-0.000 000 000 774 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 774(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
-0.000 000 000 774(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:

|-0.000 000 000 774| = 0.000 000 000 774


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: 0.000 000 000 774.

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) 0.000 000 000 774 × 2 = 0 + 0.000 000 001 548;
  • 2) 0.000 000 001 548 × 2 = 0 + 0.000 000 003 096;
  • 3) 0.000 000 003 096 × 2 = 0 + 0.000 000 006 192;
  • 4) 0.000 000 006 192 × 2 = 0 + 0.000 000 012 384;
  • 5) 0.000 000 012 384 × 2 = 0 + 0.000 000 024 768;
  • 6) 0.000 000 024 768 × 2 = 0 + 0.000 000 049 536;
  • 7) 0.000 000 049 536 × 2 = 0 + 0.000 000 099 072;
  • 8) 0.000 000 099 072 × 2 = 0 + 0.000 000 198 144;
  • 9) 0.000 000 198 144 × 2 = 0 + 0.000 000 396 288;
  • 10) 0.000 000 396 288 × 2 = 0 + 0.000 000 792 576;
  • 11) 0.000 000 792 576 × 2 = 0 + 0.000 001 585 152;
  • 12) 0.000 001 585 152 × 2 = 0 + 0.000 003 170 304;
  • 13) 0.000 003 170 304 × 2 = 0 + 0.000 006 340 608;
  • 14) 0.000 006 340 608 × 2 = 0 + 0.000 012 681 216;
  • 15) 0.000 012 681 216 × 2 = 0 + 0.000 025 362 432;
  • 16) 0.000 025 362 432 × 2 = 0 + 0.000 050 724 864;
  • 17) 0.000 050 724 864 × 2 = 0 + 0.000 101 449 728;
  • 18) 0.000 101 449 728 × 2 = 0 + 0.000 202 899 456;
  • 19) 0.000 202 899 456 × 2 = 0 + 0.000 405 798 912;
  • 20) 0.000 405 798 912 × 2 = 0 + 0.000 811 597 824;
  • 21) 0.000 811 597 824 × 2 = 0 + 0.001 623 195 648;
  • 22) 0.001 623 195 648 × 2 = 0 + 0.003 246 391 296;
  • 23) 0.003 246 391 296 × 2 = 0 + 0.006 492 782 592;
  • 24) 0.006 492 782 592 × 2 = 0 + 0.012 985 565 184;
  • 25) 0.012 985 565 184 × 2 = 0 + 0.025 971 130 368;
  • 26) 0.025 971 130 368 × 2 = 0 + 0.051 942 260 736;
  • 27) 0.051 942 260 736 × 2 = 0 + 0.103 884 521 472;
  • 28) 0.103 884 521 472 × 2 = 0 + 0.207 769 042 944;
  • 29) 0.207 769 042 944 × 2 = 0 + 0.415 538 085 888;
  • 30) 0.415 538 085 888 × 2 = 0 + 0.831 076 171 776;
  • 31) 0.831 076 171 776 × 2 = 1 + 0.662 152 343 552;
  • 32) 0.662 152 343 552 × 2 = 1 + 0.324 304 687 104;
  • 33) 0.324 304 687 104 × 2 = 0 + 0.648 609 374 208;
  • 34) 0.648 609 374 208 × 2 = 1 + 0.297 218 748 416;
  • 35) 0.297 218 748 416 × 2 = 0 + 0.594 437 496 832;
  • 36) 0.594 437 496 832 × 2 = 1 + 0.188 874 993 664;
  • 37) 0.188 874 993 664 × 2 = 0 + 0.377 749 987 328;
  • 38) 0.377 749 987 328 × 2 = 0 + 0.755 499 974 656;
  • 39) 0.755 499 974 656 × 2 = 1 + 0.510 999 949 312;
  • 40) 0.510 999 949 312 × 2 = 1 + 0.021 999 898 624;
  • 41) 0.021 999 898 624 × 2 = 0 + 0.043 999 797 248;
  • 42) 0.043 999 797 248 × 2 = 0 + 0.087 999 594 496;
  • 43) 0.087 999 594 496 × 2 = 0 + 0.175 999 188 992;
  • 44) 0.175 999 188 992 × 2 = 0 + 0.351 998 377 984;
  • 45) 0.351 998 377 984 × 2 = 0 + 0.703 996 755 968;
  • 46) 0.703 996 755 968 × 2 = 1 + 0.407 993 511 936;
  • 47) 0.407 993 511 936 × 2 = 0 + 0.815 987 023 872;
  • 48) 0.815 987 023 872 × 2 = 1 + 0.631 974 047 744;
  • 49) 0.631 974 047 744 × 2 = 1 + 0.263 948 095 488;
  • 50) 0.263 948 095 488 × 2 = 0 + 0.527 896 190 976;
  • 51) 0.527 896 190 976 × 2 = 1 + 0.055 792 381 952;
  • 52) 0.055 792 381 952 × 2 = 0 + 0.111 584 763 904;
  • 53) 0.111 584 763 904 × 2 = 0 + 0.223 169 527 808;
  • 54) 0.223 169 527 808 × 2 = 0 + 0.446 339 055 616;

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:


0.000 000 000 774(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0000 0101 1010 00(2)

6. Positive number before normalization:

0.000 000 000 774(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0000 0101 1010 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:


0.000 000 000 774(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0000 0101 1010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0000 0101 1010 00(2) × 20 =

1.1010 1001 1000 0010 1101 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):
1.1010 1001 1000 0010 1101 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. 101 0100 1100 0001 0110 1000 =

101 0100 1100 0001 0110 1000


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) =
101 0100 1100 0001 0110 1000


Decimal number -0.000 000 000 774 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 0100 1100 0001 0110 1000

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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