Convert -11.851 100 000 100 111 101 100 16 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

-11.851 100 000 100 111 101 100 16(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. Start with the positive version of the number:

|-11.851 100 000 100 111 101 100 16| = 11.851 100 000 100 111 101 100 16

2. First, convert to the binary (base 2) the integer part: 11.
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;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

11(10) =


1011(2)


4. Convert to the binary (base 2) the fractional part: 0.851 100 000 100 111 101 100 16.

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) 0.851 100 000 100 111 101 100 16 × 2 = 1 + 0.702 200 000 200 222 202 200 32;
  • 2) 0.702 200 000 200 222 202 200 32 × 2 = 1 + 0.404 400 000 400 444 404 400 64;
  • 3) 0.404 400 000 400 444 404 400 64 × 2 = 0 + 0.808 800 000 800 888 808 801 28;
  • 4) 0.808 800 000 800 888 808 801 28 × 2 = 1 + 0.617 600 001 601 777 617 602 56;
  • 5) 0.617 600 001 601 777 617 602 56 × 2 = 1 + 0.235 200 003 203 555 235 205 12;
  • 6) 0.235 200 003 203 555 235 205 12 × 2 = 0 + 0.470 400 006 407 110 470 410 24;
  • 7) 0.470 400 006 407 110 470 410 24 × 2 = 0 + 0.940 800 012 814 220 940 820 48;
  • 8) 0.940 800 012 814 220 940 820 48 × 2 = 1 + 0.881 600 025 628 441 881 640 96;
  • 9) 0.881 600 025 628 441 881 640 96 × 2 = 1 + 0.763 200 051 256 883 763 281 92;
  • 10) 0.763 200 051 256 883 763 281 92 × 2 = 1 + 0.526 400 102 513 767 526 563 84;
  • 11) 0.526 400 102 513 767 526 563 84 × 2 = 1 + 0.052 800 205 027 535 053 127 68;
  • 12) 0.052 800 205 027 535 053 127 68 × 2 = 0 + 0.105 600 410 055 070 106 255 36;
  • 13) 0.105 600 410 055 070 106 255 36 × 2 = 0 + 0.211 200 820 110 140 212 510 72;
  • 14) 0.211 200 820 110 140 212 510 72 × 2 = 0 + 0.422 401 640 220 280 425 021 44;
  • 15) 0.422 401 640 220 280 425 021 44 × 2 = 0 + 0.844 803 280 440 560 850 042 88;
  • 16) 0.844 803 280 440 560 850 042 88 × 2 = 1 + 0.689 606 560 881 121 700 085 76;
  • 17) 0.689 606 560 881 121 700 085 76 × 2 = 1 + 0.379 213 121 762 243 400 171 52;
  • 18) 0.379 213 121 762 243 400 171 52 × 2 = 0 + 0.758 426 243 524 486 800 343 04;
  • 19) 0.758 426 243 524 486 800 343 04 × 2 = 1 + 0.516 852 487 048 973 600 686 08;
  • 20) 0.516 852 487 048 973 600 686 08 × 2 = 1 + 0.033 704 974 097 947 201 372 16;
  • 21) 0.033 704 974 097 947 201 372 16 × 2 = 0 + 0.067 409 948 195 894 402 744 32;
  • 22) 0.067 409 948 195 894 402 744 32 × 2 = 0 + 0.134 819 896 391 788 805 488 64;
  • 23) 0.134 819 896 391 788 805 488 64 × 2 = 0 + 0.269 639 792 783 577 610 977 28;
  • 24) 0.269 639 792 783 577 610 977 28 × 2 = 0 + 0.539 279 585 567 155 221 954 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...)


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:

0.851 100 000 100 111 101 100 16(10) =


0.1101 1001 1110 0001 1011 0000(2)


6. Positive number before normalization:

11.851 100 000 100 111 101 100 16(10) =


1011.1101 1001 1110 0001 1011 0000(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left so that only one non zero digit remains to the left of it:

11.851 100 000 100 111 101 100 16(10) =


1011.1101 1001 1110 0001 1011 0000(2) =


1011.1101 1001 1110 0001 1011 0000(2) × 20 =


1.0111 1011 0011 1100 0011 0110 000(2) × 23


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): 3


Mantissa (not normalized):
1.0111 1011 0011 1100 0011 0110 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


3 + 2(8-1) - 1 =


(3 + 127)(10) =


130(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;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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) =


130(10) =


1000 0010(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 011 1101 1001 1110 0001 1011 0000 =


011 1101 1001 1110 0001 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) =
1000 0010


Mantissa (23 bits) =
011 1101 1001 1110 0001 1011


Number -11.851 100 000 100 111 101 100 16 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
1 - 1000 0010 - 011 1101 1001 1110 0001 1011

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 1

      30
    • 0

      29
    • 0

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 0

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 1

      0

More operations of this kind:

-11.851 100 000 100 111 101 100 17 = ? ... -11.851 100 000 100 111 101 100 15 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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