000 002 416 64;
  • 13) 0.000 002 416 64 × 2 = 0 + 0.000 004 833 28;
  • 14) 0.000 004 833 28 × 2 = 0 + 0.000 009 666 56;
  • 15) 0.000 009 666 56 × 2 = 0 + 0.000 019 333 12;
  • 16) 0.000 019 333 12 × 2 = 0 + 0.000 038 666 24;
  • 17) 0.000 038 666 24 × 2 = 0 + 0.000 077 332 48;
  • 18) 0.000 077 332 48 × 2 = 0 + 0.000 154 664 96;
  • 19) 0.000 154 664 96 × 2 = 0 + 0.000 309 329 92;
  • 20) 0.000 309 329 92 × 2 = 0 + 0.000 618 659 84;
  • 21) 0.000 618 659 84 × 2 = 0 + 0.001 237 319 68;
  • 22) 0.001 237 319 68 × 2 = 0 + 0.002 474 639 36;
  • 23) 0.002 474 639 36 × 2 = 0 + 0.004 949 278 72;
  • 24) 0.004 949 278 72 × 2 = 0 + 0.009 898 557 44;
  • 25) 0.009 898 557 44 × 2 = 0 + 0.019 797 114 88;
  • 26) 0.019 797 114 88 × 2 = 0 + 0.039 594 229 76;
  • 27) 0.039 594 229 76 × 2 = 0 + 0.079 188 459 52;
  • 28) 0.079 188 459 52 × 2 = 0 + 0.158 376 919 04;
  • 29) 0.158 376 919 04 × 2 = 0 + 0.316 753 838 08;
  • 30) 0.316 753 838 08 × 2 = 0 + 0.633 507 676 16;
  • 31) 0.633 507 676 16 × 2 = 1 + 0.267 015 352 32;
  • 32) 0.267 015 352 32 × 2 = 0 + 0.534 030 704 64;
  • 33) 0.534 030 704 64 × 2 = 1 + 0.068 061 409 28;
  • 34) 0.068 061 409 28 × 2 = 0 + 0.136 122 818 56;
  • 35) 0.136 122 818 56 × 2 = 0 + 0.272 245 637 12;
  • 36) 0.272 245 637 12 × 2 = 0 + 0.544 491 274 24;
  • 37) 0.544 491 274 24 × 2 = 1 + 0.088 982 548 48;
  • 38) 0.088 982 548 48 × 2 = 0 + 0.177 965 096 96;
  • 39) 0.177 965 096 96 × 2 = 0 + 0.355 930 193 92;
  • 40) 0.355 930 193 92 × 2 = 0 + 0.711 860 387 84;
  • 41) 0.711 860 387 84 × 2 = 1 + 0.423 720 775 68;
  • 42) 0.423 720 775 68 × 2 = 0 + 0.847 441 551 36;
  • 43) 0.847 441 551 36 × 2 = 1 + 0.694 883 102 72;
  • 44) 0.694 883 102 72 × 2 = 1 + 0.389 766 205 44;
  • 45) 0.389 766 205 44 × 2 = 0 + 0.779 532 410 88;
  • 46) 0.779 532 410 88 × 2 = 1 + 0.559 064 821 76;
  • 47) 0.559 064 821 76 × 2 = 1 + 0.118 129 643 52;
  • 48) 0.118 129 643 52 × 2 = 0 + 0.236 259 287 04;
  • 49) 0.236 259 287 04 × 2 = 0 + 0.472 518 574 08;
  • 50) 0.472 518 574 08 × 2 = 0 + 0.945 037 148 16;
  • 51) 0.945 037 148 16 × 2 = 1 + 0.890 074 296 32;
  • 52) 0.890 074 296 32 × 2 = 1 + 0.780 148 592 64;
  • 53) 0.780 148 592 64 × 2 = 1 + 0.560 297 185 28;
  • 54) 0.560 297 185 28 × 2 = 1 + 0.120 594 370 56;

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 59(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1000 1000 1011 0110 0011 11(2)

6. Positive number before normalization:

0.000 000 000 59(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1000 1000 1011 0110 0011 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 59(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1000 1000 1011 0110 0011 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1000 1000 1011 0110 0011 11(2) × 20 =

1.0100 0100 0101 1011 0001 111(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):
1.0100 0100 0101 1011 0001 111


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. 010 0010 0010 1101 1000 1111 =

010 0010 0010 1101 1000 1111


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) =
010 0010 0010 1101 1000 1111


Decimal number -0.000 000 000 59 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 010 0010 0010 1101 1000 1111

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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